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

$$dF={\displaystyle \frac{GmdM}{s^2}}\cos \alpha$$
Mass =$$dM=\sigma 2\pi R\sin \theta Rd\theta$$
The force from the entire spherical shell is given by
$$F=2\pi G\sigma m{R}^2\underset{\theta =0}{\overset{\pi }{\int }}{\displaystyle \frac{\cos \alpha \sin \theta d\theta }{s^2}}$$ -----(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
$2sds=2Rr\sin \theta d\theta$
$\Rightarrow \sin \theta d\theta ={\displaystyle \frac{sds}{Rr}}$ ------------(2)
Now using the cosine formula for the external angle
$$R^2=r^2+s^2-2rs\cos \alpha$$
$$\cos \alpha ={\displaystyle \frac{r^2+s^2-R^2}{2rs}}$$ ------------(3)
$$\cos \alpha ={\displaystyle \frac{r^2+s^2-R^2}{2rs}}$$ ------------(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
$$F=-2\pi G\sigma m{R}^2\underset{s=R-r}{\overset{s=R+r}{\int }}{\displaystyle \frac{\left(r^2+s^2-R^2\right)sds}{s^2.2rs.Rr}}$$
$$\Rightarrow F={\displaystyle \frac{-\pi G\sigma m{R}^2}{r^2}}\underset{s=R-r}{\overset{s=R+r}{\int }}\left(1+{\displaystyle \frac{r^2-R^2}{s^2}}\right)ds$$
Using the area density expression $$\Rightarrow \sigma ={\displaystyle \frac{M}{4\pi R^2}}$$ the above equation reduces to
$$F={\displaystyle \frac{-GmM}{4R{r}^2}}\underset{s=R-r}{\overset{s=R+r}{\int }}\left(1+{\displaystyle \frac{r^2-R^2}{s^2}}\right)ds$$
$$\Rightarrow F={\displaystyle \frac{-GmM}{4R{r}^2}}{\left[s-\left(r^2-R^2\right){\displaystyle \frac{1}{s}}\right]}_{s=R-r}^{s=R+r}$$
$$\Rightarrow F={\displaystyle \frac{-GmM}{4R{r}^2}}\left[\left(R+r\right)-\left(R-r\right)-\left(r^2-R^2\right)\left({\displaystyle \frac{1}{R+r}}-{\displaystyle \frac{1}{R-r}}\right)\right]$$
$$\Rightarrow F={\displaystyle \frac{-GmM}{4R{r}^2}}\left[\left(2r\right)+\left(-2r\right)\right]\Rightarrow F=0$$

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.