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. $\frac{R_E{v}^2}{g{R}_E-v^2}$
• B. $\frac{R_E{v}^2}{g{R}_E+v^2}$
• C. $\frac{R_E{v}^2}{2g{R}_E-v^2}$
• D. $\frac{R_E{v}^2}{2g{R}_E+v^2}$

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
$<br/>E_s=\frac{1}{2}m{v}^2-\frac{G{M}_{E}m}{R_E}<br/>$
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
$<br/>E_h=-\frac{G{M}_{E}m}{\left(R_E+h\right)}<br/>$
According to law of conservation of energy,
$<br/>\begin{array}{l}<br/>E_s=E_h\\ <br/>\therefore \frac{1}{2}m{v}^2-\frac{G{M}_{E}m}{R_E}=-\frac{G{M}_{E}m}{R_E+h}\\ <br/>\frac{1}{2}{v}^2-\frac{G{M}_E}{R_E}=-\frac{G{M}_E}{R_E+h}\\ <br/>=\frac{g{R}_E^2}{R_E}-\frac{g{R}_E^2}{R_E+h}\left(\therefore G=\frac{G{M}_E}{R_E^2}\right)\\ <br/>=g{R}_E\left(1-\frac{R_E}{R_E+h}\right)=g{R}_E\left(\frac{h}{R_E+h}\right)\\ <br/>v^2\left(R_E+h\right)=2g{R}_{E}h\\ <br/>v^2{R}_E=2g{R}_{E}h-v^{2}h\\ <br/>R_E{v}^2=h\left(2g{R}_E-v^2\right)\\ <br/>h=\frac{R_E{v}^2}{2g{R}_E-v^2}<br/>\end{array}<br/>$