Question

A rocket is fired with a speed of $5.6\, km \,s ^{-1}$ from the earth's surface. How far from the earth's surface does the rocket go before returning to the earth? (Escape speed of rocket $\left.=11.2\, km \,s ^{-1}\right)$

• A. R / 4
• B. R / 3
• C. R / 2
• D. 4 R / 3

Answer 1

Answer: (B)

$U_{i}+K_{i}=U_{f}+K_{f}$
$\frac{-G M m}{R}+\frac{1}{2} m v^{2}=-\frac{G M m}{(R+h)}+0$
$\frac{1}{2} v^{2}=G M\left[\frac{1}{R}-\frac{1}{R+h}\right]$
$v^{2}=2 g R \cdot\left[1-\frac{R}{R+h}\right]$
Now $\sqrt{2 g R}=11.2 kms ^{-1}$
As, $\left(\frac{5.6}{11.2}\right)^{2}=1-\left(\frac{R}{R+h}\right)$ or $\frac{R}{R+h}=1-\left(\frac{1}{2}\right)^{2}$
$\frac{R+h}{R}=\frac{4}{3} ; h=\frac{R}{3}$