Escape velocity of a body `1 kg` mass on a planet is `100 ms^(-1)`. Gravitational potential energy of the body at that planet is

Options

A. `-5000 J`
B. `-1000 J`
C. `-2400 J`
D. `5000 J`

Explanation

To find the gravitational potential energy of a body with a mass of 1 kg on a planet where the escape velocity is 100 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship between escape velocity and gravitational potential energy**:
The escape velocity \( v_e \) from the surface of a planet is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

2. **Square the escape velocity**:
Given that the escape velocity \( v_e = 100 \, \text{m/s} \), we can square this value:
\[ v_e^2 = (100 \, \text{m/s})^2 = 10000 \, \text{m}^2/\text{s}^2 \]

3. **Relate escape velocity to gravitational potential energy**:
From the escape velocity formula, we can express \( \frac{GM}{R} \):
\[ \frac{GM}{R} = \frac{v_e^2}{2} = \frac{10000}{2} = 5000 \, \text{m}^2/\text{s}^2 \]

4. **Calculate gravitational potential energy**:
The gravitational potential energy \( U \) of a mass \( m \) at a distance \( R \) from the center of a planet is given by:
\[ U = -\frac{GMm}{R} \]
Substituting \( \frac{GM}{R} = 5000 \, \text{m}^2/\text{s}^2 \) and \( m = 1 \, \text{kg} \):
\[ U = -\left(5000 \, \text{m}^2/\text{s}^2\right) \times (1 \, \text{kg}) = -5000 \, \text{J} \]

5. **Final result**:
The gravitational potential energy of the body at that planet is:
\[ U = -5000 \, \text{J} \]