The mass of the earth is 81 times that of the moon and the radius of the earth is 3.5 times that of the moon. The ratio of the escape velocity on the surface of earth to that on the surface of moon will be

Options

A. 0.2
B. 2.57
C. 4.81
D. 0.39

Answer

To find the ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon, we can use the formula for escape velocity, which is given by:

\[ V = \sqrt{2gR} \]

where \( g \) is the acceleration due to gravity and \( R \) is the radius of the celestial body.

### Step 1: Define the variables
Let:
- \( M_m \) = mass of the Moon
- \( M_e \) = mass of the Earth = \( 81 M_m \)
- \( R_m \) = radius of the Moon
- \( R_e \) = radius of the Earth = \( 3.5 R_m \)

### Step 2: Calculate the acceleration due to gravity for Earth and Moon
The acceleration due to gravity \( g \) is given by:

\[ g = \frac{GM}{R^2} \]

For the Earth:
\[ g_e = \frac{G M_e}{R_e^2} = \frac{G (81 M_m)}{(3.5 R_m)^2} \]

For the Moon:
\[ g_m = \frac{G M_m}{R_m^2} \]

### Step 3: Substitute the values
Substituting \( M_e \) and \( R_e \) into the equation for \( g_e \):

\[ g_e = \frac{G (81 M_m)}{(3.5 R_m)^2} = \frac{81 G M_m}{12.25 R_m^2} \]

### Step 4: Find the ratio of \( g_e \) to \( g_m \)
Now, we can find the ratio \( \frac{g_e}{g_m} \):

\[ \frac{g_e}{g_m} = \frac{\frac{81 G M_m}{12.25 R_m^2}}{\frac{G M_m}{R_m^2}} = \frac{81}{12.25} \]

### Step 5: Simplify the ratio
Calculating \( \frac{81}{12.25} \):

\[ \frac{81}{12.25} = \frac{81}{\frac{49}{4}} = \frac{81 \times 4}{49} = \frac{324}{49} \approx 6.61 \]

### Step 6: Calculate the escape velocity ratio
Now, we can find the ratio of escape velocities \( \frac{V_e}{V_m} \):

\[ \frac{V_e}{V_m} = \sqrt{\frac{g_e}{g_m} \cdot \frac{R_e}{R_m}} \]

Substituting the values we have:

\[ \frac{V_e}{V_m} = \sqrt{\frac{81}{12.25} \cdot \frac{3.5 R_m}{R_m}} = \sqrt{\frac{81 \cdot 3.5}{12.25}} = \sqrt{\frac{283.5}{12.25}} \approx \sqrt{23.14} \approx 4.81 \]

### Final Answer
Thus, the ratio of the escape velocity on the surface of the Earth to that on the surface of the Moon is approximately:

\[ \frac{V_e}{V_m} \approx 4.81 \]