Question

The kinetic energy of a satellite in its orbit around Earth is \( E \). What should be the kinetic energy of the satellite to escape Earth's gravity?

• A. \( 4E \)
• B. \( 2E \)
• C. \( \sqrt{2}E \)
• D. \( E \)

Hint:

Escape velocity is given by \( v_e = \sqrt{2} v_0 \), where \( v_0 \) is the orbital velocity.

Answer 1

The Correct Option is B

Step 1: {Escape velocity formula}
\[ v_e = \sqrt{2} v_0 \] Step 2: {Find kinetic energy for escape}
\[ KE_{{escape}} = \frac{1}{2} M v_e^2 = \frac{1}{2} M (2 v_0^2) \] \[ = 2E \] Thus, the correct answer is \( 2E \).

Answer 2

Step 1: For a satellite in a stable circular orbit, the gravitational force provides the centripetal force.
The total mechanical energy \( E_{\text{total}} \) of the satellite is:
\( E_{\text{total}} = -\frac{G M m}{2r} \), where:
- \( G \) is the gravitational constant
- \( M \) is the mass of Earth
- \( m \) is the mass of the satellite
- \( r \) is the orbital radius

Step 2: The kinetic energy of the satellite in orbit is:
\( E = \frac{G M m}{2r} \)

Step 3: To escape Earth, the satellite must have zero total mechanical energy:
That means it must gain enough kinetic energy to cancel out the negative potential energy:
So, required kinetic energy for escape = \( \frac{G M m}{r} \)

Step 4: Compare this with orbital kinetic energy:
Orbital kinetic energy = \( \frac{G M m}{2r} = E \)
Escape kinetic energy = \( \frac{G M m}{r} = 2E \)

Final Answer: \( 2E \)