Question

In the figure a spherical body of mass ${m_1}$ is at a distance a from one end of a uniform rod of length L and mass ${m_2}$. The gravitational force between them is

A. $\dfrac{{G{m_1}{m_2}}}{{{a^2}}}$
B. $\dfrac{{G{m_1}{m_2}}}{{a(a + L)}}$
C. $\dfrac{{G{m_1}{m_2}}}{{{{[a + (L/2)]}^2}}}$
D. $\dfrac{{G{m_1}{m_2}}}{{2{{(a/L)}^2}}}$

Answer 1

Sphere can be treated as a point on its centre because the rod is outside of the sphere, now it’s a question of gravitational force by a rod on a point and in order to calculate it.we will have to use integration.

Complete step by step answer:
Since rod is outside of the sphere we can treat sphere as point mass and it will look like

Now, lets take an elemental dx length of the rod at a distance x from the left end

Total length of the rod is L and mass is ${m_2}$
So mass of elemental length dx will be $\dfrac{{{m_2}dx}}{L}$.
Gravitational force due to this dx on ${m_1}$ will be $\dfrac{{G{m_1}{m_2}dx}}{{L{{(a + L - x)}^2}}}$.
Net force will be $\int\limits_0^L {\dfrac{{G{m_1}{m_2}dx}}{{L{{(a + L - x)}^2}}}} = - \left. {\dfrac{{G{m_1}{m_2}}}{{L(a + L - x)}}_0^L} \right] = \dfrac{{G{m_1}{m_2}}}{{L(a + L)}} - \dfrac{{G{m_1}{m_2}}}{{L(a + L - L)}} = - \dfrac{{G{m_1}{m_2}}}{{(a)(a + L)}}$.

So, the correct answer is “Option B”.

Note:
Here in this question rod is outside of the sphere so we can consider it as a point mass now since rod is not a symmetrical object we cannot consider it as a point mass nor can we take centre of mass of the rod as the point mass for the rod (some students do make this mistake by thinking that they can take the centre of mass of the rod as a single point and then solve it which is incorrect because all the elemental particle of the rod apply force of different magnitude on the rod and in an unsymmetrical way.