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BHU PMT BHU PMT (Screening) Solved Paper-2011

  • 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 Kepler's law \[{{T}^{2}}\propto {{r}^{3}}\] \[{{5}^{2}}\propto {{r}^{3}}\]                                       ...(i) \[{{(T')}^{2}}\propto {{(4)}^{3}}\]                                            ...(ii) From Eqs. (i) and (ii), we have \[\frac{25}{{{(T')}^{2}}}=\frac{{{r}^{3}}}{64{{r}^{3}}}\]                 \[T'=\sqrt{1600}\]                 \[T'=40\,h\]


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