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 \[K.E.=\frac{1}{2}m\upsilon _{0}^{2}\] ...(i) where \[{{\upsilon }_{0}}=\sqrt{\left( \frac{GM}{R} \right)}\] Now putting the value of\[{{\upsilon }_{0}}\]in Eq (i), we get \[K.E.=\frac{1}{2}m{{\left( \sqrt{\left( \frac{GM}{R} \right)} \right)}^{2}}\] \[=\frac{1}{2}\frac{mGM}{R}\] Hence, \[K.E.\propto \frac{1}{R}\]
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