Question

Distance between the two masses is reduced to one-fourth of the initial value, force of gravitation between them increases to of the initial value.

• A. $4$ times
• B. $8$ times
• C. $16$ times

Answer 1

The correct option is C $16$ times

Let $M_1$ and $M_2$ be the masses of two objects and $R$ be the distance between them.
The gravitational force, F = $\frac{G{M}_1{M}_2}{R^2}$, where $G$ is the universal gravitational constant.
Since $M_1,M_2$ and $G$ are constants,
$F\propto \frac{1}{R^2}$
So, when distance between them is reduced to $\frac{R}{4}$, the force of attraction between them increases to $16\text{times}$ of the initial value.