Gravitation — Practice & Self Evaluation

Which of the following statements is true regarding Newton's law of universal gravitation?

1. The gravitational force between two masses is inversely proportional to the product of their masses.
2. The gravitational force between two masses is directly proportional to the square of the distance between them.
3. The gravitational force between two masses is directly proportional to the product of their masses.
4. The gravitational force between two masses is independent of the distance between them.

Answer

The gravitational force between two masses is directly proportional to the product of their masses.

Reason — Newton's universal law of gravitation states that the gravitational force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them, acting along the line joining them. Thus

$\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$

Options 1, 2 and 4 contradict this relation, since the force is not inversely proportional to the product of the masses, not directly proportional to the square of the distance, and certainly not independent of the distance.

If the distance between two masses is doubled, the gravitational force between them becomes:

1. four times greater
2. four times smaller
3. twice as large
4. half as large

Answer

four times smaller

Reason — By Newton's law of gravitation $\text F \propto \dfrac{1}{\text r^2}$. If the distance is doubled, r becomes 2r, so

$\dfrac{\text F'}{\text F} = \left(\dfrac{\text r}{2\text r}\right)^2 = \dfrac{1}{4}$

Hence the force is reduced to one-fourth of its original value, that is, it becomes four times smaller. This is why the force-distance relationship is called the inverse-square law.

The value of the gravitational constant 'G' does vary with:

1. distance
2. mass
3. distance and mass
4. None of these

Answer

None of these

Reason — The gravitational constant G is the constant of proportionality in $\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$. Its value is the same for all pairs of bodies, at all places, at all distances of separation, and it is also independent of the presence of other bodies or of the properties of the intervening medium. For this reason G is called a 'universal constant', and it does not vary with distance or with mass.

The force-distance relationship in Newton's Law of Universal Gravitation is also referred to as the:

1. inverse-exponent law
2. exponential-decay law
3. inverse-cube law
4. inverse-square law

Answer

inverse-square law

Reason — Since the gravitational force varies as $\dfrac{1}{\text r^2}$, doubling the distance between two objects results in a force which is four times smaller. A relationship of this kind is termed the 'inverse-square law'. An inverse-cube law would give $\text F \propto \dfrac{1}{\text r^3}$, which is not the case for gravitation.

At which location is the acceleration due to gravity the least?

1. Equator
2. Poles
3. Centre of the Earth
4. Tropics

Answer

Centre of the Earth

Reason — At a depth h below the earth's surface the acceleration due to gravity varies as

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

At the centre of the earth h = R_e, so that

$\text g' = \text g\left(1 - \dfrac{\text R_e}{\text R_e}\right) = 0$

Thus the acceleration due to gravity is exactly zero at the centre of the earth, which is the least value it can have anywhere.

Among the three locations on the surface, g is least at the equator (about 9.78 m s^-2) and greatest at the poles (about 9.83 m s^-2), the tropics lying in between. This is because the earth is an oblate spheroid, so that $\text g \propto \dfrac{1}{\text R_e^2}$ gives a smaller value where the radius is larger, and because the rotation of the earth reduces g by the term R_eω²cos²λ, which is maximum at the equator and zero at the poles. Even the smallest of these surface values, however, is far greater than zero.

Note: The printed answer key marks option 1, considering only locations on the earth's surface. However, since g = 0 at the centre of the earth, option 3 is the correct answer as given.

The value of acceleration due to gravity decreases with:

1. increase in altitude
2. decrease in altitude
3. decrease in latitude
4. increase in the core temperature of the Earth

Answer

increase in altitude

Reason — The acceleration due to gravity at a height h above the earth's surface is

$\text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$

so that g' < g. As we go above the surface of the earth, the acceleration due to gravity goes on decreasing. The value of g in fact increases with increasing latitude, and it does not depend on the core temperature of the earth at all.

The gravitational field emanating from any material object:

1. is always infinite
2. exists only in the presence of any external body
3. is zero at an infinite distance from the source object
4. is always finite and independent of the presence of any external body

Answer

is always finite and independent of the presence of any external body

Reason — The gravitational field due to a mass is the region of space around it in which any other mass experiences an attractive gravitational force. Theoretically this field extends up to an infinite distance, but at large distances its strength becomes negligibly small, though never exactly zero. The field of an object exists whether or not another object is nearby; the presence of an external body is required only for the gravitational force to be exerted, not for the field to exist.

Which of the following is true of two objects of different masses falling freely near the surface of the moon?

1. They both have different accelerations.
2. They have the same velocities at any instant.
3. They experience forces of the same magnitude.
4. They change their inertia

Answer

They have the same velocities at any instant.

Reason — The acceleration due to gravity at the surface of the moon,

$\text g_m = \dfrac{\text{GM}_m}{\text R_m^2}$

is free from the mass of the falling body. Hence two bodies of different masses fall with the same acceleration, and since the moon has no atmosphere there is no buoyancy or viscous drag to disturb this. Starting together from rest, both therefore have the same velocity at any instant and reach the surface simultaneously.

The forces on them are not the same, since F = mg_m is proportional to the mass, so the heavier body experiences the greater gravitational force. Their inertia also does not change, inertia being a constant property of a body.

The Earth's atmosphere is held by the:

1. Wind
2. Clouds
3. Earth's magnetic field
4. Gravity

Answer

Gravity

Reason — The molecules of the atmosphere are held to the earth by the earth's gravitational attraction. Because the escape velocity at the earth's surface (11.2 km s^-1) is much greater than the average thermal speed of the atmospheric molecules, these molecules are generally unable to reach the speed necessary to escape into space, and the earth is able to retain a thick atmosphere. The moon, whose escape velocity is only about 2.4 km s^-1, cannot retain an atmosphere for this reason.

Which of the following factors does the acceleration due to gravity on the Earth depend upon?

1. Mass of the body.
2. Mass of the Earth.
3. The volume of the body
4. Shape and size of the body

Answer

Mass of the Earth.

Reason — The acceleration due to gravity at the earth's surface is

$\text g = \dfrac{\text{GM}_e}{\text R_e^2}$

This expression contains the mass M_e and the radius R_e of the earth, but is free from the mass of the body. Hence g depends on the mass of the earth and not upon the mass, volume, shape or size of the body placed on it.

Gravitational force on a particle of mass m inside a spherical shell of mass M and radius R is:

1. $\dfrac{GM}{R^2}$

2. $\dfrac{GM}{R}$

3. $\dfrac{GM}{R^2}$

4. Zero

Answer

Zero

Reason — By Newton's shell theorem, a spherically symmetric shell of mass exerts no net gravitational force on a particle located anywhere inside the shell. Every small mass element of the shell pulls the particle in one direction, but there is always a corresponding arrangement of mass elements on the opposite side pulling it the other way, and because of the symmetry of the shell these contributions cancel out exactly. Hence the gravitational force on a particle inside a spherical shell is zero, whatever its position inside.

Note: Options 1 and 3 are printed identically as $\dfrac{\text{GM}}{\text R^2}$ in the textbook. This does not affect the answer, since the correct option is 4.

A satellite is in a circular orbit at a certain distance from the Earth. If its speed is increased slightly, it will:

1. move to a higher orbit.
2. move to a lower orbit.
3. continue in the same orbit.
4. escape from earth's gravity.

Answer

move to a higher orbit.

Reason — When the speed of the satellite is increased slightly, its kinetic energy increases and the balance between the gravitational force and the centripetal force required for the circular orbit is disturbed. The satellite therefore no longer moves in a circular orbit but enters an elliptical orbit, moving away from the earth to a higher altitude at its apogee while coming closer to the earth at its perigee. In this way the new orbit has a greater average distance from the earth than the original circular orbit.

An artificial satellite is in an orbit where it is experiencing atmospheric drag. What happens to the satellite's orbit over time?

1. The satellite will gradually spiral inwards towards the Earth.
2. The satellite's orbit will remain stable.
3. The satellite's speed will increase.
4. The satellite will gradually spiral outwards away from the Earth.

Answer

The satellite will gradually spiral inwards towards the Earth.

Reason — Atmospheric drag is a resistive force which acts opposite to the direction of motion of the satellite and therefore does negative work on it. This causes the satellite to lose energy continuously. As its total energy decreases (becomes more negative), the radius of its orbit becomes smaller, resulting in a gradual decrease in altitude. The satellite therefore spirals inward towards the earth.

Which of the following is true about geostationary satellites?

1. They rotate around the Earth twice in 24 h.
2. They appear stationary relative to a fixed point on Earth.
3. They orbit the Earth in a highly elliptical path.
4. They have the same orbital speed as low Earth orbit satellites.

Answer

They appear stationary relative to a fixed point on Earth.

Reason — A geostationary satellite is placed in an orbit such that its period of revolution is exactly equal to the period of the axial motion of the earth, that is, 24 h, and its direction of rotation is the same as that of the earth, in an orbital plane coplanar with the equatorial plane. Being synchronous with the earth's spin, it appears stationary over a fixed point on the earth's equator. Its orbit is circular, not highly elliptical, and its orbital speed (about 3.1 km s^-1) is much smaller than that of a low earth orbit satellite (about 8 km s^-1).

Which of the following is not a characteristic of geostationary satellites?

1. They orbit the Earth in 24 h.
2. They are positioned above the equator.
3. They can cover the entire Earth's surface.
4. They have a circular orbit.

Answer

They can cover the entire Earth's surface.

Reason — A geostationary satellite remains fixed over one point on the equator and therefore provides continuous coverage of the same geographical region only. It has limited effectiveness near the poles, which is why polar regions rely on polar-orbiting satellites. Covering the entire earth's surface is a characteristic of polar satellites, which sweep over the whole globe as the earth rotates beneath their orbit. Orbiting in 24 h, being positioned above the equator, and having a circular orbit are all genuine characteristics of geostationary satellites.

For a satellite to be in a geostationary orbit, it must:

1. be positioned above the equator
2. have an orbital period of 12 h
3. be at an altitude of 100 km
4. travel in a polar orbit

Answer

be positioned above the equator

Reason — The conditions for a satellite to be geostationary are that its period of revolution (or angular velocity) should be the same as that of the earth, that its direction of rotation should be the same as that of the earth from west to east, and that its orbital plane should be coplanar with the equatorial plane of the earth. Hence it must be positioned above the equator. Its period must be 24 h and not 12 h, its altitude must be about 35,830 km and not 100 km, and a polar orbit would not keep it stationary over any point.

A satellite in orbit around Earth is kept in orbit by:

1. the centripetal force provided by earth's gravitational pull.
2. the satellite's engines.
3. a balance between the satellite's speed and the atmospheric drag.
4. a tether to Earth (a connection to Earth/tied to Earth).

Answer

the centripetal force provided by earth's gravitational pull.

Reason — When a satellite revolves in a circular orbit around the earth, the gravitational force of attraction between the earth and the satellite provides exactly the centripetal force required for its circular motion,

$\dfrac{\text{GM}_e \text m}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$

No engine, tether or atmospheric drag is involved; this is why a satellite continues to revolve in its orbit without using any fuel.

