Question

If distance between the two masses is reduced to one-fourth of the initial value, then force of gravitation between them becomes ____ of the initial value.

• A. $4$ times
• B. $8$ times
• C. $16$ times
• D. $2$ 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,
$F\propto \frac{1}{R^2}$

Let the force when distance is reduced to one-fourth be $F^{\prime }$
$\therefore \frac{F^{\prime }}{F}=\frac{1/(\frac{R}{4}{)}^2}{1/R^2}$
$\frac{F^{\prime }}{F}=\frac{16/R^2}{1/R^2}$
$\frac{F^{\prime }}{F}=16⟹F^{\prime }=16F$
Hence, the gravitational force becomes $16$ times of the initial value.