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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2010

  • question_answer
    Time period of revolution of a satellite around a planet of radius R is T. The period of revolution around another planet whose radius is 3R is

    A)  T                                            

    B)  3T

    C)  9T                                         

    D)  \[3\sqrt{3}T\]

    Correct Answer: D

    Solution :

    According to Kepler's law of planetary motion \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{R}_{1}}}{{{R}_{2}}} \right)}^{3/2}}={{\left( \frac{R}{3R} \right)}^{3/2}}\]                \[\Rightarrow \]               \[{{T}_{2}}=3\sqrt{3}{{T}_{1}}\]


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