Weightlessness experienced in an orbiting satellite is due to:

1. absence of gravity.
2. zero orbital velocity.
3. free fall of the satellite and its occupants.
4. lack of atmosphere.

Answer

free fall of the satellite and its occupants.

Reason — Inside an orbiting satellite the gravitational force on a spaceman supplies exactly the centripetal force needed for his circular motion, so the reactionary force R of the base of the satellite on him becomes zero,

$\dfrac{\text{GM}_e \text m'}{\text r^2} - \text R = \dfrac{\text m'\text v_o^2}{\text r} \quad \Rightarrow \quad \text R = 0$

Since weight is felt only through this reaction, he feels his weight to be zero. Gravity is certainly present at that height, and the satellite has a large orbital velocity; the weightlessness is entirely due to the satellite and its occupants being in a state of free fall.

The total energy of a satellite in a stable orbit is:

1. positive
2. zero
3. negative
4. infinity

Answer

negative

Reason — For a satellite revolving close to the earth's surface, the potential energy and the kinetic energy are

$\text U = -\dfrac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text K = \dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

so the total energy is

$\text E = \text U + \text K = -\dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

The kinetic energy is less in magnitude than the potential energy, so the total energy is negative. This negative total energy indicates that the satellite is bound to its orbit, and additional energy must be supplied to free it.

Which of the following correctly expresses the condition for a stable orbit?

1. The satellite's speed is equal to the escape velocity
2. The centrifugal force is less than the gravitational force.
3. The centripetal force is equal to the gravitational force.
4. The satellite has zero angular momentum.

Answer

The centripetal force is equal to the gravitational force.

Reason — A satellite remains in a stable circular orbit only when the gravitational force of the earth on it supplies precisely the centripetal force needed for that orbit,

$\dfrac{\text{GM}_e \text m}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$

If the satellite's speed were equal to the escape velocity it would leave the orbit altogether, and a satellite moving in an orbit certainly does not have zero angular momentum.

Which of the following best describes Kepler's first law?

1. The square of the orbital period of a planet is proportional to the cube of the semi-major axis.
2. The orbit of a planet is an ellipse with the Sun at one focus.
3. The speed of a planet is fastest at perihelion.
4. The orbit of a planet is a perfect circle.

Answer

The orbit of a planet is an ellipse with the Sun at one focus.

Reason — Kepler's first law is the law of orbits, which states that all planets move around the sun in elliptical orbits having the sun at one focus of the orbit. Option 1 is the statement of Kepler's third law (the law of periods) and option 3 is a consequence of the second law (the law of areas). A perfect circle is only a special case, and is not what the first law asserts.

According to Kepler's second law, the line joining a planet and the Sun sweeps out equal areas in:

1. equal time intervals
2. varying time intervals
3. equal distances
4. varying distances

Answer

equal time intervals

Reason — Kepler's second law is the law of areas, according to which a line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant. As a consequence the planet's velocity is maximum when it is nearest the sun and minimum when it is farthest from the sun.

The ratio of the square of the period of revolution to the cube of the semi-major axis is:

1. a constant.
2. varying with time.
3. varying with distance from the Sun.
4. proportional to the orbital velocity.

Answer

a constant.

Reason — By Kepler's law of periods, the square of the period of revolution of a planet is directly proportional to the cube of its mean distance from the sun,

$\text T^2 \propto \text r^3 \quad \text{or} \quad \text T^2 = \text K\text r^3$

so that $\dfrac{\text T^2}{\text r^3} = \text K$. The constant K is the same for all the planets and does not change with time, with distance from the sun, or with orbital velocity.

In a satellite, the astronauts experience weightlessness because:

1. The gravitational force of Earth is balanced by that of the Sun.
2. The satellite is moving in a vacuum.
3. The satellite is in free fall towards Earth.
4. There is no gravitational force acting on them.

Answer

The satellite is in free fall towards Earth.

Reason — The satellite and everything inside it are in a state of free fall towards the earth, since the gravitational force is entirely used up in providing the centripetal force for the circular orbit. The reactionary force of the floor of the satellite on the astronaut therefore becomes zero, and he feels weightless. If he stands on a spring balance, the balance will read zero.

Which of the following conditions best describes weightlessness?

1. No gravitational forces are acting.
2. Gravitational forces are present, but there is no normal force.
3. The net force is equal to the weight.
4. The object is stationary in space.

Answer

Gravitational forces are present, but there is no normal force.

Reason — The weight of a body is felt due to the reactionary force applied on the body by some other body in contact with it. When this reaction, that is the normal force, becomes zero, we feel as if our weight has also become zero, and this is called the state of weightlessness. Gravitational forces continue to act throughout; it is only the absence of the normal reaction that produces the sensation of weightlessness.

Which situation would result in a condition closest to weightlessness on Earth?

1. Jumping off a tall building.
2. Skydiving before deploying a parachute.
3. Standing on a high mountain.
4. Swimming underwater.

Answer

Skydiving before deploying a parachute.

Reason — Before the parachute is deployed, the skydiver is falling freely under gravity with no upward reaction acting on him, and this free fall is the condition closest to weightlessness. Jumping off a tall building would also produce free fall, but only very briefly and it is not a controlled situation. Standing on a high mountain and swimming underwater both involve a normal or buoyant reaction acting on the body, so weight continues to be felt.

The time period of a satellite in a circular orbit around the Earth depends on:

1. its mass.
2. the radius of the orbit.
3. the speed of the satellite.
4. the mass of the Earth.

Answer

the radius of the orbit.

Reason — The period of revolution of a satellite in an orbit of radius r = R_e + h is

$\text T = 2\pi\sqrt{\dfrac{(\text R_e + \text h)^3}{\text{GM}_e}}$

This expression is free from the mass of the satellite. It shows that the period depends only upon the height of the satellite above the earth's surface, that is, on the radius of its orbit. Two satellites of different masses revolving in the same orbit therefore have the same period.

The escape velocity on the Moon compared to that on Earth is:

1. higher because the moon is smaller.
2. lower because the moon's gravitational pull is weaker.
3. the same as earth's due to the same gravitational constant.
4. dependent on the satellite's mass.

Answer

lower because the moon's gravitational pull is weaker.

Reason — The escape velocity from the surface of a body of mass M and radius R is

$\text v_e = \sqrt{\dfrac{2\text{GM}}{\text R}}$

which depends only on the mass and radius of that body, not on the mass of the escaping object. The moon has a much smaller mass than the earth, so its gravitational pull is weaker and its escape velocity is only about 2.4 km s^-1, against 11.2 km s^-1 for the earth.

If g is the acceleration due to gravity at Earth's surface and R_e is the radius of Earth, the acceleration due to gravity g' at a height h above the Earth's surface is given by:

1. $g' = g\left(1 + \dfrac{h}{R_e}\right)^{-2}$

2. $g' = g\left(1 + \dfrac{h}{R_e}\right)^{2}$

3. $g' = g\left(1 + \dfrac{R_e}{h}\right)^{-2}$

4. $g' = g\left(1 + \dfrac{h}{R_e}\right)^{\frac{1}{2}}$

Answer

$g' = g\left(1 + \dfrac{h}{R_e}\right)^{-2}$

Reason — If g is the acceleration due to gravity at the earth's surface, then at a height h above the surface

$\text g = \dfrac{\text{GM}_e}{\text R_e^2} \quad \text{and} \quad \text g' = \dfrac{\text{GM}_e}{(\text R_e + \text h)^2}$

Dividing the second by the first,

$\dfrac{\text g'}{\text g} = \dfrac{\text R_e^2}{(\text R_e + \text h)^2} = \dfrac{1}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$

$\text g' = \text g\left(1 + \dfrac{\text h}{\text R_e}\right)^{-2}$

A planet's orbital period is 8 years. What is the semi-major axis of its orbit? (Assume the Sun as the central body and T² = kr³).

1. 2 AU
2. 4 AU
3. 8 AU
4. 16 AU

Answer

4 AU

Reason — By Kepler's third law, T² = k r³. Comparing the planet with the earth, for which T_e = 1 year and r_e = 1 AU,

$\left(\dfrac{\text T_p}{\text T_e}\right)^2 = \left(\dfrac{\text r_p}{\text r_e}\right)^3$

Substituting T_p = 8 years,

$\left(\dfrac{8}{1}\right)^2 = \left(\dfrac{\text r_p}{1}\right)^3 \quad \Rightarrow \quad \text r_p^3 = 64$

$\text r_p = 4\ \text{AU}$

Assuming the mass of Saturn to be constant, if its radius were to increase, then the acceleration due to gravity at its surface would:

1. increase
2. decrease
3. remain unchanged
4. first increase and then stay constant beyond a certain size

Answer

decrease

Reason — The acceleration due to gravity at the surface of a planet is

$\text g = \dfrac{\text{GM}}{\text r^2}$

Here G and the mass M of Saturn are constants, so $\text g \propto \dfrac{1}{\text r^2}$. As the radius r increases, the acceleration due to gravity at its surface falls off with the square of the radius, and therefore decreases.

For which of the following does the graph denote the variation of force of gravity 'F' along a distance 'r'?

1. Solid sphere
2. Spherical shell
3. Plate
4. Point

Answer

Spherical shell

Reason — The graph shows that the force of gravity is zero up to a distance R and then falls off as $\dfrac{1}{\text r^2}$ beyond R.

By Newton's shell theorem, a spherically symmetric shell of mass exerts no net gravitational force on a particle located anywhere inside it, so F = 0 for r < R. For a point outside the shell, the shell behaves like a point mass located at its centre, so

$\text F = \dfrac{\text{GMm}}{\text r^2} \quad \Rightarrow \quad \text F \propto \dfrac{1}{\text r^2}$

This is exactly the variation shown, so the graph is that of a spherical shell.

For which of the following does the graph denote the variation of force of gravity 'F' along a distance 'r'?

1. Solid sphere
2. Spherical shell
3. Plate
4. Point

Answer

Solid sphere

Reason — The graph shows that the force of gravity increases linearly with r up to r = R and then decreases as $\dfrac{1}{\text r^2}$.

For a point inside a solid sphere of uniform density, only the mass enclosed within the radius r contributes to the force, and

$\text F = \dfrac{\text{GMm}}{\text R^3}\text r \quad \Rightarrow \quad \text F \propto \text r$

so the force is zero at the centre and rises linearly to a maximum at the surface. For a point outside, the whole mass acts as if concentrated at the centre and

$\text F = \dfrac{\text{GMm}}{\text r^2} \quad \Rightarrow \quad \text F \propto \dfrac{1}{\text r^2}$

This is the variation shown, so the graph is that of a solid sphere.

A satellite is moving in a stable orbit around Earth. If the satellite's velocity is suddenly increased by a small amount in the direction of its motion, which of the following occurs?

