A) \[-1/2M_{v}^{2}\]
B) \[1/2M_{v}^{2}\]
C) \[3/2M_{v}^{2}\]
D) \[M_{v}^{2}\]
Correct Answer: A
Solution :
Kinetic energy of the satellite\[=\frac{1}{2}M{{v}^{2}}\] Potential energy of the satellite\[(U)=-\frac{G{{M}_{e}}M}{{{R}_{e}}}\] But, \[G{{M}_{e}}=gR_{e}^{2}\] \[\therefore \] \[U=-\frac{gR_{e}^{2}M}{{{R}_{e}}}=-g{{R}_{e}}M\] Orbital velocity of a satellite \[v=\sqrt{g{{R}_{e}}}\] \[\therefore \] \[U=-M{{v}^{2}}\] \[\therefore \]Total energy of the satellite\[=\frac{1}{2}M{{v}^{2}}-M{{v}^{2}}\] \[=-\frac{1}{2}M{{v}^{2}}\]
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