Question

A body is projected up with a velocity $2$ times the escape velocity $v_e$ from the surface of the earth. The velocity of the body at a point far away the earth's attraction is

• A. $v_{es}$
• B. $\sqrt{7}{v}_{es}$
• C. $\sqrt{2}{v}_{es}$
• D. $\sqrt{3}{v}_{es}$

Answer 1

The correct option is D $\sqrt{3}{v}_{es}$

$v_e=(\frac{2GM}{R}{)}^{0.5}$
now, $V=2v_e=2(\frac{2GM}{R}{)}^{0.5}=(\frac{8GM}{R}{)}^{0.5}$

By conservation of energy, $\frac{-GMm}{R}+\frac{8GMm}{2R}=0+\frac{m{v}^2}{2}$

solving this, we get $v=(\frac{3\times 2GM}{R}{)}^{0.5}=3^{0.5}{v}_e$