A large sphere of mass M is fixed at one position and two identical particles each of mass m are kept on a line passing through the centre of M (see figure). The point masses are connected by a rigid

Question

A large sphere of mass M is fixed at one position and two identical particles each of mass 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 masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3 l from M , the tension in the rod is zero for m = k M 288 . The value of k is

TestHub · Accepted Answer

Due to gravitational interaction connected masses have some acceleration.

Let both small masses are moving with acceleration 'a' towards larger mass M

Force equation for mass nearer to a larger mass

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

Force equation for mass away from a larger mass

$\frac{G M m}{{(4 l)}^{2}} + \frac{G m^{2}}{l^{2}} = m a$ ...(ii)

From equation (i) & (ii)

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

$\Rightarrow \frac{M}{9} - \frac{M}{16} = m + m$

$\frac{7 M}{144} = 2 m$

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

$\Rightarrow k = 7$

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