The magnitude of gravitational potential energy of a body at a distance .r. from the center of the earth is V. Its weight at a distance .2r. from the center of the earth is

Options

A. `V/r`
B. `V/(4r)`
C. `V/(2r)`
D. `(4V)/r`

Answer

To solve the problem, we need to find the weight of a body at a distance of \(2r\) from the center of the Earth, given that its gravitational potential energy at a distance \(r\) from the center is \(V\).

### Step-by-Step Solution:

1. **Understand Gravitational Potential Energy (U)**:
The gravitational potential energy \(U\) of a body of mass \(m\) at a distance \(r\) from the center of the Earth (with mass \(M\)) is given by the formula:
\[ U = -\frac{GMm}{r} \]
where \(G\) is the gravitational constant.

2. **Relate Potential Energy at Distance \(r\) to Given Value**:
According to the problem, at distance \(r\), the potential energy is given as \(V\). Therefore, we can write:
\[ V = -\frac{GMm}{r} \]

3. **Calculate Gravitational Potential Energy at Distance \(2r\)**:
Now, we need to find the gravitational potential energy \(U'\) at a distance \(2r\):
\[ U' = -\frac{GMm}{2r} \]

4. **Express \(U'\) in Terms of \(V\)**:
We can express \(U'\) in terms of \(V\):
\[ U' = -\frac{GMm}{2r} = \frac{1}{2} \left(-\frac{GMm}{r}\right) = \frac{1}{2} V \]

5. **Determine Weight at Distance \(2r\)**:
The weight \(W\) of the body at a distance \(2r\) is given by the formula:
\[ W = \frac{GMm}{(2r)^2} = \frac{GMm}{4r^2} \]

6. **Relate Weight to Potential Energy**:
We know that:
\[ W = \frac{GMm}{4r^2} = \frac{1}{4} \left(\frac{GMm}{r^2}\right) = \frac{1}{4} \left(\frac{GMm}{r}\cdot \frac{1}{r}\right) = \frac{1}{4} \left(-\frac{GMm}{r}\cdot \frac{1}{r}\right) = \frac{1}{4} \left(-\frac{GMm}{r}\right) \cdot \frac{1}{r} \]
Since \(-\frac{GMm}{r} = V\), we can substitute:
\[ W = \frac{1}{4} \left(-\frac{GMm}{r}\right) = \frac{1}{4} V \]

### Final Answer:
Thus, the weight of the body at a distance \(2r\) from the center of the Earth is:
\[ W = \frac{1}{4} V \]