question_answer

Radius of orbit of satellite of earth is R. Its Kinetic energy is proportional to:                      [BHU M-2003]

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

B)  \[\frac{1}{\sqrt{R}}\]

C)  \[R\]                                    

D)   \[\frac{1}{{{R}^{3/2}}}\]

Correct Answer: A

Solution :

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