1. The satellite will enter an elliptical orbit with apogee at the point where velocity was increased.
2. The satellite will escape Earth's gravity.
3. The satellite will spiral outwards in a circular path.
4. The satellite's orbital period will decrease.

Answer

The satellite will enter an elliptical orbit with apogee at the point where velocity was increased.

Reason — Increasing the satellite's velocity slightly increases its kinetic energy and hence its total energy, disturbing the balance required for a circular orbit. The satellite therefore changes its path to an elliptical orbit. The point at which the velocity was increased becomes the apogee of the new orbit, since the satellite begins to move away from the earth from that point. The increase being small, the satellite does not attain the escape velocity and so it does not escape earth's gravity.

Dimensional formula for gravitational intensity is :

1. [LT^-1]
2. [LT^-2]
3. [MLT^-1]
4. [MLT^-2]

Answer

[LT^-2]

Reason — The gravitational field intensity is the force experienced by a unit mass placed at a point in the field,

$\text I = \dfrac{\text{GM}}{\text r^2} = \dfrac{\text F}{\text m}$

Its dimensional formula is therefore

$[\text I] = \dfrac{[\text{MLT}^{-2}]}{[\text M]} = [\text{LT}^{-2}]$

which is the same as that of acceleration, as it should be, since the gravitational field intensity of the earth is numerically equal to g.

The time period of a satellite orbiting close to Earth is:

1. 24 h
2. About 84 min
3. 12 h
4. 30 min

Answer

About 84 min

Reason — For a satellite orbiting very close to the earth's surface (h << R_e), the period of revolution is

$\text T = 2\pi\sqrt{\dfrac{\text R_e}{\text g}}$

Substituting R_e = 6.37 × 10⁶ m and g = 9.8 m s^-2,

$\text T = 2 \times 3.14 \times \sqrt{\dfrac{6.37 \times 10^6}{9.8}} = 5063\ \text s \approx 84\ \text{min}$

The total energy (E = K + U) of a satellite orbiting at a distance r from the centre of Earth is:

1. $E = -\dfrac{GM_em}{2R_e}$

2. $E = -\dfrac{GM_em}{R_e}$

3. $E = -\dfrac{GM_e}{2R_e}$

4. $E = -\dfrac{GM_em}{2R_e}$

Answer

$E = -\dfrac{GM_em}{2R_e}$

Reason — For a satellite of mass m revolving close to the earth in an orbit of radius R_e, the potential energy and the kinetic energy are

$\text U = -\dfrac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text K = \dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

Hence the total energy of the satellite is

$\text E = \text U + \text K = -\dfrac{\text{GM}_e \text m}{\text R_e} + \dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e} = -\dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

Note: Options 1 and 4 are printed identically as $-\dfrac{\text{GM}_e \text m}{2\text R_e}$ in the textbook, and the answer key marks option 1.

Two satellites, X and Y, have identical masses but different orbital radii around the Earth. If satellite X has an orbital radius twice that of Satellite Y, the ratio of their orbital speeds, v_x : v_y is:

1. 1 : 2
2. 1 : $\sqrt2$
3. $\sqrt2$ : 1
4. 2 : 1

Answer

1 : $\sqrt2$

Reason — The orbital speed of a satellite in an orbit of radius r is

$\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text r}} \quad \Rightarrow \quad \text v_o \propto \dfrac{1}{\sqrt{\text r}}$

which is independent of the mass of the satellite. Given that r_x = 2r_y,

$\dfrac{\text v_x}{\text v_y} = \sqrt{\dfrac{\text r_y}{\text r_x}} = \sqrt{\dfrac{\text r_y}{2\text r_y}} = \dfrac{1}{\sqrt2}$

$\text v_x : \text v_y = 1 : \sqrt2$

The ratio of the radius of a planet A to that of a planet B is 'r'. The ratio of acceleration due to gravity on the planets is 'p'. The ratio of escape velocities of the two planets is:

1. $\left(\dfrac{r}{p}\right)^{\frac{1}{2}}$

2. $(rp)^{\frac{1}{2}}$

3. $(rp)^{-\frac{1}{2}}$

4. $(p/r)^{-\frac{1}{2}}$

Answer

$(rp)^{\frac{1}{2}}$

Reason — The escape velocity from the surface of a planet of radius R on which the acceleration due to gravity is g is

$\text v_e = \sqrt{2\text{gR}}$

For the two planets A and B,

$\dfrac{\text v_{e(A)}}{\text v_{e(B)}} = \sqrt{\dfrac{2\text g_A \text R_A}{2\text g_B \text R_B}} = \sqrt{\dfrac{\text g_A}{\text g_B}} \times \sqrt{\dfrac{\text R_A}{\text R_B}}$

Given $\dfrac{\text R_A}{\text R_B} = \text r$ and $\dfrac{\text g_A}{\text g_B} = \text p$,

$\dfrac{\text v_{e(A)}}{\text v_{e(B)}} = \sqrt{\text r \times \text p} = (\text{rp})^{1/2}$

If the gravitational constant G were to suddenly increase by 10%, the escape velocity from Earth would:

1. increase by about 10%
2. increase by about 5%
3. increase by about 20%
4. remain the same.

Answer

increase by about 5%

Reason — The escape velocity from the earth's surface is

$\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}} \quad \Rightarrow \quad \text v_e \propto \sqrt{\text G}$

If G is increased by 10%, the new value is G' = 1.10 G, so the new escape velocity is

$\text v_e' = \text v_e \times \sqrt{1.10}$

The percentage increase in the escape velocity is therefore

$\dfrac{\text v_e' - \text v_e}{\text v_e} \times 100 = (\sqrt{1.10} - 1) \times 100 \\[1em] = (1.0488 - 1) \times 100 = 4.88$

If the radius of the Earth were to shrink by 1%, keeping its mass constant, the gravitational acceleration on its surface would:

1. increase by approximately 2%
2. decrease by approximately 2%
3. remain unchanged
4. decrease by approximately 1%.

Answer

increase by approximately 2%

Reason — The acceleration due to gravity at the earth's surface is

$\text g = \dfrac{\text{GM}_e}{\text R_e^2} \quad \Rightarrow \quad \text g \propto \dfrac{1}{\text R_e^2}$

Taking the fractional change on both sides, since the mass is kept constant,

$\dfrac{\Delta \text g}{\text g} = -2\dfrac{\Delta \text R_e}{\text R_e}$

The radius shrinks by 1%, that is $\dfrac{\Delta \text R_e}{\text R_e} = -1$, so

$\dfrac{\Delta \text g}{\text g} = -2 \times (-1$

Hence g increases by approximately 2%.

Assertion (A): The gravitational force between two masses is inversely proportional to the square of the distance between them.

Reason (R): Gravitational force follows an inverse-square law as per Newton's law of universal gravitation.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: By Newton's universal law of gravitation, the force of attraction between two material particles is inversely proportional to the square of the distance between them, $\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$.

Reason (R) is also correct: This dependence on $\dfrac{1}{\text r^2}$ is precisely what is termed the inverse-square law, and it is a part of the statement of Newton's law of universal gravitation.

Since the inverse-square dependence stated in the Reason is the very relation asserted in (A), the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Gravitational force is a conservative force.

Reason (R): The work done by the gravitational force on a moving object is path-independent.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Gravitational force is a central force, that is, it acts along the line joining the centres of two bodies, and central forces are conservative in nature. Hence gravitational force is an example of a conservative force.

Reason (R) is also correct: A conservative force is one for which the work done in moving a body from one position to another depends only on the initial and the final positions and not on the path followed, that is, the work done is path-independent.

Path-independence of the work done is the defining property of a conservative force, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The gravitational force between two objects is independent of the medium between them.

Reason (R): Gravitational force depends only on the masses of the objects and the distance between them.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The gravitational forces between two bodies are completely independent of the presence of other bodies and of the properties of the intervening medium. If the force of attraction between two bodies in air is F, it remains the same when they are placed in water, the separation being unchanged.

Reason (R) is also correct: By Newton's law of gravitation, $\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$, the force depends only on the masses of the two bodies and on the distance between them.

Since the expression for the force contains only the masses and the separation, and no term for the medium, the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The gravitational force between two particles in a vacuum is zero.

Reason (R): The gravitational force acts along the line joining the masses.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: Gravitational force exists even in a vacuum and does not require any medium to act. Two particles placed in vacuum attract each other with the force $\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$, which is certainly not zero.

Reason (R) is correct as a statement of fact: Gravitational force is a central force, that is, it acts along the line joining the centres of the two masses.

Therefore, both assertion and reason are false.

Note: The answer key is incorrect. The assertion is false, but the reason is true. Therefore, none of the given options is correct.

Assertion (A): The acceleration due to gravity decreases with height above the Earth's surface.

Reason (R): The value of g is inversely proportional to the square of the distance from the centre of the Earth.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: As we go above the surface of the earth, the acceleration due to gravity goes on decreasing, since $\text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$, which is less than g.

Reason (R) is also correct: At a height h the distance from the earth's centre becomes (R_e + h), and $\text g' = \dfrac{\text{GM}_e}{(\text R_e + \text h)^2}$, so that $\text g' \propto \dfrac{1}{\text r^2}$.

As the height increases the distance from the earth's centre increases, and by the inverse-square dependence the value of g must fall. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): If earth stops rotating about its axis, the value of g is the same at all points on the Earth's surface.

Reason (R): The Earth is a perfect sphere.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: The variation of g over the earth's surface arises from two separate causes, the rotation of the earth and its oblate shape. Stopping the rotation removes only the term R_eω²cos²λ. The earth would still be an oblate spheroid, and since $\text g = \dfrac{\text{GM}_e}{\text R_e^2}$, that is $\text g \propto \dfrac{1}{\text R_e^2}$, the value of g would still be greater at the poles, where the radius is smaller, than at the equator. Hence g would not be the same at all points on the surface.

Reason (R) is false: The earth is not a perfect sphere. It is an oblate spheroid, slightly flattened at the poles and bulging at the equator, its equatorial diameter being about 21 km greater than the polar diameter.

Therefore, both assertion and reason are false.

Note: The textbook’s explanation shows that variations in g are caused by both the earth’s rotation and its flattened shape. Since some variation would remain even if the earth stopped rotating, the assertion is false. Therefore, option 4 is correct as the options stand.

Assertion (A): The value of g decreases as we move from the equator to the poles.

Reason (R): The radius of the Earth is greater at the pole than at the equator.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: The value of g actually increases as we move from the equator to the poles, both because the distance from the earth's centre decreases and because the centrifugal effect of rotation, $\text R_e\omega^2\cos^2\lambda$, becomes smaller as the latitude increases.

Reason (R) is also false: The radius of the earth is greater at the equator than at the poles, the equatorial radius being about 6378 km against the polar radius of about 6357 km.

Therefore, both assertion and reason are false.

Assertion (A): The value of g decreases inside the Earth as we move towards the centre.

Reason (R): The gravitational force inside the Earth decreases linearly with distance from the Earth's centre.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: At a depth h below the surface, $\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$, so that g' < g. As we go below the surface of the earth the acceleration due to gravity goes on decreasing, and it becomes zero at the centre.

