A uniform metal sphere of radius ‘a’ and mass ‘M’ is surrounded by a thin uniform spherical shell of equal mass and radius 4a. The centre of the shell falls on the surface of the inner sphere. Find the gravitational field at the points ${\text{p}}_{\text{1}}$  & ${\text{p}}_{\text{2}}$  shown in the figure

At ${\text{p}}_1=\text{a}$ 

The distance of ${\text{p}}_1$  from center is ${\text{r}}_{\text{1}}$

${\text{r}}_{\text{1}}=a+\text{4a}-a=\text{4a}$

Gravitational field  =

 Gravitational due to metal sphere at ${\text{p}}_1$ + Gravitational field due to hollow sphere at ${\text{p}}_2$

$=\frac{\text{GM}}{{\left(\text{4a}\right)}^{\text{2}}}+0=\frac{\text{GM}}{{\text{16a}}^{\text{2}}}$

$\because$ field due to solid sphere at a point outside it is $\frac{\text{GM}}{{\text{r}}^{\text{2}}}$

$\because$ field due to hollow sphere at a point inside it = 0

at${\text{p}}_2$ ,the distance of ${\text{P}}_2$ from centre is ${\text{r}}_{\text{2}}=\text{a}+4\text{a}+\text{a}=6\text{a}$

Gravitational field =

Gravitational due to metal sphere at ${\text{p}}_2$ + Gravitational field due to hollow sphere at ${\text{p}}_2$

We know field due to

(i) solid sphere at a point outside it is $=\frac{\text{GM}}{{\text{r}}^{\text{2}}}$

(ii) hollow sphere at a point outside it$=\frac{\text{GM}}{{\text{r}}^{\text{2}}}$

 $=\frac{\text{GM}}{{\left(\text{a}+\text{4a}+\text{a}\right)}^{\text{2}}}+\frac{\text{G}\text{.M}}{{\left(\text{4a}+\text{a}\right)}^{\text{2}}}$  

$=\frac{\text{GM}}{{\left(\text{6a}\right)}^{\text{2}}}+\frac{\text{GM}}{{\left(\text{5a}\right)}^{\text{2}}}=\frac{\text{GM}}{{\text{a}}^{\text{2}}}\left[\frac{1}{36}+\frac{1}{25}\right]$  

$=\frac{\text{GM}}{{\text{a}}^{\text{2}}}\left[\frac{25+36}{900}\right]=\frac{\text{61GM}}{{\text{900a}}^{\text{2}}}$