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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2008

  • question_answer
    The time period of a satellite of earth is 5 h. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become

    A)  10 h                                      

    B)  80 h

    C)  40 h                      

    D)         20 h

    Correct Answer: C

    Solution :

    According to Keplers law                 \[{{T}^{2}}\propto {{r}^{3}}\] or            \[{{5}^{2}}\propto {{r}^{3}}\]                                                      ... (i) and        \[{{(T)}^{2}}\propto {{(4r)}^{3}}\]                                             ... (ii) From Eqs. (i) and (ii)                 \[\frac{25}{{{(T)}^{2}}}=\frac{{{r}^{3}}}{64{{r}^{3}}}\]                 \[T=\sqrt{1600}\]                 \[=40\,\,h\]


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