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: Understanding the Question:
An object is thrown upwards with a speed much greater than the escape velocity. We must find its residual velocity (interstellar velocity) once it completely escapes Earth's gravitational field (i.e., reaches infinity).

Step 2: Key Formula or Approach:
Use the principle of Conservation of Mechanical Energy:
Total Energy at Earth's surface = Total Energy at infinity
$$K_i + U_i = K_f + U_f$$
At infinity, the gravitational potential energy $U_f$ becomes exactly zero.

Step 3: Detailed Explanation:
Initial Kinetic Energy: $K_i = \frac{1}{2}m v_i^2$
Initial Potential Energy: $U_i = -\frac{GMm}{R}$
Final Kinetic Energy: $K_f = \frac{1}{2}m v_f^2$
Final Potential Energy: $U_f = 0$
Setting up the equation:
$$\frac{1}{2}m v_i^2 - \frac{GMm}{R} = \frac{1}{2}m v_f^2$$
We know that the formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$, which means that $\frac{GMm}{R} = \frac{1}{2}m V_e^2$.
Substitute this into our energy equation:
$$\frac{1}{2}m v_i^2 - \frac{1}{2}m V_e^2 = \frac{1}{2}m v_f^2$$
Cancel out the $\frac{1}{2}m$ from all terms:
$$v_i^2 - V_e^2 = v_f^2$$
The problem states the initial velocity is thrice the escape velocity ($v_i = 3V_e$):
$$(3V_e)^2 - V_e^2 = v_f^2$$ $$9V_e^2 - V_e^2 = v_f^2$$ $$8V_e^2 = v_f^2$$
Take the square root to find $v_f$:
$$v_f = \sqrt{8} V_e = 2\sqrt{2}V_e$$

Step 4: Final Answer:
The velocity upon escaping is $2\sqrt{2}V_e$, matching option (C).