Question

If three uniform spheres, each having mass $M$ and radius $R$, are kept in such a way that each touch the other two, the magnitude of the gravitational force on any sphere due to the other two is

• A. $\frac{2G{M}^2}{R^2}$
• B. $\frac{2G{M}^2}{4R^2}$
• C. $\frac{G{M}^2}{4R^2}$
• D. $\frac{\sqrt{3}G{M}^2}{4R^2}$

Answer 1

The correct option is D $\frac{\sqrt{3}G{M}^2}{4R^2}$

For calculation of force at a point outside, a uniform spherical mass can be considered as a point mass placed at its center.

Hence, we can consider the setup to be $3$ point masses placed at the vertices of an equilateral triangle of side length $2R$.


Force experienced by sphere $1$ due to sphere $2$ is given by

$F_1=\frac{G{M}^2}{(2R{)}^2}$

Similarly, $F_2=\frac{G{M}^2}{(2R{)}^2}$

So, net force $F_{net}$ experienced by sphere $1$ is given by

${\overrightarrow{F}}_{net}=\overrightarrow{F_1}+\overrightarrow{F_2}$

$\Rightarrow |{\overrightarrow{F}}_{net}|=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos {60}^{\circ }}$

$\Rightarrow |{\overrightarrow{F}}_{net}|=\sqrt{3}{F}_1(\because F_1=F_2)$

$\Rightarrow |{\overrightarrow{F}}_{net}|=\frac{\sqrt{3}G{M}^2}{4R^2}$

Therefore, option $\left(d\right)$ is correct.

Key Concept - For uniform spherical masses, gravitational force can be calculated assuming them to be point masses placed at their centers.