JCECE Medical JCECE Medical Solved Paper-2007

  • 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: Kinetic energy of satellite is half of its potential energy, Potential energy of satellite \[U=-\frac{G{{M}_{e}}m}{{{R}_{e}}}\] where\[{{R}_{e}}\]is radius of earth, \[{{M}_{e}}\]the mass of earth,\[m\]the mass of satellite and G the gravitational constant. \[|U|\,=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Kinetic energy of satellite \[K=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\] Thus, \[\frac{K}{|U|}=\frac{1}{2}\frac{G{{M}_{e}}m}{{{R}_{e}}}\times \frac{{{R}_{e}}}{G{{M}_{e}}m}\] \[=\frac{1}{2}\]


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