A planet having mass `9 Me` and radius 4R_e, where Me and Re are mass and radius of earth respectively , has escape velocity in km/s given by:
(Given escape velocity on earth `V_e =11.2 times 10^3 m.s)

Options

A. `33.6`
B. `67.2`
C. `11.2`
D. `16.8`

Answer

To find the escape velocity of a planet with mass \(9 M_e\) and radius \(4 R_e\), we can use the formula for escape velocity:

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

Where:
- \(V_e\) is the escape velocity,
- \(G\) is the gravitational constant,
- \(M\) is the mass of the planet,
- \(R\) is the radius of the planet.

### Step 1: Substitute the mass and radius of the planet
Given:
- Mass of the planet \(M = 9 M_e\)
- Radius of the planet \(R = 4 R_e\)

Substituting these values into the escape velocity formula, we have:

\[ V'_{e} = \sqrt{\frac{2G(9 M_e)}{4 R_e}} \]

### Step 2: Simplify the expression
We can simplify the expression inside the square root:

\[ V'_{e} = \sqrt{\frac{18 G M_e}{4 R_e}} = \sqrt{\frac{9 G M_e}{2 R_e}} \]

### Step 3: Relate to Earth's escape velocity
The escape velocity on Earth is given by:

\[ V_e = \sqrt{\frac{2GM_e}{R_e}} = 11.2 \times 10^3 \, \text{m/s} \]

Now, we can express \(V'_{e}\) in terms of \(V_e\):

\[ V'_{e} = \sqrt{9} \cdot \sqrt{\frac{2GM_e}{R_e}} \cdot \frac{1}{\sqrt{2}} = 3 \cdot V_e \cdot \frac{1}{\sqrt{2}} \]

### Step 4: Calculate the escape velocity
Now substituting \(V_e\):

\[ V'_{e} = 3 \cdot 11.2 \times 10^3 \cdot \frac{1}{\sqrt{2}} \, \text{m/s} \]

Calculating \(\frac{1}{\sqrt{2}} \approx 0.707\):

\[ V'_{e} = 3 \cdot 11.2 \times 10^3 \cdot 0.707 \, \text{m/s} \]

Calculating \(3 \cdot 11.2 \cdot 0.707\):

\[ V'_{e} \approx 3 \cdot 7.9 \times 10^3 \approx 23.7 \times 10^3 \, \text{m/s} \]

### Step 5: Convert to km/s
To convert from m/s to km/s, we divide by \(1000\):

\[ V'_{e} \approx 23.7 \, \text{km/s} \]

### Final Answer
Thus, the escape velocity of the planet is approximately:

\[ \boxed{23.7 \, \text{km/s}} \]

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