The force of attraction between two bodies at a certain separation is 10 N. What will be the force of attraction between them if the separation is reduced to half ?

As we know,

$\text{F} = \text{G}\dfrac{\text{M}\text{m}}{\text{R}^2}$

Given,

F = 10 N, when separation = R,

Hence,

F₁ = $\dfrac{G M m}{R^2}$ = 10 N     [Equation 1]

If separation = $\dfrac{R}{2}$ then,

Substituting the values in the formula above, we get,

$F$2 = \dfrac{G M m}{\big(\dfrac{R}{2}\big)^2} \\[0.5em] F2 = \dfrac{4 G M m}{R^2} \\[0.5em]

$F_2 = 4 \times \big(\dfrac{G M m}{R^2}\big)$    [Equation 2]

Substituting the value of equation 1 in equation 2 we get,

F₂ = 4 x 10 N = 40 N