The mass of earth is 9 times of that of Mars. The radius of the earth is twice that of Mars. The escape velocity on the earth is `12 "kms"^(-1)` , then the escape velocity on the Mars is

Options

A. 5.65 `"kms"^(-1)`
B. 6 `"kms"^(-1)`
C. 7 `"kms"^(-1)`
D. 8 `"kms"^(-1)`

Answer

To find the escape velocity on Mars given the mass and radius relationships with Earth, we can use the formula for escape velocity:

\[ v = \sqrt{\frac{2GM}{R}} \]

where \( G \) is the universal gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the planet.

### Step 1: Define the masses and radii
Let:
- Mass of Earth \( M_E = 9M_M \) (where \( M_M \) is the mass of Mars)
- Radius of Earth \( R_E = 2R_M \) (where \( R_M \) is the radius of Mars)

### Step 2: Write the escape velocity for Earth
The escape velocity for Earth \( v_E \) is given as:
\[ v_E = \sqrt{\frac{2GM_E}{R_E}} \]
Substituting the known values:
\[ v_E = 12 \text{ km/s} \]

### Step 3: Write the escape velocity for Mars
The escape velocity for Mars \( v_M \) can be written as:
\[ v_M = \sqrt{\frac{2GM_M}{R_M}} \]

### Step 4: Relate \( v_M \) to \( v_E \)
Using the relationships for mass and radius:
\[ M_E = 9M_M \quad \text{and} \quad R_E = 2R_M \]
We can express \( v_M \) in terms of \( v_E \):
\[ v_M = \sqrt{\frac{2G( \frac{M_E}{9})}{\frac{R_E}{2}}} \]
Substituting \( M_E \) and \( R_E \):
\[ v_M = \sqrt{\frac{2G( \frac{9M_M}{9})}{\frac{2R_M}{2}}} \]
This simplifies to:
\[ v_M = \sqrt{\frac{2GM_M}{R_M}} \cdot \sqrt{\frac{2}{1}} = \sqrt{2} \cdot v_E \]

### Step 5: Substitute \( v_E \) into the equation
Now substituting \( v_E = 12 \text{ km/s} \):
\[ v_M = \sqrt{2} \cdot 12 \text{ km/s} \]
Calculating this gives:
\[ v_M = 12 \cdot 1.414 \approx 16.97 \text{ km/s} \]

### Step 6: Final Calculation
To find the escape velocity on Mars:
\[ v_M = \frac{12 \sqrt{2}}{3} \approx 5.65 \text{ km/s} \]

Thus, the escape velocity on Mars is approximately **5.65 km/s**.