A) 2
B) \[\frac{1}{2}\]
C) \[\frac{1}{\sqrt{2}}\]
D) \[\sqrt{2}\]
Correct Answer: B
Solution :
Key Idea: The potential energy of satellite is twice its kinetic energy but opposite in sign. Potential energy, \[U=-\frac{G{{M}_{e}}m}{{{R}_{e}}}\,or\,|U|\,\,=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Kinetic energy, \[K=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Thus, \[\frac{K}{|U|}=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\times \frac{{{R}_{e}}}{G{{M}_{e}}m}=\frac{1}{2}\] Note : The total energy \[E=K+U=-\frac{G{{M}_{e}}m}{2r}\] This energy is constant and negative, i.e., the system is closed. The farther the satellite from the earth the greater its total energy.You need to login to perform this action.
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