If $\mathrm{V}$ is the gravitational potential due to sphere of uniform density on it's surface, then it's value at the center of sphere will be:-
Solution
The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:
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$V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right)$
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At the surface of the sphere (r = R), the gravitational potential is:
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$V = \frac{GM}{R}$
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Now, let's find the gravitational potential at the center of the sphere (r = 0):
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$V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R}$
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The gravitational potential at the center of the sphere is $\frac{3}{2}$ times the potential at the surface of the sphere. Therefore:
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$V(0) = \frac{3V}{2}$
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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