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JEE Main & Advanced AIEEE Solved Paper-2003

  • 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     AIEEE  Solved  Paper-2003

    A) 10 h       

    B)       80 h       

    C)       40 h       

    D)       20 h

    Correct Answer: C

    Solution :

    According to Kepler's  law,                     \[{{T}^{2}}\propto {{r}^{3}}\] where, T is time period and r is radius of the orbit follower by sattelites or                \[{{5}^{2}}\propto {{r}^{3}}\]                                      ... (i)                     \[{{(T')}^{2}}\propto {{(4r)}^{3}}\]                            ... (ii) From Eqs. (i) and (ii),     \[\frac{25}{{{(T')}^{2}}}=\frac{{{r}^{3}}}{64{{r}^{3}}},T'=\sqrt{1600}\]                     T? = 40 h


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