Question 57: Gravitation

Question

A large spherical mass M is fixed at one position and two identical 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 = 3l from M the tension in the rod is zero for m = $$k\left( {{M \over {288}}} \right)$$. The value of k is

Accepted Answer

The acceleration $\vec{a}$ of the point masses are equal because they are connected by a massless rigid rod.
Consider the situation when tension in the rod is zero. The gravitational forces on the two point masses are shown in the figure. The forces $f_1=\frac{G M m}{r^2}$ and $f_3=\frac{G M m}{(r+l)^2}$ are due to the attraction by the larger mass $M$. The force $f_2=\frac{G m m}{l^2}$ is due to mutual attraction between the two point masses. Apply Newton's second law on the two point masses to get
$$
\frac{G M m}{r^2}-\frac{G m m}{l^2}=m a
$$ ....... (1)
$$
\frac{G M m}{(r+l)^2}+\frac{G m m}{l^2}=m a
$$ ......... (2)
From eqn. (1) and (2), we get
$$
\begin{aligned}
& \frac{G M}{9 l^2}-\frac{G m}{l^2}=\frac{G M}{16 l^2}+\frac{G m}{l^2} \\
& \frac{M}{9}-\frac{M}{16}=m+m \Rightarrow \frac{7 M}{144}=2 m
\end{aligned}
$$
$$
m=\frac{7 M}{288}=k\left(\frac{M}{288}ight)
$$
$$ 	herefore $$ k = 7

Gravitation

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