The escape velocity corresponding to a planet of mass `M` and radius `R` is `50 km s^(-1)`. If the planet's mass and radius were `4M` and `R`, respectively, then the corresponding escape velocity would be

Options

A. `100 kms^(-1)`
B. `50 kms^(-1)`
C. `200 kms^(-1)`
D. `25kms^(-1)`

Explanation

To solve the problem, we need to find the escape velocity of a planet with mass \(4M\) and radius \(R\), given that the escape velocity for a planet with mass \(M\) and radius \(R\) is \(50 \, \text{km/s}\).

### Step-by-Step Solution:

1. **Recall the formula for escape velocity**:
The escape velocity \(v_e\) is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is the radius of the planet.

2. **Substitute the known values**:
For the original planet with mass \(M\) and radius \(R\), we know:
\[ v_e = 50 \, \text{km/s} \]
Therefore, we can write:
\[ 50 = \sqrt{\frac{2GM}{R}} \]

3. **Calculate the escape velocity for the new planet**:
Now, we consider the new planet with mass \(4M\) and radius \(R\). The escape velocity \(v_e'\) for this planet can be calculated as:
\[ v_e' = \sqrt{\frac{2G(4M)}{R}} = \sqrt{\frac{8GM}{R}} \]

4. **Factor out the escape velocity of the original planet**:
We can express \(v_e'\) in terms of the original escape velocity:
\[ v_e' = \sqrt{4} \cdot \sqrt{\frac{2GM}{R}} = 2 \cdot v_e \]

5. **Substitute the value of \(v_e\)**:
Since \(v_e = 50 \, \text{km/s}\), we substitute this value into the equation:
\[ v_e' = 2 \cdot 50 = 100 \, \text{km/s} \]

6. **Conclusion**:
Therefore, the escape velocity for the planet with mass \(4M\) and radius \(R\) is:
\[ \boxed{100 \, \text{km/s}} \]