Question

A rocket is fired with a speed of $5.6km{s}^{-1}$ 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 $=11.2km{s}^{-1})$

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

Answer 1

Answer: (B)

$U_i+K_i=U_f+K_f$
$\frac{-GMm}{R}+\frac{1}{2}m{v}^2=-\frac{GMm}{(R+h)}+0$
$\frac{1}{2}{v}^2=GM\left[\frac{1}{R}-\frac{1}{R+h}\right]$
$v^2=2gR\cdot \left[1-\frac{R}{R+h}\right]$
Now $\sqrt{2gR}=11.2km{s}^{-1}$
As, ${\left(\frac{5.6}{11.2}\right)}^2=1-\left(\frac{R}{R+h}\right)$ or $\frac{R}{R+h}=1-{\left(\frac{1}{2}\right)}^2$
$\frac{R+h}{R}=\frac{4}{3};h=\frac{R}{3}$