A) \[\frac{1}{R}\]
B) \[\frac{1}{\sqrt{R}}\]
C) R
D) \[\frac{1}{{{R}^{3/2}}}\]
Correct Answer: A
Solution :
Kinetic energy of the satellite \[KE=\frac{1}{2}\,mv_{0}^{2}\] where, \[{{v}_{0}}=\sqrt{\left( \frac{GM}{R} \right)}\] Now, putting the value of \[{{v}_{0}}\] in Eq. (i), we get \[KE=\frac{1}{2}\,m\,{{\left( \sqrt{\left( \frac{GM}{R} \right)} \right)}^{2}}\] \[=\frac{1}{2}\,\frac{mGM}{R}\] Hence, \[KE\propto \,\frac{1}{R}\]
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