Reason (R) is also correct: At a depth h, by Newton's shell theorem the outer spherical shell exerts no force, and only the inner solid sphere of radius (R_e − h) attracts the body. Hence the mass contributing to the gravitational pull decreases as we go deeper, and the force varies linearly with the distance from the centre.

Because the effective attracting mass decreases linearly with the distance from the centre, the value of g must also decrease as we move inward. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The value of g at a height h above the Earth is greater than that at the Earth's surface.

Reason (R): The gravitational force is directly proportional to the height above the Earth's surface.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: The value of g at a height h above the earth is less than that at the earth's surface, since $\text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$ is smaller than g.

Reason (R) is also false: The gravitational force is inversely proportional to the square of the distance from the earth's centre, and is not directly proportional to the height above the earth's surface.

Therefore, both assertion and reason are false.

Assertion (A): The value of g at the poles is slightly higher than at the equator.

Reason (R): The Earth's polar radius is smaller than the equatorial radius.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The value of g is about 9.83 m s^-2 at the poles and about 9.78 m s^-2 at the equator, so g is slightly higher at the poles.

Reason (R) is also correct: The earth is an oblate spheroid, so its polar radius is smaller than its equatorial radius.

Since $\text g = \dfrac{\text{GM}_e}{\text R_e^2}$, that is $\text g \propto \dfrac{1}{\text R_e^2}$, a smaller radius at the poles gives a larger value of g there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Orbital velocity of a satellite depends on its mass.

Reason (R): Orbital velocity is derived from balancing gravitational force and centripetal force, which both depend on the mass of the satellite.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: The orbital velocity of a satellite, $\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text r}}$, is free from the mass of the satellite. Hence it does not depend on the satellite's mass.

Reason (R) is also false: Although the orbital velocity is obtained by balancing the gravitational force against the required centripetal force, the mass m of the satellite appears on both sides of $\dfrac{\text{GM}_e \text m}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$ and cancels out. The result therefore depends only on the mass of the central body and the radius of the orbit.

Therefore, both assertion and reason are false.

Assertion (A): The orbital velocity of a satellite is higher for orbits closer to the Earth.

Reason (R): Gravitational force is stronger at lower altitudes.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The orbital velocity $\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text r}}$ varies as $\dfrac{1}{\sqrt{\text r}}$, so a satellite in an orbit closer to the earth (smaller r) has a higher orbital velocity.

Reason (R) is also correct: The gravitational force $\dfrac{\text{GM}_e \text m}{\text r^2}$ is stronger at lower altitudes, since the distance from the earth's centre is smaller.

A stronger gravitational pull at a lower altitude demands a larger centripetal force, and hence a higher orbital velocity, to maintain the orbit. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The orbital velocity of a satellite decreases with the height of its orbit.

Reason (R): The gravitational force decreases as the distance from the Earth increases.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since $\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text R_e + \text h}}$, the orbital velocity decreases as the height h of the orbit increases.

Reason (R) is also correct: The gravitational force varies inversely as the square of the distance, so it decreases as the distance from the earth increases.

At higher altitudes the weaker gravitational pull requires a smaller centripetal force, so a lower orbital velocity is sufficient to maintain the orbit. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Satellites in geostationary orbit have the highest orbital velocity among all satellites.

Reason (R): Geostationary satellites orbit close to the Earth's surface.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are false.

Explanation

Assertion (A) is false: Geostationary satellites are in high orbits, at a height of about 35,830 km, and have a comparatively low orbital velocity of about 3.1 km s^-1, against about 8 km s^-1 for satellites in low earth orbits.

Reason (R) is also false: Geostationary satellites do not orbit close to the earth's surface; they are placed far above it in the parking orbit above the equator.

Therefore, both assertion and reason are false.

Assertion (A): Orbital velocity is independent of the mass of the satellite.

Reason (R): Orbital velocity depends only on the mass of the central body (around which, the satellite revolves) and the radius of the orbit.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The mass m of the satellite cancels out in $\dfrac{\text{GM}_e \text m}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$, giving $\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text r}}$, which is independent of the satellite's mass.

Reason (R) is also correct: The resulting expression contains only the mass M_e of the central body around which the satellite revolves and the radius r of the orbit.

Because the orbital velocity is determined entirely by the central mass and the orbital radius, it cannot depend on the mass of the satellite. This is why two satellites of different masses revolving in the same orbit have the same speed. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Escape velocity is the speed needed to break free from the Earth's gravitational pull.

Reason (R): Escape velocity from the Earth is dependent on the direction of projection.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body without any further propulsion.

Reason (R) is false: The escape velocity, $\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}}$, is independent of the angle, that is the direction, of projection. It depends only on the mass and radius of the earth.

Therefore, assertion is true but reason is false.

Assertion (A): The escape velocity from a planet is greater than the orbital velocity for a satellite around the same planet.

Reason (R): Escape velocity is the minimum speed required to leave the gravitational field without further propulsion.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: For a body near the earth's surface, $\text v_e = \sqrt{2\text{gR}_e}$ and $\text v_o = \sqrt{\text{gR}_e}$, so that $\text v_e = \sqrt2\text v_o$. The escape velocity is therefore greater than the orbital velocity.

Reason (R) is also correct: Escape velocity is the minimum speed required to leave the gravitational field of the planet without further propulsion, whereas the orbital velocity is only the speed needed to remain bound in a circular orbit.

Since escaping the field altogether demands more energy than merely remaining in a bound orbit, the escape velocity must exceed the orbital velocity. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Gravitational pull is stronger for more massive planets.

Reason (R): Escape velocity decreases with an increase in the mass of the planet.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: By Newton's law of gravitation the force of attraction is directly proportional to the mass of the attracting body, so a more massive planet exerts a stronger gravitational pull.

Reason (R) is false: Since $\text v_e = \sqrt{\dfrac{2\text{GM}}{\text R}}$, the escape velocity increases with the mass of the planet, because a stronger gravitational pull requires a higher speed to escape.

Therefore, assertion is true but reason is false.

Assertion (A): The escape velocity from the Earth is higher at the poles than at the equator.

Reason (R): The Earth's radius is smaller at the poles than at the equator.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The escape velocity $\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}}$ varies as $\dfrac{1}{\sqrt{\text R_e}}$, so it is higher at the poles, where the radius is smaller.

Reason (R) is also correct: The earth is an oblate spheroid, so its radius is smaller at the poles than at the equator.

A smaller radius at the poles gives a larger value of $\dfrac{2\text{GM}_e}{\text R_e}$ and hence a higher escape velocity there. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The escape velocity from the surface of the Moon is less than that from the Earth.

Reason (R): The Moon has less mass and a smaller radius than the Earth.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The escape velocity from the moon's surface is about 2.4 km s^-1, which is much less than the 11.2 km s^-1 required at the earth's surface.

Reason (R) is also correct: The moon has a considerably smaller mass and a smaller radius than the earth.

With a much smaller mass, the moon's gravitational pull is weaker, and $\text v_e = \sqrt{\dfrac{2\text{GM}}{\text R}}$ works out smaller for the moon in spite of its smaller radius. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Geosynchronous satellites appear stationary with respect to an observer on Earth.

Reason (R): These satellites complete one orbit around the Earth in 24 h.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: A geosynchronous satellite is synchronous with the earth's spin, so it appears stationary with respect to an observer on the earth.

Reason (R) is also correct: These satellites have a period of revolution exactly equal to the period of the axial motion of the earth, that is, 24 h.

Because the satellite completes one orbit in exactly the time the earth takes to complete one rotation, it stays above the same point and appears stationary. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): A geostationary satellite must orbit the Earth at an altitude of approximately 35,786 km.

Reason (R): At this altitude, the satellite's orbital period matches the Earth's rotational period.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Using $\text h = \left(\dfrac{\text T^2 \text{gR}_e^2}{4\pi^2}\right)^{1/3} - \text R_e$ with T = 24 h, the height of a geostationary satellite works out to about 35,830 km above the earth's surface.

Reason (R) is also correct: It is at exactly this altitude that the period of revolution of the satellite becomes equal to the rotational period of the earth.

The height is determined by the requirement that the satellite's period should match the earth's rotational period, so the Reason correctly explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Geostationary satellites orbit the Earth above the equator.

Reason (R): Geostationary satellites can be used for polar region communication.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: One of the conditions for a satellite to be geostationary is that its orbital plane should be coplanar with the equatorial plane of the earth. Hence geostationary satellites orbit above the equator.

Reason (R) is false: Precisely because they are positioned over the equator, geostationary satellites have limited effectiveness near the poles. Polar regions therefore rely on polar-orbiting satellites for communication and weather monitoring.

Therefore, assertion is true but reason is false.

Assertion (A): A satellite is geostationary if it remains fixed relative to a point on the Earth's surface.

Reason (R): All geosynchronous satellites are geostationary satellites.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: A satellite is called geostationary if it appears stationary over a fixed point on the earth's equator, that is, it remains fixed relative to a point on the earth's surface.

Reason (R) is false: All geostationary satellites are geosynchronous, but the converse is not true. A geosynchronous satellite need not have an equatorial orbit, and only those with an equatorial orbital plane are geostationary.

Therefore, assertion is true but reason is false.

Assertion (A): A geostationary satellite has a constant velocity relative to the Earth's surface.

Reason (R): The satellite's orbital velocity matches the rotational speed of the Earth.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Since a geostationary satellite moves exactly in step with the earth's rotation, its position relative to a point on the earth's surface does not change, so its velocity relative to that surface is constant.

Reason (R) is also correct: The satellite's angular velocity, and hence its orbital motion, matches the rotational speed of the earth.

Because the satellite's orbital motion is synchronised with the earth's rotation, it maintains a constant position, and hence a constant velocity, relative to a point on the surface. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): Kepler's first law describes the shape of planetary orbits as ellipses with the Sun at one focus.

Reason (R): According to Kepler's first law, planets move in circular orbits with the Sun at the centre.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: Kepler's first law, the law of orbits, states that all planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

Reason (R) is false: Kepler's first law asserts elliptical orbits with the sun at one focus, not circular orbits with the sun at the centre. A circle is only a special limiting case of an ellipse.

Therefore, assertion is true but reason is false.

Assertion (A): Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Reason (R): This law implies that a planet moves faster when it is closer to the Sun and slower when it is farther from the Sun.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Kepler's second law, the law of areas, states that a line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant.

Reason (R) is also correct: When the planet is nearest the sun its velocity is maximum, and when it is farthest from the sun its velocity is minimum.

For equal areas to be swept in equal times, the planet must move faster when the radius vector is short, that is near the sun, and slower when it is long. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): According to Kepler's third law, the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

Reason (R): Kepler's third law applies to all planets orbiting the Sun and shows the relationship between the orbital period and the distance from the Sun.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: Kepler's third law, the law of periods, states that the square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun, T² = K r³.

