A) \[\frac{1}{2}m{{v}^{2}}\]
B) \[-\frac{1}{2}m{{v}^{2}}\]
C) \[-m{{v}^{2}}\]
D) \[\frac{3}{2}m{{v}^{2}}\]
Correct Answer: B
Solution :
Kinetic energy of satellite, \[KE=\frac{1}{2}m{{v}^{2}}\] where \[v=\sqrt{\frac{GM}{r}}\] Potential energy of satellite, \[PE=\frac{-GMm}{r}=-m{{v}^{2}}\] \[\therefore \]Total energy\[=KE+PE\] \[=\frac{1}{2}m{{v}^{2}}-m{{v}^{2}}=-\frac{1}{2}m{{v}^{2}}\]
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