3.2 Gravitational field due to a uniform solid sphere

  Gravitation

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

• At a point on the surface of sphere i.e. $r=R$,
  

$$E = \frac{{GM}}{{{R^2}}}$$ or $$\overrightarrow E = - \frac{{GM}}{{{R^2}}}\widehat R$$

• At a point inside the sphere i.e. $r<R$,

$$E = \frac{{GM}}{{{R^3}}}r$$ or $$\overrightarrow E = - \frac{{GM}}{{{R^3}}}\overrightarrow r $$

Proof:

At a point outside the sphere

Force on mass $m$ at point $P$ due to a body of mass $M$ is, $$F = \frac{{GMm}}{{{r^2}}}$$Gravitational field is, $$E = \frac{F}{m}$$ or $$E = \frac{{GM}}{{{r^2}}}$$
Note: Outside the sphere, we can assume it as point mass.

At a point on the surface of sphere

Force on mass $m$ due to a sphere of mass $M$ is, $$F = \frac{{GMm}}{{{R^2}}}$$
Gravitational field is, $$E = \frac{F}{m}$$$$E = \frac{{GM}}{{{R^2}}}$$

At a point inside the sphere

The density of the sphere is, $$\rho = \frac{M}{{\left( {\frac{4}{3}\pi {R^3}} \right)}} = \frac{{3M}}{{4\pi {R^3}}}$$
The mass of sphere of radius $r$ is, $$M' = \frac{4}{3}\pi {r^3}\rho $$$$M' = \frac{4}{3}\pi {r^3} \times \frac{{3M}}{{4\pi {R^3}}}$$$$M' = M{\left( {\frac{r}{R}} \right)^3}$$
Force on mass $m$ due to mass $M$ is, $$F = \frac{{GM'm}}{{{r^2}}}$$ or $$F = \frac{{Gm}}{{{r^2}}}\left( {\frac{{M{r^3}}}{{{R^3}}}} \right)$$ or $$F = \frac{{GMm}}{{{R^3}}}r$$
So, the gravitational field is,
$$E = \frac{{GM}}{{{R^3}}}r$$
Note:

The gravitational field at yellow area will be due to blue portion only. Red portion will not produce any gravitational field at yellow area.

The gravitational field due to a uniform sphere varies as,
$$E = \frac{{GM}}{{{r^2}}}\quad {\text{or}}\quad E \propto \frac{1}{{{r^2}}}\quad {\text{for}}\quad r \geqslant R$$ $$E = \frac{{GM}}{{{R^3}}}r\quad {\text{or}}\quad E \propto r\quad {\text{for}}\quad r \leqslant R$$
$E$ vs $r$ graph can be drawn using the above equation we get,