Reason (R) is also correct: The constant K is the same for all the planets, so the law relates the orbital period of every planet to its distance from the sun.

Since the relation T² = K r³ holds with the same constant for all planets, it is exactly the statement made in the Assertion. The Reason therefore explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

Assertion (A): The gravitational potential energy of an object increases as it moves away from the Earth.

Reason (R): Gravitational potential energy is inversely proportional to the distance from the Earth's centre.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The gravitational potential energy is $\text U = -\dfrac{\text{GM}_e \text m}{\text r}$. As r increases, the magnitude of U decreases, that is, U becomes less negative. Hence the potential energy increases as the object moves away from the earth.

Reason (R) is false: It is the magnitude of the potential energy that is inversely proportional to the distance from the earth's centre. Since U itself is negative, an inverse relation of the kind stated would wrongly imply that U decreases as r increases.

Therefore, assertion is true but reason is false.

Assertion (A): A satellite in a higher orbit has a longer period of revolution than one in a lower orbit.

Reason (R): The orbital period is proportional to the square of the radius of orbit.

1. If both assertion and reason are true and reason is the correct explanation of assertion.
2. If both assertion and reason are true but reason is not the correct explanation of assertion.
3. If assertion is true but reason is false.
4. If both assertion and reason are false.

Answer

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The period of revolution of a satellite is $\text T = 2\pi\sqrt{\dfrac{(\text R_e + \text h)^3}{\text{GM}_e}}$, which increases with the height of the orbit. Hence a satellite in a higher orbit has a longer period of revolution.

Reason (R) is false: By Kepler's third law the square of the orbital period is proportional to the cube of the radius of the orbit, T² ∝ r³, and not the period to the square of the radius.

Therefore, assertion is true but reason is false.

How many times large is electrostatic force between two electrons, to gravitational force between them?

Answer

The electrostatic force between two electrons is about 10⁴³ times the gravitational force between them.

Gravitational forces are the weakest forces existing in nature. This is why the gravitational attraction between ordinary bodies is never felt in daily life, while their electrostatic interaction is easily observed.

The mass of the moon is nearly 1% of the mass of the earth. What will be the gravitational force of the earth on the moon, in comparison to the gravitational force of the moon on the earth?

Answer

Both will be equal.

By Newton's law of gravitation the gravitational forces between two bodies form an action-reaction pair, that is, the forces exerted by two bodies on each other are equal in magnitude but oppositely directed. Hence the force of the earth on the moon and that of the moon on the earth are equal in magnitude, however different their masses may be.

For two bodies situated in vacuum, the value of G is 6.67 × 10^-11 N m² kg^-2. What will it be for bodies situated in a dense medium?

Answer

It will remain 6.67 × 10^-11 N m² kg^-2.

The gravitational constant G is a universal constant. Its value is the same for all pairs of bodies at all places and at all distances of separation, and gravitational forces are completely independent of the properties of the intervening medium. Hence the value of G is unaffected by the presence of a dense medium.

Does the value of g increase or decrease below the earth?

Answer

The value of g decreases below the earth's surface.

At a depth h below the surface only the inner solid sphere of radius (R_e − h) attracts the body, the outer spherical shell exerting no force. Hence

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

so that g' < g, and g becomes zero at the centre of the earth, where h = R_e.

The value of g on the moon is 1/6th of that on the earth. If a body is taken from the earth to the moon, then what will be the change in its (i) weight, (ii) inertial mass and (iii) gravitational mass?

Answer

(i) Weight : The weight of the body will become 1/6th of that on the earth, since W = mg and the value of g on the moon is 1/6th of that on the earth.

(ii) Inertial mass : There will be no change. Inertial mass is a measure of the inertia of the body and is a constant property of the body itself.

(iii) Gravitational mass : There will be no change. The gravitational mass is also a constant property of the body and does not depend on the place where the body is taken.

When a clock controlled by a pendulum is taken from the plane to a mountain, it becomes slow; but a wrist-watch controlled by a spring remains unaffected. Explain the reason of this difference in the behaviour of the two watches.

Answer

The time-period of a simple pendulum is $\text T = 2\pi\sqrt{\dfrac{\text l}{\text g}}$, that is, $\text T \propto \dfrac{1}{\sqrt{\text g}}$.

On a mountain the value of g is less than that on the plane, because the acceleration due to gravity decreases with altitude. Hence the time-period of the pendulum increases, and the pendulum clock becomes slow.

A wrist-watch, on the other hand, is controlled by a spring, and its time-period depends upon the elastic properties of the spring and not upon the value of g. Hence it remains unaffected by the variation of g.

If the earth stops rotating about its axis, what will be the effect on the value of 'g'? Will this effect be same at all places?

Answer

If the earth stops rotating about its axis, the value of g will increase at all places except at the poles.

The apparent value of the acceleration due to gravity is

$\text g' = \text g - \text R_e\omega^2\cos^2\lambda$

On stopping the rotation, ω becomes zero and the term R_eω²cos²λ vanishes.

No, the effect will not be the same at all places. The increase depends upon the latitude λ. It will be maximum at the equator (λ = 0°) and zero at the poles (λ = 90°), since there cos λ = 0 and the rotation makes no contribution.

Why is earth flat at the poles?

Answer

The earth is flat at the poles, that is, it is an oblate spheroid, due to the rotation of the earth about its polar axis.

Because of this rotation a centrifugal effect acts outwards from the axis of rotation. This effect is greatest at the equator and zero at the poles, so over a long time the earth has bulged at the equator and become flattened at the poles. The equatorial diameter is about 21 km greater than the polar diameter.

Why is the weight of a body at the equator less than the weight of the same body at the poles?

Answer

The weight of a body is W = mg, and the value of g is smaller at the equator than at the poles.

This is due to two causes. Firstly, the earth is an oblate spheroid, so its equatorial radius is greater than its polar radius, and since $\text g \propto \dfrac{1}{\text R_e^2}$, the value of g is smaller where the radius is larger. Secondly, the rotation of the earth reduces g by the term R_eω²cos²λ, which is maximum at the equator and zero at the poles.

Hence the same body weighs less at the equator than at the poles.

Where from does a satellite get centripetal force for moving around its planet?

Answer

A satellite gets the necessary centripetal force from the gravitational force of attraction exerted on it by its planet.

When the satellite revolves in a circular orbit of radius r, this gravitational force supplies exactly the centripetal force required,

$\dfrac{\text{GMm}}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$

This is why a satellite continues to revolve in its orbit without using any fuel.

The centripetal force on a satellite revolving around the earth is F. What is the gravitational force due to earth on it? Net force?

Answer

The gravitational force due to the earth on the satellite is F, and the net force on the satellite is also F.

The gravitational force of the earth on the satellite is the only force acting on it, and this force itself provides the required centripetal force. Hence both the gravitational force and the net force on the satellite are equal to F, directed towards the centre of the earth.

The earth is acted upon by the gravitational force of attraction due to the sun. Then why does the earth not fall towards sun?

Answer

The earth does not fall towards the sun because it is revolving around the sun in a stable orbit.

The gravitational force of attraction of the sun on the earth is exactly the centripetal force needed to keep the earth moving in its orbit. This force acts along the line joining the earth and the sun, that is, perpendicular to the direction of the earth's velocity at every instant. It therefore only changes the direction of the earth's motion and does not pull the earth inward to the sun.

"Two artificial satellites revolve in a particular orbit around the earth at the same speed, irrespective of the difference between their masses". Explain this statement.

Answer

The orbital velocity of a satellite revolving in an orbit of radius r is

$\text v_o = \sqrt{\dfrac{\text{GM}_e}{\text r}}$

In deriving this, the mass m of the satellite appears on both sides of $\dfrac{\text{GM}_e \text m}{\text r^2} = \dfrac{\text{mv}_o^2}{\text r}$ and cancels out. Hence the expression is free from the mass of the satellite and depends only on the mass of the earth and the radius of the orbit.

Therefore two artificial satellites revolving in the same orbit must have the same speed, irrespective of the difference between their masses.

If mass of the moon were twice its present value but the orbit would have been the same as at present, then would its period of revolution around the earth have been changed?

Answer

No, its period of revolution would not change.

The period of revolution of a satellite in an orbit of radius r is

$\text T = 2\pi\sqrt{\dfrac{\text r^3}{\text{GM}_e}}$

This expression contains the mass M_e of the earth and the radius of the orbit, but is free from the mass of the moon. Hence doubling the mass of the moon, the orbit remaining the same, would leave its period of revolution unchanged.

How much is the angular velocity, in radian per hour, of a geostationary satellite?

Answer

A geostationary satellite completes one revolution in 24 h, the same as the period of the axial motion of the earth. Hence its angular velocity is

$\omega = \dfrac{2\pi}{\text T} = \dfrac{2\pi}{24} = \dfrac{\pi}{12}\ \text{radian per hour}$

Hence, the angular velocity of a geostationary satellite is $\dfrac{\pi}{12}$ radian per hour.

An artificial satellite is revolving around the earth at a height 100 km from the earth's surface. If a packet is released from the satellite, what will happen to it? Will it reach the earth?

Answer

No, the packet will not reach the earth. It will also continue to revolve around the earth along with the satellite.

When the packet is released, it already possesses the same orbital velocity as the satellite at that height. Since the orbital velocity is independent of the mass of the body, the packet continues to move in the same orbit with the same speed, and it therefore behaves as a satellite of the earth itself.

Can we determine the weight of a body inside an artificial satellite?

Answer

No, the weight of a body cannot be determined inside an artificial satellite.

Inside the satellite every body is in a state of weightlessness, since the artificial satellite is like a freely-falling body. The gravitational force is entirely used up in providing the centripetal force, so the reactionary force on the body becomes zero. If the body is placed on a spring balance inside the satellite, the balance will read zero.

On the planet Venus a body has − 7.5 × 10⁶ J of gravitational potential energy. Work out the energy required for the body to escape from the planet.

Answer

Given,

• Gravitational potential energy of the body on Venus, U = − 7.5 × 10⁶ J

To escape from the planet, the body must be taken to infinity, where its gravitational potential energy is zero. Hence the energy required is

$\text E = \text U_\infty - \text U = 0 - (-7.5 \times 10^6)$

$\text E = + 7.5 \times 10^6\ \text J$

Hence, the energy required for the body to escape from Venus is + 7.5 × 10⁶ J.

What is meant by escape velocity?

Answer

Escape velocity is the minimum velocity that an object needs in order to escape the gravitational pull of a celestial body, without any further propulsion.

If a body is launched from the earth's surface with this speed, it will continue moving away from the earth, overcoming the earth's gravitational pull, and will not fall back or enter an orbit. Its value at the earth's surface is

$\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}} = \sqrt{2\text{gR}_e} = 11.2\ \text{km s}^{-1}$

The escape velocity from earth for a piece of 1 g is 11.2 km s^-1. What would it be for a piece of 10 g?

Answer

It will remain 11.2 km s^-1.

