Question

A rocket is fired vertically with a speed of $5km{s}^{-1}$ from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth ? Mass of the earth $=6.0\times {10}^{24}kg$; mean radius of the radius of the Earth $=6.4\times {10}^{6}m$; $G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

Velocity of the rocket, $v=5\times {10}^{3}m/s$
Mass of the Earth, $M_e=6\times {10}^{24}kg$
Radius of the Earth, $R_e=6.4\times {10}^6$
Height reached by rocket mass $m=h$

At the surface of the Earth,
$\text{Total energy of the rocket = Kinetic energy + Potential energy}$
$=\frac{1}{2}m{v}^2+(-\frac{G{M}_{e}m}{R_e})$

At the highest point $h$,
$v=0⟹\text{Kinetic energy}=0$

$\text{Potential energy}=-\frac{G{M}_{e}m}{R_e+h}$

Total energy of the rocket $=-\frac{G{M}_{e}m}{R_e+h}$

By law of conservation of energy, total energy at surface = total energy at height h

$\frac{1}{2}m{v}^2-\frac{G{M}_{e}m}{R_e}=-\frac{G{M}_{e}m}{R_e+h}$

$v^2/2=\frac{G{M}_{e}h}{R_e(R_e+h)}$

Substituting $g=G{M}_e/R_e^2$ and rearranging the terms,

$h=\frac{R_e{v}^2}{2g{R}_e-v^2}$

$=\frac{6.4\times {10}^6\times (5\times {10}^3{)}^2}{2\times 9.8\times 6.4\times {10}^6-(5\times {10}^3{)}^2}$

$=1.6\times {10}^{6}m$

Height achieved by the rocket with respect to the earth is

$R_e+h=6.4\times {10}^6+1.6\times {10}^6=8.0{10}^{6}m$