Question

A particle is projected from the surface of the Earth with a velocity equal to twice the escape velocity. When the particle is far from the Earth, its speed will be:

• A. \( v_e \)
• B. \( 2v_e \)
• C. \( \sqrt{3} v_e \)
• D. \( \sqrt{2} v_e \)

Hint:

When a particle is projected with a velocity greater than escape velocity, use energy conservation to find the final velocity at infinity.

Answer 1

The Correct Option is C

Step 1: Using Energy Conservation The total mechanical energy of the particle is: \[ E = \frac{1}{2} m v^2 - \frac{GMm}{R} \] For escape velocity: \[ v_e = \sqrt{\frac{2GM}{R}} \] Step 2: Applying Given Condition The initial velocity given is \( v = 2v_e \): \[ E = \frac{1}{2} m (4 v_e^2) - \frac{GMm}{R} \] Substituting \( v_e^2 = \frac{2GM}{R} \): \[ E = 2GMm/R - GMm/R = GMm/R \] At infinity, kinetic energy remains: \[ \frac{1}{2} m v_{\infty}^2 = GMm/R \] Solving for \( v_{\infty} \): \[ v_{\infty} = \sqrt{3} v_e \] Thus, the correct answer is option (3).

Answer 2

Given:
Initial velocity of the particle, \( u = 2 v_e \), where \( v_e \) is the escape velocity.

Step 1: Recall the escape velocity formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

Step 2: Use conservation of energy. The total mechanical energy at the surface:
\[ E_i = \frac{1}{2} m u^2 - \frac{GMm}{R} \] At a very large distance, the potential energy is zero and the particle will have speed \( v \), so:
\[ E_f = \frac{1}{2} m v^2 \] Conservation of energy implies:
\[ E_i = E_f \] \[ \frac{1}{2} m u^2 - \frac{GMm}{R} = \frac{1}{2} m v^2 \]

Step 3: Substitute \( u = 2 v_e \) and \( v_e^2 = \frac{2GM}{R} \):
\[ \frac{1}{2} m (2 v_e)^2 - \frac{GM m}{R} = \frac{1}{2} m v^2 \] \[ \frac{1}{2} m 4 v_e^2 - \frac{1}{2} m v_e^2 = \frac{1}{2} m v^2 \] \[ 2 m v_e^2 - m v_e^2 = \frac{1}{2} m v^2 \] \[ m v_e^2 = \frac{1}{2} m v^2 \]

Step 4: Cancel \( m \) and multiply both sides by 2:
\[ 2 v_e^2 = v^2 \] \[ v = \sqrt{2} v_e \]

Step 5: Check carefully; the above simplification missed subtracting potential energy properly.
Recalculate Step 3 precisely:
\[ \frac{1}{2} m u^2 - \frac{GM m}{R} = \frac{1}{2} m v^2 \] \[ \frac{1}{2} m (2 v_e)^2 - \frac{GM m}{R} = \frac{1}{2} m v^2 \] \[ \frac{1}{2} m 4 v_e^2 - \frac{1}{2} m v_e^2 = \frac{1}{2} m v^2 \] Because \( v_e^2 = \frac{2GM}{R} \), so potential energy \( -\frac{GM m}{R} = -\frac{1}{2} m v_e^2 \) (correcting the previous step).

Step 6: Substitute:
\[ 2 m v_e^2 - \frac{1}{2} m v_e^2 = \frac{1}{2} m v^2 \] \[ \left(2 - \frac{1}{2}\right) m v_e^2 = \frac{1}{2} m v^2 \] \[ \frac{3}{2} m v_e^2 = \frac{1}{2} m v^2 \] Cancel \( \frac{1}{2} m \) from both sides:
\[ 3 v_e^2 = v^2 \] \[ v = \sqrt{3} v_e \]

Therefore, the speed of the particle far from the Earth is:
\[ \boxed{ \sqrt{3} v_e } \]