Question

A large spherical mass M is fixed at one position and two identical point masses m are kept on a line passing through the centre of M. The point masses are connected by a rigid massless rod of length $l$ and this assembly is free to move along the line connecting them. All three masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance $r=3l$ from M, the tension in the rod is zero for $m=k\left(\frac{M}{288}\right)$. The value of k is

Answer 1

Force on left mass m $F_1=\frac{GMm}{(3l{)}^2}-\frac{Gmm}{(l{)}^2}$

Force on right mass m $F_2=\frac{GMm}{(4l{)}^2}+\frac{Gmm}{(l{)}^2}$

When tension in the light rod is zero, $F_1=F_2$

$\therefore \frac{GMm}{(3l{)}^2}-\frac{Gmm}{(l{)}^2}=\frac{GMm}{(4l{)}^2}+\frac{Gmm}{(l{)}^2}$

$\Rightarrow \frac{7GMm}{144l^2}=\frac{2G{m}^2}{l^2}$

$\Rightarrow m=\frac{7M}{288}$

$\Rightarrow k=7$