As shown in the figure, two spherical cavities are made in the uniform solid sphere of radius $R$. The boundaries of the cavities touch at the centre of the sphere. The centers of the cavities and the sphere lie on the $X$-axis. The mass of the solid sphere before the cavities were created was $M$. The gravitational force on a point mass $m$ at a distance $d$ away from the centre of the solid sphere is

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Aicharya · Accepted Answer

Given situation is as shown

Radius of sphere given $=R$ Mass of sphere $=M$ Density of sphere, $d=\frac{M}{\frac{4}{3} \pi R^3}$ Radius of each of cavity $=R / 2$ Mass of each of portion removed to create a cavity $=$ density $\times$ volume $ =\frac{M}{\frac{4}{3} \pi R^3} \times \frac{4}{3} \pi\left(\frac{R}{2}\right)^3=\frac{M}{8} $ Now, force of gravity on $m$, $F$ = force due to complete sphere of mass $M$ - force of mass of cavity of sphere centre at $A$ - force of mass of cavity of sphere centre at $B$. $\begin{aligned} & =\frac{G M m}{d^2}=\frac{G M^{\prime} m}{\left(d+\frac{R}{2}\right)^2}-\frac{G M^{\prime} m}{\left(d-\frac{R}{2}\right)^2} \\ & \Rightarrow F=\frac{G M m}{d^2}-\frac{G M m}{8\left(d+\frac{R}{2}\right)^2}-\frac{G M m}{8\left(d-\frac{R}{2}\right)^2} \\ & =G M m\left(\frac{1}{d^2}-\frac{1}{8\left(d+\frac{R}{2}\right)^2}-\frac{1}{8\left(d-\frac{R}{2}\right)^2}\right) \\ & =\frac{G M m}{d^2}\left(1-\frac{1}{8\left(1+\frac{R}{2 d}\right)^2}-\frac{1}{8\left(1-\frac{R}{2 d}\right)^2}\right)\end{aligned}$

As shown in the figure, two spherical cavities are made in the uniform solid sphere of radius $R$. The…

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