Question

30 A satellite is revolving around the earth in a circular orbit of radius equal to diameter of earth.The minimum % increase in the speed of that satellite so that it escapes from earths gravity is

Answer 1

$$\begin{gathered} r=2R \\ workdonetoescapethesatelliteisgivenby: \\ W={\int }_{2R}^{\infty }F.dr \\ F\hspace{0.17em}=\frac{G{M}_{e}m}{r^2} \\ W={\int }_{2R}^{\infty }\frac{G{M}_{e}m}{r^2}dr \\ W=\frac{Gm{M}_e}{r}=\frac{G{M}_{e}m}{2R}=kineticenergyofthestatellite \\ \frac{G{M}_{e}m}{2R}=\frac{1}{2}m{v_e}^2 \\ where{v}_e=escapevelocity \\ {v}_e=\sqrt{\frac{GM}{R}} \\ whensatellitewasrevolvinfitsvelocityisgivenby: \\ \frac{m{v}^2}{r}=\frac{G{M}_{e}m}{r^2} \\ v=\sqrt{\frac{GM}{2R}} \\ increaseinvelocityisgivenby: \\ \frac{v_e-v}{v}=\frac{\sqrt{\frac{GM}{R}}-\sqrt{\frac{GM}{2R}}}{\sqrt{\frac{GM}{2R}}}=\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}=\sqrt{2}-1=0.414 \\ \%increase=0.414\times 100=1=41.4\% \end{gathered}$$