question_answer

Radius of orbit of satellite of earth is\[R\]. Its kinetic energy is proportional to

A) \[\frac{1}{R}\]                                  

B) \[\frac{1}{\sqrt{R}}\]

C) \[R\]                     

D)        \[\frac{1}{{{R}^{3/2}}}\]

Correct Answer: A

Solution :

Gravitational force provides the required centripetal force. The gravitational force provides the required centripetal force in orbit of earth. \[\therefore \]  \[\frac{G{{M}_{e}}M}{{{R}^{2}}}=\frac{mv_{0}^{2}}{R}\]                           \[{{v}_{0}}=\sqrt{\frac{G{{M}_{e}}}{R}}\] kinetic energy   \[=\frac{1}{2}mv_{0}^{2}\] \[\therefore \]        \[KE=\frac{1}{2}m{{\left( \frac{G{{M}_{e}}}{R} \right)}^{2/2}}\]                              \[=\frac{1}{2}\frac{mG{{M}_{e}}}{R}\]                       \[KE\propto \frac{1}{R}\]