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$\Rightarrow\ \text{U}=\text{mgr}$
$\text{mg}=\frac{\text{U}}{\text{r}}$
(weight of the body).$\frac{\text{G}\frac{4}{3}\pi\text{R}^3\rho\text{m}}{\text{R}^2}=\text{m R}\frac{4\pi^2}{\text{T}^2}$
$\rho\text{T}^2=\frac{3\pi}{\text{G}}=\text{a constant}.$
$\text{M}=(\text{p})\times\frac{4}{3}\pi\text{R}^3$ Therefore, gravitational force between hollow shell and point mass is $\text{F}=\frac{\text{GMm}}{\text{r}^2}$ where M is the mass of the hollow spherical shell and m is the mass of point mass. Therefore, the mass is distributed on the surface of the sphere only, then F = 0 for $0<\text{r}<\text{R}$ (i.e., force inside the shell is zero)
And $\text{F}=\frac{\text{GM}}{\text{r}^2}$ for $\text{r}>\text{R}$
The variation of F versus r is shown in the diagram. Force is maximum at the surface of shell and it is zero if r tends to infinity.
