question_answer

                For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:

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.