Question

A body is projected from earth's surface with thrice the escape velocity from the surface of the earth. What will be its velocity when it will escape the gravitational pull?

• A. $2V_e$
• B. $4V_e$
• C. $2\sqrt{2}V_e$
• D. $\frac{V_e}{2}$

Hint:

There is a direct shortcut formula for interstellar velocity: $v_{final} = \sqrt{v_{initial}^2 - V_e^2}$. If you project a body with velocity $n V_e$, its final velocity at infinity will always be $\sqrt{n^2 - 1} \times V_e$.

Answer 1

The Correct Option is C

Step 1: Understand the question.
A body is thrown up from the Earth with three times the escape speed. We must find its leftover speed once it is far away, fully free of Earth's gravity.

Step 2: Use energy conservation.
No energy is lost, so the total energy at the surface equals the total energy far away:
\[ K_i + U_i = K_f + U_f \]
Far away, the gravity potential energy is zero, so $U_f = 0$.

Step 3: Write the energies.
The starting kinetic energy is $\tfrac{1}{2}m v_i^2$ and starting potential energy is $-\dfrac{GMm}{R}$. So:
\[ \frac{1}{2}m v_i^2 - \frac{GMm}{R} = \frac{1}{2}m v_f^2 \]

Step 4: Bring in escape speed.
Escape speed satisfies $V_e = \sqrt{\tfrac{2GM}{R}}$, which means $\dfrac{GMm}{R} = \tfrac{1}{2}m V_e^2$. Substitute this:
\[ \frac{1}{2}m v_i^2 - \frac{1}{2}m V_e^2 = \frac{1}{2}m v_f^2 \]

Step 5: Cancel and substitute.
Cancel $\tfrac{1}{2}m$ from each term:
\[ v_i^2 - V_e^2 = v_f^2 \]
The body is thrown with $v_i = 3V_e$:
\[ (3V_e)^2 - V_e^2 = v_f^2 \quad\Rightarrow\quad 9V_e^2 - V_e^2 = 8V_e^2 \]

Step 6: Take the square root.
\[ v_f = \sqrt{8}\,V_e = 2\sqrt{2}\,V_e \]
This matches option (3).
\[ \boxed{v_f = 2\sqrt{2}\,V_e} \]