JEE Main & Advanced Physics Gravitation / गुरुत्वाकर्षण Question Bank Motion of Satellite

  • question_answer
    Periodic time of a satellite revolving above Earth?s surface at a height equal to R, radius of Earth, is [g is acceleration due to gravity at Earth?s surface]                                 [MP PMT 2002]

    A)             \[2\pi \sqrt{\frac{2R}{g}}\]

    B)               \[4\sqrt{2}\pi \sqrt{\frac{R}{g}}\]

    C)             \[2\pi \sqrt{\frac{R}{g}}\]

    D)               \[8\pi \sqrt{\frac{R}{g}}\]

    Correct Answer: B

    Solution :

                    \[T=2\pi \sqrt{\frac{{{(R+h)}^{3}}}{g{{R}^{2}}}}=2\pi \sqrt{\frac{{{(2R)}^{3}}}{g{{R}^{2}}}}=4\sqrt{2\pi }\sqrt{\frac{R}{g}}\]


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