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 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 $m = k (\frac{M}{288})$ . The value of k is __

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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 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 $m = k (\frac{M}{288})$ . The value of k is __

Infinity Learn · 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 Newton's 2nd law of motion for point mass nearer to larger mass, $F_{1} - F = m a$
$\frac{G M m}{{(3 l)}^{2}} - \frac{G m^{2}}{l^{2}} = m a ........... (i)$
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Again using 2nd law of motion for another mass $F_{2} + F = m a$
$\frac{G M m}{{(4 l)}^{2}} + \frac{G m^{2}}{l^{2}} = m a ....... (i i)$
From eqn. (0 and (ii), we get
$\frac{G M}{9 l^{2}} - \frac{G m}{l^{2}} = \frac{G M}{16 l^{2}} + \frac{G m}{l^{2}}$
$\frac{M}{9} - \frac{M}{16} = m + m \Rightarrow \frac{7 m}{144} = 2 m$
$m = \frac{7 M}{288} = k (\frac{M}{288}) ∴ k = 7$

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