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:
($$G=gravitational $$ $$constant$$)

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

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: