Question

A rocket is fired vertically from the surface of the earth with a speed $v$. How far from the earth does the rocket go before returning to the earth?
(where $R_E$ is the radius of the earth and $g$ is acceleration due to gravity)

• A. $\frac{R_{E}v^{2}}{gR_{E}-v^{2}}$
• B. $\frac{R_{E}v^{2}}{gR_{E}+v^{2}}$
• C. $\frac{R_{E}v^{2}}{2gR_{E}-v^{2}}$
• D. $\frac{R_{E}v^{2}}{2gR_{E}+v^{2}}$

Answer 1

Answer: (C)

Let the rocket reaches a height $h$ from the surface of earth. Total energy at the surface of the earth is
$
E_{s}=\frac{1}{2} m v^{2}-\frac{G M_{E} m}{R_{E}}
$
where $m$ and $M_{E}$ are the masses of rocket and earth respectively. At highest point, the velocity of the rocket becomes zero.
$\therefore $ Total energy at the highest point is
$
E_{h}=-\frac{G M_{E} m}{\left(R_{E}+h\right)}
$
According to law of conservation of energy,
$
\begin{array}{l}
E_{s}=E_{h} \\
\therefore \frac{1}{2} m v^{2}-\frac{G M_{E} m}{R_{E}}=-\frac{G M_{E} m}{R_{E}+h} \\
\frac{1}{2} v^{2}-\frac{G M_{E}}{R_{E}}=-\frac{G M_{E}}{R_{E}+h} \\
=\frac{g R_{E}^{2}}{R_{E}}-\frac{g R_{E}^{2}}{R_{E}+h}\left(\therefore G=\frac{G M_{E}}{R_{E}^{2}}\right) \\
=g R_{E}\left(1-\frac{R_{E}}{R_{E}+h}\right)=g R_{E}\left(\frac{h}{R_{E}+h}\right) \\
v^{2}\left(R_{E}+h\right)=2 g R_{E} h \\
v^{2} R_{E}=2 g R_{E} h-v^{2} h \\
R_{E} v^{2}=h\left(2 g R_{E}-v^{2}\right) \\
h=\frac{R_{E} v^{2}}{2 g R_{E}-v^{2}}
\end{array}
$