The KE of a satellite in its orbit around the earth is E. The KE of the satellite so as to enable to escape from gravitational pull of the earth is

Options

A. 4E
B. 2E
C. `sqrt2E`
D. `E`

Answer

To solve the problem, we need to determine the kinetic energy (KE) of a satellite that is required to escape the gravitational pull of the Earth, given that the KE of the satellite in its orbit is E.

### Step-by-Step Solution:

1. **Understanding Orbital Kinetic Energy**:
- The kinetic energy (KE) of a satellite in orbit is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- For a satellite in a circular orbit, the orbital velocity \( v_0 \) can be expressed as:
\[ v_0 = \sqrt{\frac{GM}{R}} \]
- Here, \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the distance from the center of the Earth to the satellite.

2. **Relating Orbital KE to Given E**:
- The kinetic energy in orbit can also be expressed in terms of gravitational acceleration \( g \) as:
\[ KE = \frac{1}{2} m g R \]
- Given that the orbital KE is \( E \), we can write:
\[ E = \frac{1}{2} m g R \]

3. **Calculating Escape Velocity**:
- The escape velocity \( v_e \) from the gravitational field is given by:
\[ v_e = \sqrt{2gR} \]
- The kinetic energy required to escape is then:
\[ KE_{escape} = \frac{1}{2} mv_e^2 \]
- Substituting the escape velocity into the kinetic energy formula:
\[ KE_{escape} = \frac{1}{2} m (2gR) = m g R \]

4. **Relating Escape KE to E**:
- We already have \( E = \frac{1}{2} m g R \). Therefore, we can express \( KE_{escape} \) in terms of \( E \):
\[ KE_{escape} = 2E \]

5. **Final Answer**:
- Thus, the kinetic energy of the satellite required to escape from the gravitational pull of the Earth is:
\[ KE_{escape} = 2E \]

### Summary:
The kinetic energy of the satellite required to escape from the gravitational pull of the Earth is \( 2E \).