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 (see figure). 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 interact only
through their mutual gravitational interaction. When the point mass m 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

For point mass m at distance r = 31,

$\frac{GMm}{(3l{)}^2}-\frac{G{m}^2}{l}=F_1$ ...... (i)

For point mass m at distance r = 4l,

$\frac{GMm}{\left(4l^2\right)}+\frac{G{m}^2}{l^2}=F^2$ .... (ii)

When tension in the light rod is 0

So $F_1=F_2$

From (i) and (ii), we get

$\frac{GMm}{9l^2}-\frac{G{m}^2}{l^2}=\frac{GMm}{16l^2}+\frac{G{m}^2}{l^2}$

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

m= $\frac{7M}{288}=k\left(\frac{M}{288}\right)$

$\Rightarrow$ k = 7.