The maximum vertical distance through which a fully dressed astronaut can jump on the earth is 0.7 m. Estimate the maximum vertical distance through which she can jump on a planet, which has a mean density (5/9) times that of the earth and radius one quarter that of the earth.

Options

A. 2.35 m
B. 3.23 m
C. 4.10 m
D. 5.04 m

Answer

To solve the problem, we need to determine the maximum vertical distance (height) an astronaut can jump on a planet with a mean density of (5/9) times that of Earth and a radius that is one-quarter of Earth's radius. We know that the astronaut can jump 0.7 m on Earth.

### Step-by-Step Solution:

1. **Understanding the relationship between height and gravity**:
The height to which an astronaut can jump is directly related to the acceleration due to gravity (g). The formula for potential energy (PE) when jumping is:
\[ PE = mgh \]
where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is the height jumped.

2. **Acceleration due to gravity on Earth**:
On Earth, the astronaut can jump to a height of 0.7 m. Therefore, the potential energy at that height is:
\[ PE_{Earth} = mg \cdot 0.7 \]

3. **Finding the acceleration due to gravity on the new planet**:
The acceleration due to gravity \(g_p\) on the new planet can be expressed in terms of its radius and density. The formula for \(g\) is:
\[ g = \frac{GM}{R^2} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.

The mass \(M\) can be expressed in terms of density \(\rho\) and volume \(V\):
\[ M = \rho \cdot V = \rho \cdot \frac{4}{3} \pi R^3 \]
Thus,
\[ g = \frac{G \cdot \rho \cdot \frac{4}{3} \pi R^3}{R^2} = \frac{4}{3} \pi G \rho R \]

4. **Calculating the new planet's gravity**:
Given that the radius of the new planet \(R_p\) is \(\frac{1}{4} R\) (where \(R\) is Earth's radius) and the density \(\rho_p\) is \(\frac{5}{9} \rho\) (where \(\rho\) is Earth's density):
\[ g_p = \frac{4}{3} \pi G \cdot \left(\frac{5}{9} \rho\right) \cdot \left(\frac{1}{4} R\right) \]
Simplifying this gives:
\[ g_p = \frac{5}{36} g \]
where \(g\) is the acceleration due to gravity on Earth.

5. **Setting up the equation for height on the new planet**:
The potential energy on the new planet when jumping to height \(h_p\) is:
\[ PE_{Planet} = mg_p h_p \]
Setting the two potential energies equal gives:
\[ mg \cdot 0.7 = mg_p h_p \]
Canceling \(m\) from both sides:
\[ g \cdot 0.7 = g_p h_p \]

6. **Substituting for \(g_p\)**:
Substituting \(g_p = \frac{5}{36} g\) into the equation:
\[ g \cdot 0.7 = \left(\frac{5}{36} g\right) h_p \]
Canceling \(g\) from both sides:
\[ 0.7 = \frac{5}{36} h_p \]

7. **Solving for \(h_p\)**:
Rearranging gives:
\[ h_p = \frac{0.7 \cdot 36}{5} \]
Calculating this:
\[ h_p = \frac{25.2}{5} = 5.04 \text{ m} \]

### Final Answer:
The maximum vertical distance through which the astronaut can jump on the new planet is **5.04 m**.