A person jumps 4 m on the surface of the earth. How high will he be able to jump on the surface of a planet of radius 3200 km, having mass same as that of earth? (Radius of earth = 6400 km)

Options

A. 6 m
B. 5 m
C. 1 m
D. 4 m

Answer

To solve the problem, we need to determine how high a person can jump on a planet with a radius of 3200 km, given that they can jump 4 m on Earth. The mass of the planet is the same as that of Earth.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- A person jumps 4 m on Earth.
- We need to find out how high they can jump on a different planet with a radius of 3200 km, while the mass of the planet is the same as Earth's.

2. **Acceleration Due to Gravity**:
- The formula for acceleration due to gravity (g) is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

3. **Acceleration on Earth**:
- For Earth, we can denote the acceleration due to gravity as \( g_E \):
\[ g_E = \frac{GM_E}{R_E^2} \]
- Here, \( R_E = 6400 \) km is the radius of Earth.

4. **Acceleration on the New Planet**:
- For the new planet, with radius \( R_P = 3200 \) km:
\[ g_P = \frac{GM_P}{R_P^2} \]
- Since the mass of the planet \( M_P \) is the same as that of Earth \( M_E \), we can write:
\[ g_P = \frac{GM_E}{(3200 \text{ km})^2} \]

5. **Relating the Two Accelerations**:
- Since \( R_P = \frac{1}{2} R_E \), we can relate the two accelerations:
\[ g_P = \frac{GM_E}{(3200 \text{ km})^2} = \frac{GM_E}{(6400 \text{ km}/2)^2} = \frac{GM_E \cdot 4}{(6400 \text{ km})^2} = 4g_E \]
- Thus, the acceleration due to gravity on the new planet is four times that on Earth:
\[ g_P = 4g_E \]

6. **Energy Consideration**:
- The potential energy when jumping to a height \( H \) is given by:
\[ PE = mgh \]
- For Earth:
\[ PE_E = mg_E \cdot 4 \text{ m} \]
- For the new planet:
\[ PE_P = mg_P \cdot H_P \]

7. **Setting Energies Equal**:
- Since the person exerts the same energy while jumping, we can set the two potential energies equal:
\[ mg_E \cdot 4 = mg_P \cdot H_P \]
- Substituting \( g_P = 4g_E \):
\[ mg_E \cdot 4 = m(4g_E) \cdot H_P \]
- Canceling \( m \) and \( g_E \) from both sides:
\[ 4 = 4H_P \]
- Solving for \( H_P \):
\[ H_P = 1 \text{ m} \]

### Final Answer:
The person will be able to jump **1 meter** on the surface of the planet with a radius of 3200 km.