Question 106: Gravitation

Question

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:-

Accepted Answer

The gravitational potential (V) due to a sphere of uniform density at a distance r from its center is given by:
$$
V(r) = \frac{GM}{2R^3} \left(3R^2 - r^2\right)
$$
At the surface of the sphere (r = R), the gravitational potential is:
$$
V = \frac{GM}{R}
$$
Now, let's find the gravitational potential at the center of the sphere (r = 0):
$$
V(0) = \frac{GM}{2R^3} \left(3R^2 - 0^2\right) = \frac{3GM}{2R}
$$
The gravitational potential at the center of the sphere is $$\frac{3}{2}$$ times the potential at the surface of the sphere. Therefore:
$$
V(0) = \frac{3V}{2}
$$

Answer

$$\frac{3 \mathrm{~V}}{2}$$

Answer

$$\frac{\mathrm{V}}{2}$$

Answer

$$\frac{4}{3} \mathrm{~V}$$

Answer

$$\mathrm{V}$$

Gravitation

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:-

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