MGIMS WARDHA MGIMS WARDHA Solved Paper-2013

  • 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)  \[\frac{1}{{{R}^{3}}}\]

    Correct Answer: A

    Solution :

                    \[\frac{G{{M}_{e}}m}{{{R}^{2}}}=\frac{mv_{0}^{2}}{R}\] \[v_{0}^{2}=\sqrt{\frac{G{{M}_{e}}}{R}},KE=\frac{1}{2}mv_{0}^{2}\] \[KE=\frac{1}{2}m{{\left( \frac{G{{M}_{e}}}{R} \right)}^{2/2}}=\frac{1}{2}\frac{mG{{M}_{e}}}{R}\]                 \[v\propto \frac{1}{R}\]


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