A rocket has to be launched from earh in such a way that it never returns. If E is the minimum energy delivered by the rocket launcher what should be the minimum energy that the launcher should have if the same rocket has launcher from the surface of the moon ? Assume that the density of the earth and the moon are equal and that the earth's volume is `64` times the volume of the moon

Options

A. `E/(64)`
B. `E/(32)`
C. `E/4`
D. `E/(16)`

Answer

To solve the problem, we need to determine the minimum energy required for a rocket to be launched from the surface of the Moon such that it never returns, given that the same rocket requires energy \( E \) when launched from the surface of the Earth.

### Step-by-Step Solution:

1. **Understanding Escape Energy from Earth**:
The minimum energy \( E \) required for a rocket to escape Earth's gravitational field can be expressed using the formula for gravitational potential energy and kinetic energy. The total mechanical energy at the surface of the Earth is given by:
\[ E = \frac{GM_{\text{Earth}} m}{R_{\text{Earth}}} \]
where \( G \) is the gravitational constant, \( M_{\text{Earth}} \) is the mass of the Earth, \( m \) is the mass of the rocket, and \( R_{\text{Earth}} \) is the radius of the Earth.

2. **Relating the Masses and Radii of Earth and Moon**:
We know that the volume of the Earth is 64 times that of the Moon, and both have the same density. Therefore, we can express the relationship between their masses and radii as follows:
\[ V_{\text{Earth}} = 64 V_{\text{Moon}} \implies \frac{M_{\text{Earth}}}{M_{\text{Moon}}} = 64 \]
Since density \( \rho \) is mass per unit volume and is the same for both:
\[ \frac{M_{\text{Moon}}}{M_{\text{Earth}}} = \frac{1}{64} \]

3. **Finding the Radius Relationship**:
The volume of a sphere is given by \( V = \frac{4}{3} \pi R^3 \). Thus, the ratio of the volumes gives us:
\[ \frac{R_{\text{Earth}}^3}{R_{\text{Moon}}^3} = 64 \implies \frac{R_{\text{Earth}}}{R_{\text{Moon}}} = 4 \]

4. **Escape Energy from Moon**:
The escape energy \( E' \) from the Moon can be expressed similarly:
\[ E' = \frac{GM_{\text{Moon}} m}{R_{\text{Moon}}} \]

5. **Finding the Ratio of Energies**:
To find the ratio of the escape energies \( \frac{E'}{E} \):
\[ \frac{E'}{E} = \frac{GM_{\text{Moon}}}{R_{\text{Moon}}} \cdot \frac{R_{\text{Earth}}}{GM_{\text{Earth}}} \]
Substituting the mass and radius relationships:
\[ \frac{E'}{E} = \frac{G \left(\frac{1}{64} M_{\text{Earth}}\right)}{R_{\text{Moon}}} \cdot \frac{R_{\text{Earth}}}{GM_{\text{Earth}}} \]
Simplifying this gives:
\[ \frac{E'}{E} = \frac{1}{64} \cdot \frac{R_{\text{Earth}}}{R_{\text{Moon}}} \]
Since \( \frac{R_{\text{Earth}}}{R_{\text{Moon}}} = 4 \):
\[ \frac{E'}{E} = \frac{1}{64} \cdot 4 = \frac{4}{64} = \frac{1}{16} \]

6. **Final Result**:
Therefore, the minimum energy \( E' \) required for the rocket to escape from the surface of the Moon is:
\[ E' = \frac{E}{16} \]

### Conclusion:
The minimum energy that the launcher should have if the same rocket is launched from the surface of the Moon is \( \frac{E}{16} \).