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

Step 1: State the idea of energy conservation.
Gravity is a conservative force, so the total energy (kinetic plus potential) stays the same from start to the far end of the journey. We will equate energy at the start and at infinity.
Step 2: Note the launch distance and speed.
The body starts at height $0.5R$ above the surface, so its distance from Earth's centre is $r = R + 0.5R = \tfrac{3}{2}R$. It is thrown with the surface escape speed, where $v_i^{2} = \dfrac{2GM}{R}$.
Step 3: Write the starting energy.
\[ E_i = \frac{1}{2}mv_i^{2} - \frac{GMm}{r} = \frac{1}{2}m\left(\frac{2GM}{R}\right) - \frac{GMm}{\tfrac{3}{2}R} \]
Step 4: Simplify the starting energy.
\[ E_i = \frac{GMm}{R} - \frac{2GMm}{3R} = \frac{1}{3}\frac{GMm}{R} \]
Step 5: Write the energy at infinity.
Far away the potential energy is zero, so only kinetic energy remains: \[ E_f = \frac{1}{2}mv_{\infty}^{2} \] Setting $E_i = E_f$: \[ \frac{1}{2}mv_{\infty}^{2} = \frac{1}{3}\frac{GMm}{R} \]
Step 6: Solve for the final speed.
Using $\dfrac{GM}{R} = gR$: \[ v_{\infty}^{2} = \frac{2}{3}gR \;\Rightarrow\; \boxed{v_{\infty} = \sqrt{\frac{2gR}{3}}} \]