Question

If a body is projected from the surface of the earth with a velocity of $\sqrt{5}V_e$, then the velocity of the body when it escapes from the gravitational influence of the earth is
(Escape velocity of a body from the surface of the earth, $V_e = 11.2$ km s$^{-1}$)

• A. 22.4 km s$^{-1}$
• B. 11.2 km s$^{-1}$
• C. 11.2/$\sqrt{5}$ km s$^{-1}$
• D. 5.6 km s$^{-1}$

Hint:

When a body is launched with a speed greater than the escape velocity, its velocity at infinity is found using conservation of energy.

Answer 1

The Correct Option is A

The escape velocity $V_e$ is the minimum speed needed to escape Earth's gravitational field, given as 11.2 km s$^{-1}$. 
The initial velocity of the body is $\sqrt{5}V_e = \sqrt{5} \times 11.2$. 
Using conservation of energy, the total mechanical energy at the surface is: \[ E = \frac{1}{2}m (\sqrt{5}V_e)^2 - \frac{GMm}{R} = \frac{1}{2}m (5V_e^2) - \frac{GMm}{R} \] Since $V_e = \sqrt{\frac{2GM}{R}}$, we have $\frac{GM}{R} = \frac{V_e^2}{2}$. Thus: \[ E = \frac{1}{2}m (5V_e^2) - m \frac{V_e^2}{2} = m \left( \frac{5V_e^2}{2} - \frac{V_e^2}{2} \right) = m \times 2V_e^2 \] At infinity, potential energy is zero, and the total energy is just kinetic: $E = \frac{1}{2}m v^2$. Equating: \[ \frac{1}{2}m v^2 = m \times 2V_e^2 \implies v^2 = 4V_e^2 \implies v = 2V_e \] \[ v = 2 \times 11.2 = 22.4 { km s}^{-1} \] Thus, the velocity when it escapes is 22.4 km s$^{-1}$.