Question

Three uniform spheres, each having mass $m$ and radius $r$, are kept in such a way that each touches the other two. The magnitude of the gravitational force on any sphere due to the other two is

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

Answer 1

Answer: (D)

The given system may be regarded as a system of three particles located at the three vertices of an equilateral triangle of side $2r$.

Now, $F_A=F_B$

$=\frac{G{m}^2}{(2r{)}^2}=\frac{G{m}^2}{4r^2}$

$F_A$ and $F_B$ are inclined to each other at an angle of $60^{\circ }$.

If $F$ is the resultant of $F_A$ and $F_B$, then

$F=\sqrt{3}\times \frac{G{m}^2}{4r^2}$