Gravitational Potential

Practice Questions:

Question 1: easy

A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. The gravitational potential at a point situated at a/2 distance from the centre, will be :

1. -4GM/a

2. -3GM/a

3. -2GM/a

4. -GM/a

View Answer

The gravitational potential inside a spherical shell (including at the center) due to the shell is constant. The potential at a distance \( a/2 \) from the center is the sum of the potentials due to the mass at the center and the shell.

\[
V = -\frac{GM}{a/2} - \frac{GM}{a}
\]

Simplifying:

\[
V = -\frac{2GM}{a} - \frac{GM}{a} = -\frac{3GM}{a}
\]

So, the potential at \( a/2 \) is:

\[
V = -\frac{3GM}{a}
\]

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Question 2: easy

Two bodies of masses \(m\) and \(M\) are placed at distance \(d\) apart. What is the gravitational potential \(V\) at the position where the gravitational field due to them is zero?

1. \(V = -\frac{G}{d}(m + M)\)

2. \(V = -\frac{G}{d} m\)

3. \(V = -\frac{GM}{d}\)

4. \(V = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\)

View Answer

At the point where the field is zero, \(\frac{Gm}{r_1^2} = \frac{GM}{r_2^2}\), which gives \(r_1 = \frac{\sqrt{m}}{\sqrt{m} + \sqrt{M}} d\) and \(r_2 = \frac{\sqrt{M}}{\sqrt{m} + \sqrt{M}} d\). The potential is \(V = -\frac{Gm}{r_1} - \frac{GM}{r_2} = -\frac{G}{d}(\sqrt{m} + \sqrt{M})^2\).

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