The maximum vertical distance through which a full dressed astronaut can jump on the earth is 0.5m. Estimate the maximum vertical distance through which he can jump on the motion, which has a mean density 2/3 rd that of the earth and radius one-quarter that of the earth.

Options

A. 1.5 m
B. 3 m
C. 6 m
D. 7.5 m

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- The maximum vertical distance an astronaut can jump on Earth, \( h_E = 0.5 \, \text{m} \).
- The mean density of the Moon is \( \frac{2}{3} \) that of Earth, \( \rho_M = \frac{2}{3} \rho_E \).
- The radius of the Moon is \( \frac{1}{4} \) that of Earth, \( R_M = \frac{1}{4} R_E \).

### Step 2: Calculate the acceleration due to gravity on Earth and Moon
The formula for the acceleration due to gravity \( g \) at the surface of a celestial body is given by:
\[ g = \frac{4}{3} \pi G \rho R \]
Where:
- \( G \) is the universal gravitational constant,
- \( \rho \) is the density,
- \( R \) is the radius.

#### For Earth:
\[ g_E = \frac{4}{3} \pi G \rho_E R_E \]

#### For Moon:
Substituting the values for the Moon:
\[ g_M = \frac{4}{3} \pi G \left(\frac{2}{3} \rho_E\right) \left(\frac{1}{4} R_E\right) \]
Simplifying this:
\[ g_M = \frac{4}{3} \pi G \cdot \frac{2}{3} \rho_E \cdot \frac{1}{4} R_E \]
\[ g_M = \frac{2}{6} \cdot \frac{4}{3} \pi G \rho_E R_E = \frac{1}{3} g_E \]

### Step 3: Relate the jump heights on Earth and Moon
Using the kinematic equation for vertical motion, we know that:
\[ v^2 = u^2 + 2as \]
Where:
- \( v \) is the final velocity (0 at the maximum height),
- \( u \) is the initial velocity,
- \( a \) is the acceleration (negative for gravity),
- \( s \) is the displacement (height jumped).

For the astronaut jumping on Earth:
\[ 0 = u^2 - 2g_E h_E \]
Thus,
\[ u^2 = 2g_E h_E \]

For the Moon:
\[ 0 = u^2 - 2g_M h_M \]
Thus,
\[ u^2 = 2g_M h_M \]

### Step 4: Set the equations equal to each other
Since the initial velocity \( u \) is the same for both jumps:
\[ 2g_E h_E = 2g_M h_M \]
This simplifies to:
\[ g_E h_E = g_M h_M \]

### Step 5: Substitute \( g_M \) in terms of \( g_E \)
Using \( g_M = \frac{1}{3} g_E \):
\[ g_E h_E = \left(\frac{1}{3} g_E\right) h_M \]
Cancelling \( g_E \) from both sides (assuming \( g_E \neq 0 \)):
\[ h_E = \frac{1}{3} h_M \]

### Step 6: Solve for \( h_M \)
Substituting \( h_E = 0.5 \, \text{m} \):
\[ 0.5 = \frac{1}{3} h_M \]
Thus,
\[ h_M = 0.5 \times 3 = 1.5 \, \text{m} \]

### Step 7: Final result
The maximum vertical distance through which the astronaut can jump on the Moon is:
\[ \boxed{1.5 \, \text{m}} \]