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 earth $=6.4\times {10}^{6}m;G=6.67\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

Answer 1

$\text{Given}:$ $v=5km{s}^{-1},M_E=6.0\times {10}^{24}kg$
and $R_E=6.4\times {10}^{6}m$
Let's say teh rocket will go till a height $h$.

Total energy is always conserved.
$\therefore P{E}_R+K{E}_R=P{E}_h+K{E}_h$

$\Rightarrow \left(\frac{-Gm{M}_E}{R_E}\right)+\frac{1}{2}m{v}^2=\left(\frac{-Gm{M}_E}{R_E+h}\right)+\frac{1}{2}m(0{)}^2$

[$\because$ velocity will be zero at $h$]

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

$=\frac{6.4\times {10}^6\times (5\times {10}^3{)}^2}{(2\times 9.8\times 6.4\times {10}^6)-(5\times {10}^3{)}^2}=1.6\times {10}^{6}m$

So, from the earth's centre it will be
$R_E+h=6.4\times {10}^6+1.6\times {10}^6=8\times {10}^{6}m$