3.3 Gravitational field due to a uniform spherical shell

  Gravitation

• At a point outside the shell i.e. $r>R$
  $$E = \frac{{GM}}{{{r^2}}}\quad {\text{or}}\quad \overrightarrow E = - \frac{{GM}}{{{r^2}}}\widehat r$$

• At a point on the surface of the shell i.e. $r=R$
  $$E = \frac{{GM}}{{{R^2}}}\quad {\text{or}}\quad \overrightarrow E = - \frac{{GM}}{{{R^2}}}\widehat R$$

• At a point inside the shell i.e. $r<R$
  $$E=0$$

Proof:

At a point outside the shell

The force on mass $m$ due to a spherical shell of mass $M$ is given by, $$F = \frac{{GMm}}{{{r^2}}}$$

So, the gravitational field is given by, $$E = \frac{F}{m}$$ or $$E = \frac{{GM}}{{{r^2}}}$$

At a point on the shell

Force on mass $m$ due to spherical shell of mass $M$ at the surface is given by,

$$F = \frac{{GMm}}{{{R^2}}}$$
So, the gravitational field is given by, $$E = \frac{F}{m}$$ or $$E = \frac{{GM}}{{{R^2}}}$$

At a point inside the shell

The force on mass $m$ inside the spherical shell of mass $M$ is zero as no mass is present inside the shperical shell.

$$F=0$$
So, gravitational field is also zero.
$$E = \frac{F}{m} = 0$$

Note: The gravitational field due to a uniform spherical shell varies as,
$$E = \frac{{GM}}{{{r^2}}}\quad {\text{or}}\quad E \propto \frac{1}{{{r^2}}}\quad {\text{for}}\quad r \geqslant R$$
$$E = 0\quad {\text{for}}\quad r < R$$
The $E$ vs $r$ graph can be drawn using the above equation is,