question_answer

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R(R = radius of the earth), it ejects a rocket of mass  so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth): [JEE MAIN Held on 07-01-2020 Morning]

A)  x\[5m\left( {{u}^{2}}-\frac{119}{200}\frac{GM}{R} \right)\]     

B)  \[\frac{3m}{8}{{\left( u+\sqrt{\frac{5GM}{6R}} \right)}^{2}}\] C)  \[\frac{m}{20}{{\left( u-\sqrt{\frac{2GM}{3R}} \right)}^{2}}\]    D) \[\frac{m}{20}\left( {{u}^{2}}+\frac{113}{200}\frac{GM}{R} \right)\] Correct Answer: A Solution : [a]

\[\frac{1}{2}m{{u}^{2}}+\frac{-GMm}{R}=\frac{1}{2}m{{v}^{2}}+\frac{-GMm}{2R}\]

\[\Rightarrow \frac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)=\frac{-GMm}{2R}\]

\[\Rightarrow v=\sqrt{{{u}^{2}}-\frac{GM}{R}}...(i)\]

\[{{v}_{0}}=\sqrt{\frac{GM}{2R}}\therefore {{v}_{red}}=\frac{m\times v}{\left( \frac{m}{10} \right)}=10\,v\]

\[\therefore \frac{9m}{10}\times \sqrt{\frac{GM}{2R}}=\frac{m}{10}\times {{v}_{\tau }}\Rightarrow v_{\tau }^{2}=81\frac{GM}{2R}\]

\[\therefore K{{E}_{rocket}}=\frac{1}{2}\times \frac{m}{10}\times \left( ({{u}^{2}}-\frac{GM}{R})100+81\frac{GM}{2R} \right)\]

\[=\frac{m}{20}\times 100\left( {{u}^{2}}-\frac{GM}{R}+\frac{81}{200}\frac{GM}{R} \right)\]

\[=5m\left( {{u}^{2}}-\frac{119}{200}\frac{GM}{R} \right)\]