The escape velocity from the earth is

$\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}}$

This expression is free from the mass of the body. Hence the escape velocity does not depend upon the mass of the piece, and a piece of 10 g escapes with exactly the same velocity as a piece of 1 g.

What is the relationship of the orbital speed to the velocity required to send a body from the earth's surface into space, never to return?

Answer

The velocity required to send a body from the earth's surface into space, never to return, is the escape velocity v_e, and the orbital speed of a satellite close to the earth is v_o. Their values are

$\text v_e = \sqrt{2\text{gR}_e} \quad \text{and} \quad \text v_o = \sqrt{\text{gR}_e}$

Therefore

$\dfrac{\text v_o}{\text v_e} = \dfrac{\sqrt{\text{gR}_e}}{\sqrt{2\text{gR}_e}} = \dfrac{1}{\sqrt2} \quad \Rightarrow \quad \text v_e = \sqrt2\ \text v_o$

Hence, the escape velocity is $\sqrt2$ times the orbital speed.

If the kinetic energy of a satellite revolving in an orbit close to the earth happens to be doubled, will the satellite escape?

Answer

Yes, the satellite will escape.

If the kinetic energy of the satellite is doubled,

$\dfrac{1}{2}\text{mv}^2 = 2 \times \dfrac{1}{2}\text{mv}_o^2 \quad \Rightarrow \quad \text v = \sqrt2\ \text v_o$

But for a satellite revolving close to the earth, the escape velocity is exactly $\text v_e = \sqrt2\text v_o$. Hence its velocity becomes equal to the escape velocity, and the satellite will leave its orbit and escape.

The distance of a planet from sun is 40 times the distance of earth. If the masses of earth and planet be equal, then compare the gravitational forces of sun on these two planets.

Answer

Given,

• Distance of the planet from the sun = 40 times the distance of the earth from the sun
• Masses of the earth and the planet are equal

By Newton's law of gravitation the force of the sun on a body of mass m at a distance r is

$\text F = \dfrac{\text{GM}_s \text m}{\text r^2} \quad \Rightarrow \quad \text F \propto \dfrac{1}{\text r^2}$

since M_s and m are the same in both cases. Therefore

$\dfrac{\text F_{earth}}{\text F_{planet}} = \left(\dfrac{\text r_{planet}}{\text r_{earth}}\right)^2 = (40)^2 = 1600$

Hence, the gravitational forces of the sun on the earth and on the planet are in the ratio 1600 : 1.

The acceleration due to gravity on a satellite is 1.96 m s^-2, while on earth it is 9.80 m s^-2. If jumping from a height of 5 m is safe at the earth, then jumping from how much height will be safe at the satellite?

Answer

Given,

• Acceleration due to gravity on the satellite, g_s = 1.96 m s^-2
• Acceleration due to gravity on the earth, g_e = 9.80 m s^-2
• Safe height of jumping on the earth, h_e = 5 m

While jumping down from a height h, the potential energy mgh is converted into kinetic energy on landing. For the jump to be equally safe, this energy must be the same in both cases,

$\text{mg}_e \text h_e = \text{mg}_s \text h_s \quad \Rightarrow \quad \text h_s = \text h_e \times \dfrac{\text g_e}{\text g_s}$

Substituting the values,

$\text h_s = 5 \times \dfrac{9.80}{1.96} = 5 \times 5 = 25\ \text m$

Hence, jumping from a height of 25 m will be safe on the satellite.

A stone dropped from a height h strikes the earth in 1 s. If the same stone be taken to the moon and dropped from the same height h, then what time will it take in striking the surface of the moon? Acceleration due to gravity at moon is 1/6th the acceleration due to gravity at earth.

Answer

Given,

• Time taken to strike the earth, t_e = 1 s
• Acceleration due to gravity at the moon, $\text g_m = \dfrac{\text g_e}{6}$
• The stone is dropped from the same height h in both cases

For a body dropped from rest, $\text h = \dfrac{1}{2}\text{gt}^2$, so that for the same height

$\text t = \sqrt{\dfrac{2\text h}{\text g}} \quad \Rightarrow \quad \text t \propto \dfrac{1}{\sqrt{\text g}}$

Therefore,

$\dfrac{\text t_m}{\text t_e} = \sqrt{\dfrac{\text g_e}{\text g_m}} = \sqrt{\dfrac{\text g_e}{\text g_e/6}} = \sqrt6$

$\text t_m = 1 \times \sqrt6 = \sqrt6\ \text s$

Hence, the stone will take $\sqrt6$ s to strike the surface of the moon.

The distance between two bodies A and B is r. Taking the gravitational force according to the law of inverse square of r, the acceleration of the body A is a. If the gravitational force follows an inverse fourth power law, then what will be the acceleration of the body A?

Answer

Given,

• Distance between the bodies A and B = r
• Acceleration of A under the inverse square law = a

Under the inverse square law the force on A is $\text F = \dfrac{\text{Gm}_A \text m_B}{\text r^2}$, so its acceleration is

$\text a = \dfrac{\text F}{\text m_A} = \dfrac{\text{Gm}_B}{\text r^2}$

If the gravitational force follows an inverse fourth power law, the force becomes $\text F' = \dfrac{\text{Gm}_A \text m_B}{\text r^4}$, and the acceleration of A is

$\text a' = \dfrac{\text F'}{\text m_A} = \dfrac{\text{Gm}_B}{\text r^4} = \dfrac{1}{\text r^2}\left(\dfrac{\text{Gm}_B}{\text r^2}\right)$

$\text a' = \dfrac{\text a}{\text r^2}$

Hence, the acceleration of the body A would be $\dfrac{\text a}{\text r^2}$.

If the diameter of the earth becomes twice its present value but its mass remains unchanged, then how would be the weight of an object on the surface of the earth affected?

Answer

The weight will reduce to one-fourth of its present value.

The acceleration due to gravity at the earth's surface is

$\text g = \dfrac{\text{GM}_e}{\text R_e^2}$

The mass M_e remains unchanged while the diameter, and hence the radius, becomes twice, that is R_e' = 2R_e. Therefore

$\dfrac{\text g'}{\text g} = \left(\dfrac{\text R_e}{2\text R_e}\right)^2 = \dfrac{1}{4} \quad \Rightarrow \quad \text g' = \dfrac{\text g}{4}$

Since W = mg, the weight of the object also becomes one-fourth.

If the diameter of the earth becomes half its present value but its average density remains unchanged, then how would be the weight of an object on the surface of the earth affected?

Answer

The weight will be halved.

Since the average density ρ remains unchanged, the acceleration due to gravity is

$\text g = \dfrac{\text G}{\text R_e^2} \times \dfrac{4}{3}\pi \text R_e^3 \rho = \dfrac{4}{3}\pi \text{GR}_e\rho \quad \Rightarrow \quad \text g \propto \text R_e$

The diameter, and hence the radius, becomes half, so

$\text g' = \dfrac{\text g}{2}$

Since W = mg, the weight of the object on the surface is also halved.

The mass and the diameter of a planet are twice those of the earth. What will be acceleration due to gravity at the planet, if acceleration due to gravity on the earth is g?

Answer

Given,

• Mass of the planet, M_p = 2M_e
• Diameter of the planet = twice that of the earth, so R_p = 2R_e
• Acceleration due to gravity on the earth = g

The acceleration due to gravity at the surface of a planet is

$\text g = \dfrac{\text{GM}}{\text R^2}$

Therefore,

$\dfrac{\text g_p}{\text g} = \dfrac{\text M_p}{\text M_e} \times \left(\dfrac{\text R_e}{\text R_p}\right)^2 = 2 \times \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{2}$

$\text g_p = \dfrac{\text g}{2}$

Hence, the acceleration due to gravity at the planet is $\dfrac{\text g}{2}$.

The weight of a person on the earth is 600 N. The gravitational field of the moon is 1/6th of the gravitational field of the earth. (a) What will be the weight of the person on the moon? (b) If the person can jump 2 m high on the earth, how much high can he jump on the moon? (c) What is the mass of the person on the earth? On the moon? (g on earth = 10 N / kg).

Answer

Given,

• Weight of the person on the earth, W_e = 600 N
• Gravitational field of the moon $= \dfrac{1}{6}$ of that of the earth
• Height of jump on the earth = 2 m
• g on earth = 10 N kg^-1

(a) Weight on the moon : Since W = mg and the gravitational field of the moon is 1/6th that of the earth,

$\text W_m = \dfrac{\text W_e}{6} = \dfrac{600}{6} = 100\ \text N$

(b) Height of jump on the moon : While jumping, the same muscular energy mgh is expended, so

$\text{mg}_e \text h_e = \text{mg}_m \text h_m \quad \Rightarrow \quad \text h_m = \text h_e \times \dfrac{\text g_e}{\text g_m} = 2 \times 6 = 12\ \text m$

(c) Mass of the person : The mass of a body is a constant property and does not change with place,

$\text m = \dfrac{\text W_e}{\text g_e} = \dfrac{600}{10} = 60\ \text{kg}$

Hence, the weight on the moon is 100 N, the height of the jump on the moon is 12 m, and the mass is 60 kg both on the earth and on the moon.

If a man goes from the surface of the earth to a height equal to the radius of the earth, then what will be his weight relative to that on the earth? What if he goes equally below the surface of the earth?

Answer

At a height equal to the radius of the earth : The acceleration due to gravity at a height h is

$\text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$

Putting h = R_e,

$\text g' = \dfrac{\text g}{(1 + 1)^2} = \dfrac{\text g}{4}$

Hence his weight will be one-fourth of that on the earth's surface.

At a depth equal to the radius of the earth : The acceleration due to gravity at a depth h is

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

Putting h = R_e, we get g' = 0. This point is the centre of the earth, so his weight will be zero.

Calculate the distance (i) below the surface of the earth (ii) above the surface of the earth, at which the value of acceleration due to gravity becomes 1/4th the value of 'g' at the surface of the earth.

Answer

Given,

• Acceleration due to gravity at the required point $= \dfrac{\text g}{4}$

(i) Below the surface of the earth : At a depth h,

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

Putting $\text g' = \dfrac{\text g}{4}$,

$\dfrac{1}{4} = 1 - \dfrac{\text h}{\text R_e} \quad \Rightarrow \quad \dfrac{\text h}{\text R_e} = \dfrac{3}{4}$

$\text h = \dfrac{3}{4}\text R_e$

(ii) Above the surface of the earth : At a height h,

$\text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$

Putting $\text g' = \dfrac{\text g}{4}$,

$\left(1 + \dfrac{\text h}{\text R_e}\right)^2 = 4 \quad \Rightarrow \quad 1 + \dfrac{\text h}{\text R_e} = 2$

$\text h = \text R_e$

Hence, the distances are $\dfrac{3}{4}\text R_e$ below the surface and R_e above the surface.

At what depth below the surface of the earth the weight of a person will be one-fourth of his weight at the surface of the earth? Radius of the earth is 6400 km.

