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

Answer 1

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

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,
$F_A=F_B$
$\Rightarrow \frac{G{m}^2}{(2R{)}^2}=\frac{G{m}^2}{4R^2}$

$F_A\text{and}{F}_B\text{are inclined to each other at an angle of}{60}^0$.
$\text{If F is the resultant of}{F}_A\text{and}{F}_{,}\text{then}$
$F=\sqrt{3}\times \frac{G{m}^2}{4R^2}$