The kinetic energy of a satelliete in its orbit around the earht is E. What shoud be the kinetic energy of the satellite so as to enable it to escape crom the gravitational pull of the Earth?

Options

A. 4E
B. 2E
C. sqrt(2)E
D. E

Answer

To find the kinetic energy of a satellite required to escape the gravitational pull of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- The kinetic energy (KE) of the satellite in its orbit around the Earth is given as \( E \).
- We need to determine the kinetic energy required for the satellite to escape Earth's gravitational pull.

2. **Recall the Relationship Between Escape Velocity and Orbital Velocity:**
- The escape velocity (\( v_e \)) is related to the orbital velocity (\( v_0 \)) by the formula:
\[ v_e = \sqrt{2} \cdot v_0 \]

3. **Express Kinetic Energy in Terms of Velocity:**
- The kinetic energy of an object is given by the formula:
\[ KE = \frac{1}{2} m v^2 \]
- For the satellite in orbit, its kinetic energy is:
\[ KE_{\text{orbit}} = \frac{1}{2} m v_0^2 \]
- Since we know that \( KE_{\text{orbit}} = E \), we can write:
\[ E = \frac{1}{2} m v_0^2 \]

4. **Calculate the Kinetic Energy Required for Escape:**
- Now, substituting the escape velocity into the kinetic energy formula:
\[ KE_{\text{escape}} = \frac{1}{2} m v_e^2 \]
- Substituting \( v_e = \sqrt{2} \cdot v_0 \):
\[ KE_{\text{escape}} = \frac{1}{2} m (\sqrt{2} \cdot v_0)^2 \]
- Simplifying this gives:
\[ KE_{\text{escape}} = \frac{1}{2} m (2 v_0^2) = m v_0^2 \]

5. **Relate the Escape Kinetic Energy to the Given Energy \( E \):**
- Since we know \( E = \frac{1}{2} m v_0^2 \), we can express \( m v_0^2 \) in terms of \( E \):
\[ KE_{\text{escape}} = 2E \]

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

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