Gravitational potential at A, due to a uniform solid sphere of mass M and radius R is V1. If a spherical part of radius R2 is removed from the sphere as shown in figure, the gravitational potential at A due to remaining part becomes V2. Find the value of (V1V2−0.2)

V1=−GM2R V2=−GM2R−(−GM/83R/2)=GMR[112−12] GM2R−(1−612)=512GMR V1V2=12×125=65=1.2

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Gravitational potential at A, due to a uniform solid sphere of mass M and radius R is $V_{1}$ . If a spherical part of radius $\frac{R}{2}$ is removed from the sphere as shown in figure, the gravitational potential at A due to remaining part becomes $V_{2}$ . Find the value of $(\frac{V_{1}}{V_{2}} - 0.2)$

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Detailed Solution

$V_{1} = - \frac{G M}{2 R} V_{2} = - \frac{G M}{2 R} - (\frac{- G M / 8}{3 R / 2}) = \frac{G M}{R} [\frac{1}{12} - \frac{1}{2}] \frac{G M}{2 R} - (\frac{1 - 6}{12}) = \frac{5}{12} \frac{G M}{R} \frac{V_{1}}{V_{2}} = \frac{1}{2} \times \frac{12}{5} = \frac{6}{5} = 1.2$

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