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 :
\[\frac{1}{2}m{{u}^{2}}+\frac{-GMm}{R}=\frac{1}{2}m{{v}^{2}}+\frac{-GMm}{2R}\] \[\Rightarrow \frac{1}{2}m\left( {{v}^{2}}-{{u}^{2}} \right)=\frac{-GMm}{2R}\] \[\Rightarrow v=\sqrt{{{u}^{2}}-\frac{GM}{R}}...(i)\] \[{{v}_{0}}=\sqrt{\frac{GM}{2R}}\therefore {{v}_{red}}=\frac{m\times v}{\left( \frac{m}{10} \right)}=10\,v\] \[\therefore \frac{9m}{10}\times \sqrt{\frac{GM}{2R}}=\frac{m}{10}\times {{v}_{\tau }}\Rightarrow v_{\tau }^{2}=81\frac{GM}{2R}\] \[\therefore K{{E}_{rocket}}=\frac{1}{2}\times \frac{m}{10}\times \left( ({{u}^{2}}-\frac{GM}{R})100+81\frac{GM}{2R} \right)\] \[=\frac{m}{20}\times 100\left( {{u}^{2}}-\frac{GM}{R}+\frac{81}{200}\frac{GM}{R} \right)\] \[=5m\left( {{u}^{2}}-\frac{119}{200}\frac{GM}{R} \right)\]
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