Escape velocity at earth's surface is `11.2km//s`. Escape velocity at the surface of a planet having mass `100` times and radius `4` times that of earth will be

Options

A. `56km//s`
B. `112km//s`
C. `305km//s`
D. `11.2 km//s`

Answer

To find the escape velocity at the surface of a planet with a mass 100 times that of Earth and a radius 4 times that of Earth, we can use the formula for escape velocity:

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

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

### Step 1: Write the escape velocity formula for Earth
The escape velocity at the surface of Earth is given by:

\[ v_{e, \text{Earth}} = \sqrt{\frac{2GM_e}{R_e}} \]

Given that \( v_{e, \text{Earth}} = 11.2 \, \text{km/s} \).

### Step 2: Write the escape velocity formula for the new planet
For the new planet, the mass \( M \) is 100 times the mass of Earth (\( M = 100M_e \)) and the radius \( R \) is 4 times the radius of Earth (\( R = 4R_e \)). Thus, we can write the escape velocity for the new planet as:

\[ v_{e, \text{planet}} = \sqrt{\frac{2G(100M_e)}{4R_e}} \]

### Step 3: Simplify the expression
Now, we can simplify the expression:

\[ v_{e, \text{planet}} = \sqrt{\frac{200GM_e}{4R_e}} = \sqrt{\frac{50GM_e}{R_e}} \]

### Step 4: Relate it to Earth's escape velocity
We can relate this to the escape velocity of Earth:

\[ v_{e, \text{planet}} = \sqrt{50} \cdot \sqrt{\frac{2GM_e}{R_e}} = \sqrt{50} \cdot v_{e, \text{Earth}} \]

### Step 5: Calculate \( \sqrt{50} \)
Calculating \( \sqrt{50} \):

\[ \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2} \]

### Step 6: Substitute the value of \( v_{e, \text{Earth}} \)
Now substituting \( v_{e, \text{Earth}} = 11.2 \, \text{km/s} \):

\[ v_{e, \text{planet}} = 5\sqrt{2} \cdot 11.2 \, \text{km/s} \]

### Step 7: Calculate \( 5\sqrt{2} \)
Using \( \sqrt{2} \approx 1.414 \):

\[ 5\sqrt{2} \approx 5 \times 1.414 \approx 7.07 \]

### Step 8: Final calculation
Now, calculate the escape velocity:

\[ v_{e, \text{planet}} \approx 7.07 \cdot 11.2 \approx 79.424 \, \text{km/s} \]

### Conclusion
Thus, the escape velocity at the surface of the planet is approximately \( 79.4 \, \text{km/s} \).

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