Question

The escape velocity from earth is $v_{es}$. A body is projected with velocity $2v_{es}$, with what constant velocity will it move in the interplanetary space.

• A. $v_{es}$
• B. $3v_{es}$
• C. $\sqrt{3}{v}_{es}$
• D. $\sqrt{5}{v}_{es}$

Answer 1

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

Given, Escape velocity $=V_{e}s$

Radius of earth $=R$,

Mass of body $=m$,

Mass of earth $=M$,

Projection velocity of body $=2V_{es}$

Applying the law of conservation of energy

(K.E. of the body) + (Gravitational potential energy) = K.E of the body in interplanetary space

$\therefore [\frac{1}{2}.m.(2V_{es}{)}^2]+\left[\left(\frac{-GMm}{R}\right)\right]=\frac{1}{2}.m.V^2$

$\frac{4.(V_{es}{)}^2}{2}+\left(\frac{-GM}{R}\right)=\frac{1}{2}.V^2$

$2{\left(\sqrt{\frac{2GM}{R}}\right)}^2-\frac{GM}{R}=\frac{1}{2}.V^2$ $[\because V_{es}=\sqrt{\frac{2GM}{R}}]$

$2.\left(\frac{2GM}{R}\right)-\left(\frac{GM}{R}\right)=\frac{1}{2}.V^2$

$\frac{3GM}{R}=\frac{1}{2}.V^2$

$V^2=\frac{6GM}{R}$

$V=\sqrt{\frac{6GM}{R}}=\sqrt{3}.\sqrt{\frac{2GM}{R}}$ ....(i)

Now, $V_{es}=\sqrt{\frac{2GM}{R}}$.

Substitute in eqn (i),

$V=\sqrt{3}.V_{es}$,

Correct choice option C