Question

Find the gravitational force acting on a particle A inside a uniform spherical layer of matter.

Answer 1

Hint: Divide the sphere into thin spherical shells and find the force acting on the particle lying inside the sphere by a thin spherical shell and integrate within the suitable limits.

Complete step by step solution:
Let M be the mass of the uniform spherical layer of matter and m be the mass of the particle lying inside it at a distance of r from the centre. Let the sphere be divided into thin spherical shells of density $\sigma $ per unit area. Force acting on the particle by a thin spherical shell of radius R is given by

\[dF = \dfrac{{GmdM}}{{{s^2}}}\cos \alpha \]
Mass =\[dM = \sigma 2\pi R\sin \theta Rd\theta \]
The force from the entire spherical shell is given by
\[F = 2\pi G\sigma m{R^2}\int\limits_{\theta = 0}^\pi {\dfrac{{\cos \alpha \sin \theta d\theta }}{{{s^2}}}} \] -----(1)
In order to find the value of the integral we need to express in terms of. To do so we use cosine formula. Using the cosine law, we have ${s^2} = {R^2} + {r^2}\theta - 2Rr\cos \theta $

Differentiating both sides, we get
$2sds = 2Rr\sin \theta d\theta $
$ \Rightarrow \sin \theta d\theta = \dfrac{{sds}}{{Rr}}$ ------------(2)
Now using the cosine formula for the external angle
\[{R^2} = {r^2} + {s^2} - 2rs\cos \alpha \]
\[\cos \alpha = \dfrac{{{r^2} + {s^2} - {R^2}}}{{2rs}}\] ------------(3)
\[\cos \alpha = \dfrac{{{r^2} + {s^2} - {R^2}}}{{2rs}}\] ------------(3)
Now, put the values of equations (2) and (3) in equation (1). Also, for\[\theta = 0,s = R - r\]and for \[\theta = \pi ,s = r + R\]
Now (1) reduces to
\[F = - 2\pi G\sigma m{R^2}\int\limits_{s = R - r}^{s = R + r} {\dfrac{{\left( {{r^2} + {s^2} - {R^2}} \right)sds}}{{{s^2}.2rs.Rr}}} \]
\[ \Rightarrow F = \dfrac{{ - \pi G\sigma m{R^2}}}{{{r^2}}}\int\limits_{s = R - r}^{s = R + r} {\left( {1 + \dfrac{{{r^2} - {R^2}}}{{{s^2}}}} \right)ds} \]
Using the area density expression \[ \Rightarrow \sigma = \dfrac{M}{{4\pi {R^2}}}\] the above equation reduces to
\[F = \dfrac{{ - GmM}}{{4R{r^2}}}\int\limits_{s = R - r}^{s = R + r} {\left( {1 + \dfrac{{{r^2} - {R^2}}}{{{s^2}}}} \right)ds} \]
\[ \Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {s - \left( {{r^2} - {R^2}} \right)\dfrac{1}{s}} \right]_{s = R - r}^{s = R + r}\]
\[ \Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {\left( {R + r} \right) - \left( {R - r} \right) - \left( {{r^2} - {R^2}} \right)\left( {\dfrac{1}{{R + r}} - \dfrac{1}{{R - r}}} \right)} \right]\]
\[ \Rightarrow F = \dfrac{{ - GmM}}{{4R{r^2}}}\left[ {\left( {2r} \right) + \left( { - 2r} \right)} \right] \Rightarrow F = 0\]

Thus the algebraic sum of the forces acting on the particle lying inside a sphere is zero.

Note: The gravitational force acting on a particle lying inside a spherical layer of matter is always zero. This is true for all particles lying inside a sphere.