Escape velocity of a body from the surface of earth is 11.2km/sec. from the earth surface. If the mass of earth becomes double of its present mass and radius becomes half of its present radius then escape velocity will become

Options

A. `44.8 km//s`
B. `22.4 km//s`
C. `11.2 km//s` (remain unchanged)
D. `5.6 km//s`

Answer

To solve the problem, we will use the formula for escape velocity, which is given by:

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

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

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

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

### Step 2: Identify the changes in mass and radius
According to the problem:
- The mass of the Earth becomes double its present mass, so \( M' = 2M \).
- The radius of the Earth becomes half its present radius, so \( R' = \frac{R}{2} \).

### Step 3: Substitute the new values into the escape velocity formula
Now, we will substitute the new mass and radius into the escape velocity formula:

\[ v_e' = \sqrt{\frac{2G(2M)}{\frac{R}{2}}} \]

### Step 4: Simplify the expression
We can simplify the expression:

\[ v_e' = \sqrt{\frac{4GM}{R}} = \sqrt{4} \cdot \sqrt{\frac{GM}{R}} = 2 \cdot \sqrt{\frac{2GM}{R}} \]

### Step 5: Relate the new escape velocity to the original escape velocity
Since \( \sqrt{\frac{2GM}{R}} \) is the original escape velocity \( v_e \):

\[ v_e' = 2 \cdot v_e \]

### Step 6: Substitute the known value of escape velocity
Given that the original escape velocity \( v_e = 11.2 \, \text{km/s} \):

\[ v_e' = 2 \cdot 11.2 \, \text{km/s} = 22.4 \, \text{km/s} \]

### Conclusion
Thus, the new escape velocity when the mass of the Earth is doubled and the radius is halved is:

\[ \boxed{22.4 \, \text{km/s}} \]