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The gravitational force between a hollow spherical shell (of radius $R$ and uniform density) and a point mass is $F$. Show the nature of $F$ versus $r$ graph where $r$ is the distance of the point from the centre of the hollow spherical shell of uniform density.
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Verified Answer
The gravitation force inside and at the centre of hollow spherical shell is zero and on the surface is $F_0$ Consider the diagram, density of the shell is constant.
Let it is $\rho$
Mass of the shell $=($ density $) \times($ volume $)$
$$
M=(\rho) \times \frac{4}{3} \pi R^3
$$
As the density of the shell is uniform, it can be treated as a point mass placed at its centre.
So, (gravitational force between $M$ and $m$ ),
$$
F_0=\frac{G M m}{r^2}
$$
$F=0$ for $r < R$
(i.e., force inside the shell is zero)
$F_0=\frac{G M m}{r^2}$ for $r \geq R \quad(\because G M m$ is constant $)$
So, $F_0 \propto \frac{1}{r^2}$
The graph shows variation between $F$ vs $r$ :
Let it is $\rho$
Mass of the shell $=($ density $) \times($ volume $)$
$$
M=(\rho) \times \frac{4}{3} \pi R^3
$$
As the density of the shell is uniform, it can be treated as a point mass placed at its centre.
So, (gravitational force between $M$ and $m$ ),
$$
F_0=\frac{G M m}{r^2}
$$
$F=0$ for $r < R$
(i.e., force inside the shell is zero)
$F_0=\frac{G M m}{r^2}$ for $r \geq R \quad(\because G M m$ is constant $)$
So, $F_0 \propto \frac{1}{r^2}$
The graph shows variation between $F$ vs $r$ :
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