RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer The potential energy of a satellite of mass and revolving at a height equal to radius of earth from earths surface, is

    A)  \[-\,mg{{R}_{e}}\]                         

    B)  \[-0.67\,mg{{R}_{e}}\]

    C)  \[-0.5\,mg{{R}_{e}}\]                   

    D)         \[-0.33\,mg{{R}_{e}}\]

    Correct Answer: C

    Solution :

    Gravitational potential energy of a satellite at height h \[U=\frac{G{{M}_{e}}m}{{{R}_{e}}+h}\]                                                ?(1) Here \[h=R{{ & }_{e}}=\]radius of earth and        \[g=\frac{G{{M}_{e}}}{R\,_{e}^{2}}\] \[\Rightarrow \]               \[G{{M}_{e}}=R_{e}^{2}g\]                                         ...(2) \[\therefore \]  \[U=\frac{(R_{e}^{2}g)m}{{{R}_{e}}+{{R}_{e}}}=-\frac{1}{2}mg{{R}_{e}}\]                                 \[=-5.0\,mg\,{{R}_{e}}\]

adversite


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