Answer

Given,

• Weight at the depth $= \dfrac{1}{4}$ of the weight at the surface
• Radius of the earth, R_e = 6400 km

Since W = mg, the acceleration due to gravity at that depth must be $\text g' = \dfrac{\text g}{4}$. At a depth h,

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

Substituting $\text g' = \dfrac{\text g}{4}$,

$\dfrac{1}{4} = 1 - \dfrac{\text h}{\text R_e} \quad \Rightarrow \quad \dfrac{\text h}{\text R_e} = \dfrac{3}{4}$

$\text h = \dfrac{3}{4} \times 6400 = 4800\ \text{km}$

Hence, at a depth of 4800 km the weight of the person will be one-fourth of his weight at the surface.

The average distance of the sun from a planet is four times in comparison to its distance from the earth. In how many years that planet will complete one revolution around the sun?

Answer

Given,

• Distance of the planet from the sun, r_p = 4 r_e
• Period of revolution of the earth, T_e = 1 year

By Kepler's law of periods, T² ∝ r³, so that $\text T \propto \text r^{3/2}$. Therefore

$\dfrac{\text T_p}{\text T_e} = \left(\dfrac{\text r_p}{\text r_e}\right)^{3/2} = (4)^{3/2} = 8$

$\text T_p = 8 \times 1 = 8\ \text{years}$

Hence, the planet will complete one revolution around the sun in 8 years.

A force acts upon the earth revolving in a circular path about the sun. Hence, work should be done on the earth. Give comment on this statement.

Answer

The statement is not correct. No work is done on the earth.

The gravitational force of the sun on the earth is a centripetal force, directed towards the sun along the radius of the orbit. The earth's velocity at every instant is along the tangent to its circular path, that is, perpendicular to this force. Hence the angle between the force and the displacement is 90°, and

$\text W = \text F\text s\cos 90^\circ = 0$

Hence, the work done by the gravitational force of the sun on the earth is zero.

Write the various formulae for the time-period of a satellite revolving around the earth at a height h from earth's surface.

Answer

For a satellite revolving at a height h above the earth's surface, in an orbit of radius r = R_e + h, the period of revolution may be written in the following equivalent forms :

$\text T = 2\pi\sqrt{\dfrac{(\text R_e + \text h)^3}{\text{GM}_e}}$

$\text T = 2\pi\sqrt{\dfrac{(\text R_e + \text h)^3}{\text{gR}_e^2}} \qquad (\because\ \text{GM}_e = \text{gR}_e^2)$

$\text T = \sqrt{\dfrac{3\pi(\text R_e + \text h)^3}{\text G\rho \text R_e^3}}$

where ρ is the mean density of the earth.

Write the speed and period of revolution of a satellite revolving 'near' the earth.

Answer

For a satellite revolving very close to the earth's surface (h << R_e), the orbital speed and the period of revolution are

$\text v_o = \sqrt{\text{gR}_e} = \sqrt{9.8 \times (6.37 \times 10^6)} \approx 8\ \text{km s}^{-1}$

$\text T = 2\pi\sqrt{\dfrac{\text R_e}{\text g}} = 2 \times 3.14 \times \sqrt{\dfrac{6.37 \times 10^6}{9.8}} \approx 84\ \text{min}$

Hence, the speed is nearly 8 km s^-1 and the period of revolution is about 84 min.

How many hours is the periodic time of revolution around the earth of the communication satellite INSAT II B?

Answer

The periodic time of revolution of the communication satellite INSAT II B is 24 h.

INSAT II B is a geostationary satellite. For a satellite to appear stationary over a point on the earth's equator, its period of revolution must be exactly equal to the period of the axial motion of the earth, that is, 24 h.

Write the formula for the maximum height attained by a projectile.

Answer

The maximum height attained by a projectile thrown upward from the earth's surface with a velocity v is

$\text h = \dfrac{\text v^2 \text R_e}{2\text{gR}_e - \text v^2}$

where R_e is the radius of the earth and g is the acceleration due to gravity at its surface. This formula is used when the height h is not negligible in comparison with R_e.

Write down the expression for the escape velocity of a body from the surface of the earth.

Answer

The escape velocity of a body from the surface of the earth is

$\text v_e = \sqrt{\dfrac{2\text{GM}_e}{\text R_e}}$

where M_e is the mass and R_e is the radius of the earth. Since GM_e = gR_e², this may also be written as

$\text v_e = \sqrt{2\text{gR}_e}$

If the orbital speed of a satellite close to the earth be increased by 41.4%, prove that the satellite will leave its orbit and escape to infinity.

Answer

Given,

• Increase in the orbital speed = 41.4%

The orbital speed of a satellite close to the earth is v_o. On increasing it by 41.4%, the new speed is

$\text v = \text v_o + \dfrac{41.4}{100}\text v_o = \dfrac{141.4}{100}\text v_o = 1.414\ \text v_o$

$\text v = \sqrt2\ \text v_o$

But for a satellite revolving close to the earth, the escape velocity is related to the orbital speed by

$\text v_e = \sqrt2\ \text v_o$

Hence the new speed of the satellite is exactly equal to the escape velocity.

Therefore the satellite will leave its orbit and escape to infinity.

Moon travellers tie heavy weight at their back before landing on the moon. Give reason.

Answer

Moon travellers tie heavy weights at their back because the value of g on the moon is small, being about 1/6th of its value on the earth.

Because of this small value of g, their weight W = mg on the moon is only one-sixth of their weight on the earth. The reduced weight means a reduced normal reaction and therefore reduced friction between their feet and the lunar surface, so they would tend to bounce off the surface and would find it difficult to walk steadily. Tying heavy weights increases their mass, and hence their weight on the moon, allowing them to walk normally.

State Kepler's laws of planetary motion.

Answer

Kepler gave the following three laws of planetary motion :

(i) Law of Orbits : All planets move around the sun in elliptical orbits having the sun at one focus of the orbit.

(ii) Law of Areas : A line joining any planet to the sun sweeps out equal areas in equal times, that is, the areal velocity of the planet remains constant. Consequently the velocity of a planet is maximum when it is nearest the sun and minimum when it is farthest from the sun.

(iii) Law of Periods : The square of the period of revolution of any planet around the sun is directly proportional to the cube of its mean distance from the sun,

$\text T^2 \propto \text r^3$

Give the mathematical form of Kepler's third law.

Answer

Kepler's third law, the law of periods, states that if the period of a planet around the sun is T and the mean radius of its orbit is r, then

$\text T^2 \propto \text r^3 \quad \text{or} \quad \text T^2 = \text K\text r^3$

where K is a constant which is the same for all the planets and has the value 2.97 × 10^-19 s² m^-3. If the orbit is an ellipse, then r may be replaced by the semi-major axis a of the ellipse.

Thus, larger the distance of a planet from the sun, larger will be its period of revolution around the sun.

Write down Newton's law of gravitation.

Answer

Newton's law of gravitation : The gravitational force of attraction between any two material particles is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. It acts along the line joining the two particles.

If two particles of masses m₁ and m₂ are situated a distance r apart and F is the force of attraction between them, then

$\text F \propto \dfrac{\text m_1 \text m_2}{\text r^2} \quad \text{or} \quad \text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$

where G is the gravitational constant, whose value is the same for all pairs of particles in the universe, and is therefore called a universal constant.

The value of 'g' changes when we go above or below the surface of the earth. Write the necessary formulae.

Answer

Above the earth's surface : At a height h above the surface, the distance from the centre becomes (R_e + h), so

$\text g' = \dfrac{\text{GM}_e}{(\text R_e + \text h)^2} \quad \Rightarrow \quad \text g' = \dfrac{\text g}{\left(1 + \dfrac{\text h}{\text R_e}\right)^2}$

that is, g' < g. If h is negligible compared with R_e, this may be written by the binomial theorem as

$\text g' = \text g\left(1 - \dfrac{2\text h}{\text R_e}\right)$

Below the earth's surface : At a depth h, only the inner solid sphere of radius (R_e − h) attracts the body, and

$\text g' = \text g\left(1 - \dfrac{\text h}{\text R_e}\right)$

that is, g' < g. At the centre of the earth h = R_e and g' becomes zero.

Hence, the value of g decreases both on going above and on going below the surface of the earth.

Kepler's second law is based on the conservation of which physical quantity?

Answer

Kepler's second law is based on the conservation of angular momentum of the planet about the sun.

The gravitational force on a planet is a central force directed towards the sun, so it exerts no torque about the sun. The angular momentum $\vec{\text L} \space$of the planet therefore remains constant, and since the areal velocity is

$\dfrac{\text d\vec{\text A}}{\text{dt}} = \dfrac{1}{2\text m}\vec{\text L}$

a constant $\vec{\text L} \space$means a constant areal velocity, which is exactly Kepler's second law.

Define 'universal gravitational constant'.

Answer

Newton's law of gravitation is $\text F = \text G\dfrac{\text m_1 \text m_2}{\text r^2}$.

If we put m₁ = m₂ = 1 and r = 1, then G = F.

Thus, the universal gravitational constant G is numerically equal to the force with which two particles, each of unit mass and placed a unit distance apart, attract each other.

Its S.I. unit is N m² kg^-2 and its value is 6.67 × 10^-11 N m² kg^-2. It is called a universal constant because its value is the same for all pairs of bodies, at all places, at all distances of separation, and it is independent of the intervening medium.

Which of the following observations point to the equivalence of inertial and gravitational masses?

(a) Two spheres of different masses dropped from the top of a long evacuated tube reach the bottom of the tube at the same time.

(b) The time-period of a simple pendulum is independent of its mass.

(c) For a man in a closed cabin that is falling freely under gravity, gravity 'disappears'.

(d) The gravitational force on a particle inside a hollow isolated sphere is zero.

(e) An astronaut inside a spaceship orbiting around the earth feels weightless.

(f) Planets orbiting around the sun obey Kepler's third law (approximately).

(g) The gravitational force on a body due to the earth is equal and opposite to the gravitational force on the earth due to the body.

Answer

The observations (a), (b), (c), (e) and (f) point to the equivalence of inertial and gravitational masses.

The equivalence of inertial and gravitational masses is established by measuring the accelerations produced in different bodies falling freely under the earth's gravitational field. In the observations (a), (b), (c), (e) and (f) the bodies are in motion and they all point to the equivalence of inertial and gravitational masses.

For a body falling freely, the gravitational force is m_Gg and by Newton's second law this equals m_Ia. Since all bodies are found to fall with the same acceleration, it follows that m_G = m_I.

The observations (d) and (g) do not point to this equivalence. Observation (d) is a consequence of Newton's shell theorem, and observation (g) is simply a statement of Newton's third law.

Generally the path of a projectile from the earth is parabolic but it is elliptical for projectiles going to a very great height. Why?

Answer

Up to ordinary heights the change in the distance of a projectile from the centre of the earth is negligible compared with the radius of the earth. Hence the projectile moves under a nearly uniform gravitational force which is constant in both magnitude and direction, and under such a force the path of the projectile is parabolic.

