Question 209: Gravitation

Question

From a solid sphere of mass $M$ and radius $R,$ a spherical portion of radius $R/2$ is removed, as shown in the figure. Taking gravitational potential $V=0$ at $r = \infty ,$ the potential at the center of the cavity thus formed is:
($G=gravitational $ $constant$)
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l91e5lo5/91b01283-f796-40fb-b71f-474ba5b2aeda/b49c8250-47d7-11ed-9a0e-8f7a98995d79/file-1l91e5lo6.png?format=png)

Accepted Answer

Before removing the spherical portion, potential at point $P$(Center of cavity)
${V_{sphere}}\,\, = {{ - GM} \over {2{R^3}}}\left[ {3{R^2} - {{\left( {{R \over 2}} ight)}^2}} ight]$
$ = {{ - GM} \over {2{R^3}}}\left( {{{11{R^2}} \over 4}} ight) =  - 11{{GM} \over {8R}}$
Mass of removed part = ${M \over {{4 \over 3}\pi {R^3}}} 	imes {4 \over 3}\pi {\left( {{R \over 2}} ight)^2}$ = ${M \over 8}$
Due to cavity part potential at point $P$
${V_{cavity}}\,\,$ =  $ - {{G{M \over 8}} \over {2{{\left( {{R \over 2}} ight)}^3}}}\left[ {3{{\left( {{R \over 2}} ight)}^2} - {0^2}} ight]$
= $ - {{3GM} \over {8R}}$
So potential at P due to remaining part,
$ = {V_{sphere}}\,\, - \,\,{V_{cavity}}\,\,$
$ = \,\, - {{11GM} \over {8R}} - \left( { - {3 \over 8}{{GM} \over R}} ight)$
$ = {{ - GM} \over R}$
Note :
Potential inside the sphere of radius R and at a distance r from the center,
V = $ - {{GM} \over {2{R^3}}}\left[ {3{R^2} - {r^2}} ight]$
Here M = mass of the sphere, r = distance from the center of the sphere
![](https://imagex.cdn.examgoal.net/a49Jv8rcFlHMGgTxP/PGzAOvXknGnmshjXoIKxO5OwI36wd/Ma8OvyTk33ymmGKc9umGOE/image.svg)

Answer

${{ - 2GM} \over {3R}}$

Answer

${{ - 2GM} \over R}$

Answer

${{ - GM} \over {2R}}$

Answer

${{ - GM} \over R}$

Gravitation

From a solid sphere of mass M and radius R, a spherical portion of radius R/2 is removed, as shown in the figure. Taking gravitational potential V=0 at r = \infty , the potential at the center of the cavity thus formed is: (G=gravitational constant)

Choose the correct answer: