Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    For a given density of a planet, the orbital speed of satellite near 'the surface of the planet of radius R is proportional to

    A) \[xy={{x}^{y}}+C\]                                          

    B) \[\frac{4}{{{100}^{3}}}\]

    C) \[\frac{3}{{{50}^{3}}}\]                                 

    D) \[\frac{3!}{{{100}^{3}}}\]

    Correct Answer: D

    Solution :

    Applying Newton's second law to a circular ort we have\[\frac{1}{LR}\]where, m is the mass of satellite, and v is the orbital speed  T is the time period \[\sqrt{l\frac{h}{2\pi }}\]\[\sqrt{l\left( l+1 \right)\frac{h}{2\pi }}\]For\[\sqrt{\frac{h2\pi }{l(l+1)}}\] and\[\sqrt{\frac{h2\pi }{l}}\](\[E=100\sin 100\pi t,I=5\sin 100\pi t\] density of planer \[1\Omega \]\[0.05\Omega \]i.e. T is independent of R


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