A cavity of radius $R / 2$ is made inside a solid sphere of radius $R$ . The centre of the cavity is located at a distance $R / 2$ from the centre of the sphere. The gravitational force on a particle of mass ' $m$ 'at a distance $R / 2$ from the centre of the sphere on the line joining both the centres of a sphere and cavity is - (opposite to the centre of gravity)
[Here $g = G M / R^{2}$ , where $M$ is the mass of the sphere]

Question

Verified by Toppr

Answer 1

Gravitational field at mass $m$ due to full solid sphere

$\overset{ˉ}{E}_{1} = \frac{p r ˉ}{3 ε _{0}} = \frac{p R}{6 ε _{0}} . . . . . . . [ε_{0} = \frac{1}{4 π G}]$

Gravitational field at mass $m$ due to cavity $(- p)$

$\overset{ˉ}{E}_{2} = \frac{( - p ) ( R / 2 ) ^{3} r ˉ}{3 ε _{0} R ^{2}} . . . . . . . [u s i n g E = \frac{p a ^{3}}{3 ε _{0} r ^{2}}]$

$- \frac{( - p ) R ^{3}}{2 4 ε _{0} R ^{2}} = - \frac{- p R}{2 4 ε _{0}}$
Net gravitational field

$\overset{ˉ}{E} = \overset{ˉ}{E}_{1} + \overset{ˉ}{E}_{2} = \frac{p R}{6 ε _{0}} - \frac{p R}{2 4 ε _{0}} = \frac{p R}{8 ε _{0}}$

Net force on $m \to F = m \overset{ˉ}{E} = \frac{m p R}{8 ε _{0}}$

Here, $p = \frac{M}{( 4 / 3 ) π R ^{3}}$ & $ε_{0} = \frac{1}{4 π G}$

Then, $F = \frac{3 m g}{8}$

Answer 2

Gravitational field at mass

m

due to full solid sphere

\overset{ˉ}{E}_{1} = \frac{p r ˉ}{3 ε _{0}} = \frac{p R}{6 ε _{0}} . . . . . . . [ε_{0} = \frac{1}{4 π G}]

Gravitational field at mass

m

due to cavity

(- p)

\overset{ˉ}{E}_{2} = \frac{( - p ) ( R / 2 ) ^{3} r ˉ}{3 ε _{0} R ^{2}} . . . . . . . [u s i n g E = \frac{p a ^{3}}{3 ε _{0} r ^{2}}]

- \frac{( - p ) R ^{3}}{2 4 ε _{0} R ^{2}} = - \frac{- p R}{2 4 ε _{0}}

Net gravitational field

\overset{ˉ}{E} = \overset{ˉ}{E}_{1} + \overset{ˉ}{E}_{2} = \frac{p R}{6 ε _{0}} - \frac{p R}{2 4 ε _{0}} = \frac{p R}{8 ε _{0}}

Net force on

m \to F = m \overset{ˉ}{E} = \frac{m p R}{8 ε _{0}}

Here,

p = \frac{M}{( 4 / 3 ) π R ^{3}}

&

ε_{0} = \frac{1}{4 π G}

Then,

F = \frac{3 m g}{8}