For projectiles going to a very great height, however, the gravitational force decreases in inverse proportion to the square of the distance of the projectile from the centre of the earth, and its direction also changes as the projectile moves. Under such a variable force the path of the projectile becomes elliptical.

A body is taken from the centre of the earth to the moon. What will be the changes in the weight of the body?

Answer

At the centre of the earth the weight of the body will be zero, since g = 0 there.

As the body moves from the centre to the earth's surface, the value of g increases according to $\text g' = \text g\left(\dfrac{\text r}{\text R_e}\right)$, so the weight increases and becomes maximum at the surface.

On moving above the earth's surface towards the moon the value of g decreases, so the weight decreases. At one place between the earth and the moon, the gravitational forces of the earth and the moon on the body are equal and opposite, so the weight becomes zero there.

Beyond this point, on moving further towards the moon, the gravitational force of the moon on the body increases, and hence the weight goes on increasing again.

An artificial satellite revolves in its orbit around the earth without using any fuel. But an aeroplane requires fuel to fly at a certain height. Why so?

Answer

An artificial satellite revolves at a great height where the air is very much rare, so the frictional force due to air on it is nearly negligible. The gravitational force of the earth on the satellite provides exactly the centripetal force required for its circular motion, and this force does no work on the satellite since it is always perpendicular to the satellite's velocity. Hence the satellite continues to revolve in its orbit without using any fuel.

An aeroplane, on the other hand, flies at a comparatively low height where the air is dense. It therefore requires fuel continuously to do work against the frictional force due to air, and also to obtain the lift needed to stay at that height.

Is it possible to place an artificial satellite in an orbit such that it is always visible over Lucknow (or New Delhi)? Write the reason.

Answer

No, it is not possible.

For a satellite to remain always visible over a particular place, it must be a geostationary satellite, that is, it must have a period of revolution of 24 h and it must revolve in a circular orbit lying in the equatorial plane of the earth.

Lucknow (or New Delhi) is not situated on the equator, so a satellite in the equatorial plane cannot remain fixed vertically above it. Hence it is not possible to place an artificial satellite in an orbit such that it is always visible over Lucknow or New Delhi.

A missile is fired radially from the surface of earth (radius 6.4 × 10⁶ m) at a satellite, orbiting the earth. The satellite appears stationary vertically upwards from the point where the missile is launched. Its distance from the centre of the earth is 25.4 × 10⁶ m. Will the missile actually hit the satellite?

Answer

No, the missile will not hit the satellite.

A satellite appears stationary over a point on the earth's equator only when it is at such a height that its period of revolution about the earth is 24 h, the same as the period of the spin of the earth about its own axis. The satellite is therefore not actually at rest; it is moving along its orbit at the same angular rate as the earth.

The missile, after being fired towards the satellite, takes some time in covering the distance between the earth and the satellite. In this time the satellite moves along its orbit and changes its position. Hence the missile will miss the satellite.

The gravitational potential energy of a body on the surface of the earth is − 6.4 × 10⁶ J. Clarify this statement.

Answer

The gravitational potential energy of a body on the earth's surface is

$\text U = -\dfrac{\text{GM}_e \text m}{\text R_e}$

which is negative because the potential energy at infinity is taken to be zero and work is obtained in bringing the body from infinity to the earth's surface.

The statement therefore means that 6.4 × 10⁶ J of energy must be supplied to the body on the surface of the earth in order to send it away, out of the gravitational field of the earth, that is, to take it to infinity where its potential energy becomes zero.

Choose the correct alternative :

(a) If the gravitational potential energy of two mass-points infinite distance apart is taken to be zero, the gravitational potential energy of a galaxy is positive/ negative / zero.

(b) The universe on the large scale is shaped by gravitational/electromagnetic forces; on the atomic scale by gravitational/electromagnetic forces; and on the nuclear scale by gravitational / electromagnetic/ strong nuclear forces.

Answer

(a) Negative.

The gravitational potential energy of a system of two mass-points is $\text U = -\dfrac{\text{Gm}_1 \text m_2}{\text r}$, taking the potential energy at infinite separation to be zero. Since gravitational force is always attractive, every such pair in a galaxy contributes a negative term. Hence the total gravitational potential energy of a galaxy is negative.

(b) Gravitational; electromagnetic; strong nuclear.

On the large scale the universe is shaped by gravitational forces, because these are long-range forces and act between all masses, and on this scale bodies are electrically neutral. On the atomic scale the electromagnetic forces between the charged nuclei and electrons dominate. On the nuclear scale the strong nuclear force, which is the strongest of all and acts over distances of the order of nuclear dimensions, is dominant.

Among the known types of forces in nature, the gravitational force is the weakest. Why then does it play a dominant role in motion of bodies on the terrestrial, astronomical and cosmological scale?

Answer

Nuclear forces are short-range forces which act only over distances of the order of nuclear dimensions (about 10^-15 m), so they cannot act between bodies separated by large distances.

Electrical forces are long-range forces, but they act only between charged bodies. Bodies on the terrestrial, astronomical and cosmological scale, such as planets and stars, are massive but electrically neutral, so these forces play no role in their motion.

Gravitational force, though the weakest of all, is a long-range force and acts between all bodies having mass, and it can never be cancelled since it is always attractive. Hence, even though it is the weakest, only the gravitational force plays a dominant role in the motion of bodies on the earth as well as in the motion of massive bodies such as planets and stars in outer space.

The mass and the diameter of a planet are twice those of the earth. What will be the time-period of that pendulum on this planet, which is a second's pendulum on the earth?

Answer

Given,

• Mass of the planet, M_p = 2M_e
• Diameter of the planet = twice that of the earth, so R_p = 2R_e
• The pendulum is a second's pendulum on the earth, that is, T_e = 2 s

The acceleration due to gravity at the surface of a planet is $\text g = \dfrac{\text{GM}}{\text R^2}$, so

$\dfrac{\text g_p}{\text g_e} = \dfrac{\text M_p}{\text M_e} \times \left(\dfrac{\text R_e}{\text R_p}\right)^2 = 2 \times \dfrac{1}{4} = \dfrac{1}{2}$

Thus g is reduced to half on the planet. The time-period of a simple pendulum is

$\text T = 2\pi\sqrt{\dfrac{\text l}{\text g}} \quad \Rightarrow \quad \text T \propto \dfrac{1}{\sqrt{\text g}}$

Therefore,

$\dfrac{\text T_p}{\text T_e} = \sqrt{\dfrac{\text g_e}{\text g_p}} = \sqrt2$

$\text T_p = \sqrt2 \times 2 = 2\sqrt2\ \text s$

Hence, the time-period of the pendulum on this planet will be $2\sqrt2$ second.

A particle is projected vertically in upward direction from the surface of earth (radius R_e). The kinetic energy of the particle is half of the minimum escape energy. How much height from the surface of earth will it gain ?

Answer

Given,

• Radius of the earth = R_e
• Kinetic energy of projection $= \dfrac{1}{2}$ of the minimum escape energy

The escape energy of a particle of mass m on the earth's surface is $\dfrac{\text{GM}_e \text m}{\text R_e}$. Hence the kinetic energy of projection is

$\text K = \dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

The gravitational potential energies of the particle at the surface and at a height h are

$\text U = -\dfrac{\text{GM}_e \text m}{\text R_e} \quad \text{and} \quad \text U_h = -\dfrac{\text{GM}_e \text m}{\text R_e + \text h}$

The particle rises until its whole kinetic energy is converted into the increase in potential energy,

$\text U_h - \text U = \text K \\[1em] \text{GM}_e \text m\left(\dfrac{1}{\text R_e} - \dfrac{1}{\text R_e + \text h}\right) = \dfrac{1}{2}\dfrac{\text{GM}_e \text m}{\text R_e}$

$\dfrac{\text h}{\text R_e(\text R_e + \text h)} = \dfrac{1}{2\text R_e} \quad \Rightarrow \quad \dfrac{\text h}{\text R_e + \text h} = \dfrac{1}{2}$

$2\text h = \text R_e + \text h \quad \Rightarrow \quad \text h = \text R_e$

Hence, the particle will gain a height equal to the radius of the earth, R_e.

Three equal masses, each m, are placed at the vertices of an equilateral triangle ABC. What is the force acting on a mass 2 m placed at the centroid G of the triangle?

Answer

Given,

• Three equal masses, each m, at the vertices A, B and C of an equilateral triangle
• Mass placed at the centroid G = 2m

Let the distance of each vertex from the centroid be GA = GB = GC = l.

The gravitational force on the mass 2m at G due to each of the masses m at A, B and C is

$\text F = \dfrac{\text G(2\text m)(\text m)}{\text l^2}$

The three forces are therefore equal in magnitude, and they are directed from G towards A, towards B and towards C respectively. Since ABC is an equilateral triangle and G is its centroid, these three directions are equally inclined to one another at 120°.

Three equal forces acting at a point at 120° to one another form a closed triangle, so their vector sum is zero.

Hence, by symmetry, the resultant gravitational force acting on the mass 2m placed at the centroid G is zero.

Spheres of the same material and same radius r are touching each other. Show that gravitational force between them is directly proportional to r⁴.

Answer

Let two spheres of the same material, each of radius r and density ρ, be touching each other. Their masses may be assumed to be concentrated at their centres.

The mass of each sphere is

$\text M = \text{volume} \times \text{density} = \dfrac{4}{3}\pi \text r^3 \rho$

Since the spheres are touching each other, the distance between their centres is

$\text d = \text r + \text r = 2\text r$

By Newton's law of gravitation, the force of attraction between them is

$\text F = \dfrac{\text{GMM}}{\text d^2} = \dfrac{\text G\left(\dfrac{4}{3}\pi \text r^3 \rho\right) \times \left(\dfrac{4}{3}\pi \text r^3 \rho\right)}{(2\text r)^2}$

$\text F = \dfrac{\text G \times \dfrac{16}{9}\pi^2 \text r^6 \rho^2}{4\text r^2} = \dfrac{4}{9}\text G\pi^2 \rho^2 \text r^4$

Since G, π and ρ are all constants,

$\text F \propto \text r^4$

Hence, the gravitational force between the two touching spheres of the same material is directly proportional to r⁴.

The gravitational force acting on a rocket at a height h from the earth surface is $\dfrac{1}{3}$rd of the force acting on a body at sea level. What is the relation between h and R_e (radius of the earth)?

Answer

Given,

• Gravitational force on the rocket at a height h $= \dfrac{1}{3}$ of the force at sea level

The gravitational force on a body of mass m at sea level is

$\text F = \dfrac{\text{GM}_e \text m}{\text R_e^2}$

and at a height h above the surface it is

$\text F' = \dfrac{\text{GM}_e \text m}{(\text R_e + \text h)^2}$

According to the question, $\text F' = \dfrac{1}{3}\text F$, so

$\dfrac{\text{GM}_e \text m}{(\text R_e + \text h)^2} = \dfrac{1}{3} \times \dfrac{\text{GM}_e \text m}{\text R_e^2}$