A) \[-2E\]
B) \[2E\]
C) \[\frac{2E}{3}\]
D) \[-\frac{2E}{3}\]
Correct Answer: A
Solution :
We know that The potential energy of the satellite \[U=-\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ?(i) The kinetic energy of the satellite \[K=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] ...(ii) The total energy \[E=U+K=-\frac{G{{M}_{e}}m}{{{R}_{e}}}+\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] \[E=-\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] \[2E=-\frac{G{{M}_{e}}m}{{{R}_{e}}}\Rightarrow -2E=U\] So, \[PE=-2(TE)\] \[PE=-2E\]
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