Question

If a body is thrown vertically upwards from a height of 0.5 R (R is the radius of the earth) with a velocity equal to the escape velocity of a body from the surface of the earth, then the velocity of the body when it escapes from the gravitational influence of the earth is: (g is the acceleration due to gravity on the surface of the earth)

• A. \( \sqrt{2gR} \)
• B. \( \sqrt{gR} \)
• C. \( \sqrt{\frac{2gR}{3}} \)
• D. \( \sqrt{\frac{2gR}{5}} \)

Hint:

Because the launch velocity is equal to the surface escape velocity, the body has more than enough energy to escape when starting from an elevated position, meaning it will retain a non-zero residual velocity at infinity.

Answer 1

The Correct Option is C

Concept: Total mechanical energy is conserved throughout the escape trajectory. The escape velocity from the Earth's surface is given by \( v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \).

Step 1: Setting up the total initial energy equation.
The body is launched from a distance \( r = R + 0.5R = 1.5R = \frac{3}{2}R \) from the center of the Earth. The initial velocity is \( v_i = v_e = \sqrt{\frac{2GM}{R}} \). \[ E_i = \text{K.E.}_i + \text{P.E.}_i = \frac{1}{2}m v_i^2 - \frac{GMm}{r} \] Substitute the values for \( v_i^2 \) and \( r \): \[ E_i = \frac{1}{2}m \left(\frac{2GM}{R}\right) - \frac{GMm}{\frac{3}{2}R} = \frac{GMm}{R} - \frac{2GMm}{3R} = \frac{1}{3}\frac{GMm}{R} \]

Step 2: Setting up the total final energy equation at infinity.
When the body escapes completely to infinity, its potential energy drops to zero, leaving only its residual kinetic energy: \[ E_f = \frac{1}{2}m v_{\infty}^2 + 0 = \frac{1}{2}m v_{\infty}^2 \]

Step 3: Applying energy conservation to solve for \( v_{\infty} \).
\[ \frac{1}{2}m v_{\infty}^2 = \frac{1}{3}\frac{GMm}{R} \implies v_{\infty}^2 = \frac{2GM}{3R} \] Since \( \frac{GM}{R^2} = g \implies \frac{GM}{R} = gR \), we substitute this into our equation: \[ v_{\infty}^2 = \frac{2}{3}gR \implies v_{\infty} = \sqrt{\frac{2gR}{3}} \]