Question

An object is projected vertically upwards from the surface of the earth with a velocity $3$ times the escape velocity $v_e$ from earth's surface. What will be its final velocity after escape from the earth's gravitational pull?

• A. $\sqrt{2}{v}_e$
• B. $2\sqrt{2}{v}_e$
• C. $2v_e$
• D. $\frac{v_e}{\sqrt{2}}$

Answer 1

Answer: (B)

$2\sqrt{2}{v}_e$

We know,

$V_e\text{(on surface of earth)}=\sqrt{\frac{2GM}{R}}$

Given,

$v_i=3v_e$

Let mass of object=$m$,

Apply energy conservation,

$\frac{1}{2}m{v}_i+\frac{-GMm}{R}=\frac{1}{2}m{v}_f^2+0$ (as $R_{final}=\infty )$

$\frac{1}{2}m\times 9\frac{2GM}{R}-\frac{GMm}{R}=\frac{1}{2}m{v}_f^2$

$\frac{8GMm}{R}=\frac{1}{2}m{v}_f^2$

$v_f=\sqrt{\frac{16GM}{R}}$

$v_f=\sqrt{8}\times \sqrt{\frac{2GM}{R}}=\sqrt{8}{v}_e$

Hence option $\text{B}$ is the correct answer.