What will be the escape speed from earth if the mass of earth is doubled keeping the same size ?

Explanation

To find the escape speed from Earth when its mass is doubled while keeping the same size (radius), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Escape Velocity**:
The escape velocity \( v \) from a celestial body is given by the formula:
\[ v = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the body, and \( R \) is the radius of the body.

2. **Identify the Initial Conditions**:
Let the initial mass of the Earth be \( M \) and the radius be \( R \). The initial escape velocity \( v_0 \) can be expressed as:
\[ v_0 = \sqrt{\frac{2GM}{R}} \]
Given that the initial escape velocity from Earth is approximately \( 11.2 \, \text{km/s} \).

3. **Consider the New Conditions**:
If the mass of the Earth is doubled, the new mass \( M' \) will be:
\[ M' = 2M \]
The radius remains the same, so \( R' = R \).

4. **Calculate the New Escape Velocity**:
The new escape velocity \( v' \) can be calculated using the new mass:
\[ v' = \sqrt{\frac{2G(2M)}{R}} = \sqrt{\frac{4GM}{R}} = \sqrt{2} \cdot \sqrt{\frac{2GM}{R}} = \sqrt{2} \cdot v_0 \]

5. **Substitute the Initial Escape Velocity**:
Now substituting the value of \( v_0 \):
\[ v' = \sqrt{2} \cdot 11.2 \, \text{km/s} \]

6. **Calculate the Numerical Value**:
Using the approximate value of \( \sqrt{2} \approx 1.414 \):
\[ v' \approx 1.414 \cdot 11.2 \, \text{km/s} \approx 15.79 \, \text{km/s} \]

### Final Answer:
The escape speed from Earth, if its mass is doubled while keeping the same size, is approximately \( 15.79 \, \text{km/s} \).

---