A high jumper can jump 2.0 m on the earth With the same effort how high will be able to jump on a planet whose density is one-third and radius one-fourth those of the earth?

Options

A. 4m
B. 8 m
C. 12 m
D. 24 m

Explanation

To solve the problem, we need to find out how high a high jumper can jump on a planet with different density and radius compared to Earth.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The high jumper can jump 2.0 m on Earth.
- The density of the new planet is one-third that of Earth: \( \rho_p = \frac{1}{3} \rho_e \).
- The radius of the new planet is one-fourth that of Earth: \( R_p = \frac{1}{4} R_e \).

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

3. **Calculate the Mass of the New Planet**:
The mass \( M \) of a planet can be expressed in terms of its density \( \rho \) and volume \( V \):
\[ M = \rho \cdot V \]
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi R^3 \]
Therefore, the mass of the new planet is:
\[ M_p = \rho_p \cdot V_p = \frac{1}{3} \rho_e \cdot \frac{4}{3} \pi \left(\frac{1}{4} R_e\right)^3 \]

4. **Substituting the Values**:
\[ M_p = \frac{1}{3} \rho_e \cdot \frac{4}{3} \pi \cdot \frac{1}{64} R_e^3 = \frac{4}{192} \pi \rho_e R_e^3 = \frac{1}{48} \cdot \left(\rho_e \cdot \frac{4}{3} \pi R_e^3\right) = \frac{1}{48} M_e \]
where \( M_e \) is the mass of Earth.

5. **Calculate the Gravitational Acceleration on the New Planet**:
Using the mass calculated:
\[ g_p = \frac{G M_p}{R_p^2} = \frac{G \cdot \frac{1}{48} M_e}{\left(\frac{1}{4} R_e\right)^2} = \frac{G \cdot \frac{1}{48} M_e}{\frac{1}{16} R_e^2} = \frac{16 G M_e}{48 R_e^2} = \frac{1}{3} g_e \]
where \( g_e \) is the acceleration due to gravity on Earth.

6. **Relate Jump Heights**:
The jumper's potential energy when jumping to a height \( h \) is given by:
\[ PE = mgh \]
On Earth:
\[ PE_e = mg_e \cdot 2.0 \]
On the new planet:
\[ PE_p = mg_p \cdot h_p \]
Since the jumper exerts the same effort, we can equate the potential energies:
\[ mg_e \cdot 2.0 = mg_p \cdot h_p \]
Simplifying gives:
\[ 2.0 g_e = h_p g_p \]
Substituting \( g_p = \frac{1}{3} g_e \):
\[ 2.0 g_e = h_p \cdot \frac{1}{3} g_e \]
Dividing both sides by \( g_e \):
\[ 2.0 = \frac{1}{3} h_p \]
Solving for \( h_p \):
\[ h_p = 2.0 \cdot 3 = 6.0 \, \text{m} \]

### Final Answer:
The high jumper will be able to jump **6.0 meters** on the planet.