Question

Four point masses each of mass $m$ are at four corner of a cube of side length $a$. Then

• A. the gravitational potential energy of system is $=-\frac{3\sqrt{2}G{m}^2}{a}$
• B. the gravitational potential energy of system is $=\frac{4\sqrt{2}G{m}^2}{a}$
• C. the magnitude of force on one of the particle due to the remaining masses is $=\frac{\sqrt{6}G{m}^2}{a^2}$
• D. the magnitude of force on one of the particle due to other is $\sqrt{\frac{3}{2}}\frac{G{m}^2}{a^2}$

Answer 1

Answer: (A), (D)

The correct options are
A the gravitational potential energy of system is $=-\frac{3\sqrt{2}G{m}^2}{a}$
D the magnitude of force on one of the particle due to other is $\sqrt{\frac{3}{2}}\frac{G{m}^2}{a^2}$

Gravitational potential energy of system is given by
$U=-\frac{GMm}{r}$
Here, $M=m$
and $r=a\sqrt{2}$
$U=-6\frac{G{m}^2}{a\sqrt{2}}=-\frac{3\sqrt{2}G{m}^2}{a}$
So, option $A$ matches correctly.
$\overrightarrow{F}\text{on}A=$\frac{-Gm^2}{(a\sqrt2)^3}\left[a\hat i+a\hat k-a\hat j+a \hat k+a\hat i-a \hat j\right]|F|=\frac{Gm^2}{a^3 2\sqrt 2}\left[\sqrt{4a^2+4a^2+4a^2}\right]=\sqrt{\frac{3}{2}}\frac{Gm^2}{a^2}$

So, option 4 matches.