Question

ENERGY OF AN ORBITING SATELLITE
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

The correct option is C $\frac{R_E{v}^2}{2g{R}_E-v^2}$

Let the rocket reaches a height h from the surface of the 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 the highest point, the velocity of the rocket becomes zero.
$\therefore$ Total energy at the highest point is $E_h=-\frac{G{M}_{E}m}{(R_E+h)}$
According to law of conservation of energy, $E_s=E_h$
$\therefore \frac{1}{2}m{v}^2-\frac{G{M}_{E}m}{R_E}=-\frac{G{M}_{E}m}{R_E+h}$

$\frac{1}{2}{v}^2-\frac{G{M}_E}{R_E}=-\frac{G{M}_E}{R_E+h}$

$=\frac{g{R}_E^2}{R_E}-\frac{g{R}_E^2}{R_E+h}(\because g=\frac{G{M}_E}{R_E^2})$

$=g{R}_E(1-\frac{R_E}{R_E+h})=g{R}_E(1-\frac{h}{R_E+h})$

$v^2{R}_E+h=2g{R}_{E}h$

$v^2{R}_E=2g{R}_{E}h-v^{2}h$

$R_E{v}^2=h(2g{R}_E-v^2)$

$h=\frac{R_E{v}^2}{2g{R}_E-v^2}$