Question

A rocket is fired vertically with a speed of 5 $km{s}^{-1}$ 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 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, $v=5km/s=5\times {10}^{3}m/s$
Mass of the Earth,
$M_e=6.0\times {10}^{24}kg$
Radius of the Earth,
$R_e=6.4\times {10}^{6}m$
Height reached by rocket mass, m = h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
$=\frac{1}{2}m{v}^2+\left(\frac{-G{M}_{e}m}{R_e}\right)$
At highest point h,
v = 0
And, Potential energy $=-\frac{G{M}_{e}m}{R_e+h}$
Total energy of the rocket
$=0+(-\frac{G{M}_{e}m}{R_e})=-\frac{G{M}_{e}m}{R_e+h}\frac{1}{2}{v}^2=G{M}_e(\frac{1}{R_e}-\frac{1}{R_e+h})$
$=G{M}_e\left(\frac{R_e+h-R_e}{R_e(R_e+h)}\right)$
$\frac{1}{2}{v}^2=\frac{G{M}_{e}h}{R_e(R_e+h)}\times \frac{R_e}{R_e}\frac{1}{2}\times v^2=\frac{g{R}_{e}h}{R_e+h}$
Where $g=\frac{GM}{R_e^2}=9.8m/s^2$
(Acceleration due to gravity on the Earth's surface)
$\therefore v^2(R_e+h)=2g{R}_{e}h$
$v^2{R}_e=h(2g{R}_e-v^2)$
$h=\frac{R_e{v}^2}{2g{R}_e-v^2}$
$=\frac{6.4\times {10}^6\times (5\times {10}^3{)}^2}{2\times 9.8\times 6.4\times {10}^6-(5\times {10}^3{)}^2}$
$h=\frac{6.4\times 25\times {10}^{12}}{100.44\times {10}^6}=1.6\times {10}^{6}m$
Height achieved by the rocket with respect to the centre of the Earth
$=R_e+h$
$=6.4\times {10}^6+1.6\times {10}^6$
$=8.0\times {10}^{6}m$