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 I 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((M/288)). The value of k is

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 I 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((M/288)). The value of k is (A) 5 (B) 6 (C) 7 (D) 8

Tardigrade Admin · Accepted Answer

Both the point masses are connected by a light rod so they have same acceleration.
Suppose each point mass is moving with acceleration a towards larger mass

M

.
Using Newtons

2^{n d}

law of motion for point mass nearer to larger mass,

F_{1} - F = ma

\frac{GM m}{( 3 l ) ^{2}} - \frac{G m ^{2}}{l ^{2}} = ma \dots (i)

Again using

2^{n d}

law of motion for the other mass

F_{2} + F = ma

\frac{GM m}{( 4 l ) ^{2}} - \frac{G m ^{2}}{l ^{2}} = ma \dots (ii)

From eqn.

(i)

and

(ii)

, we get

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

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

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

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

∴ k = 7

$5$

$6$

$7$

$8$