Question

A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (11−E1). A particle of mass m' is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r < x < 2r, (b) 2r < x < 2R and (c) x > 2R.
Figure

Answer 1

Explanation:

It is given that a particle of mass m' is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell.

(a) $r<x<2 r$

For a thin portion of solid sphere of mass ‘m’, we can write, $m=\frac{m}{\left(\frac{4}{3}\right) \pi r^{2}} *\left(\frac{4}{3}\right) \pi x^{3}=\frac{m x^{3}}{r^{3}}$ .

Thus, on integration, $\int d m=\frac{m x^{3}}{r^{3}}$  .

Therefore, the magnitude of the resultant gravitational force on this particle due to the sphere and the shell, $F=\frac{G m x}{r^{2}}$ .  

(b) $2 r<x<2 R$

In this case, gravitational force is only due to the sphere and not due to the shell  

Thus, the resultant gravitational force is, $F=\frac{G m m^{\prime}}{(x-r)^{2}}$.

(c) $x>2 R$

Here, the gravitational force is due to both sphere and the shell.

We know that, gravitational force due to shell is, $F=\frac{C M m^{\prime}}{(x-R)^{2}}$.

The gravitational force due to sphere is, $F=\frac{G m m^{\prime}}{(x-r)^{2}}$.

Thus, the resultant gravitational force is =  $\frac{G m m^{\prime}}{(x-r)^{2}}+\frac{G M m^{\prime}}{(x-R)^{2}}$.