Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    The ratio of the energy required to raise a satellite upto a height h above the earth to the kinetic energy of the satellite into the orbit is

    A) \[x=\frac{\pi }{2},\]                                        

    B) \[-\frac{1}{6}\]

    C) \[-\frac{1}{24}\]                               

    D) \[-\frac{1}{48}\]

    Correct Answer: C

    Solution :

     Energy required to raise a satellite upto a height h,\[\mu ,\]                       ...(i) \[\frac{2}{a}\] energy required to put in orbit \[\frac{5}{12}\] \[\frac{1}{5}\] as \[\frac{2}{5}\]orbital speed \[\text{ }\!\!\theta\!\!\text{ }\] \[{{\mu }_{s}}.\]\[\tan \theta \ge 3{{\mu }_{s}}\]                                             ?(ii) From Eqs. (i) and (ii), we get\[\tan \theta >3{{\mu }_{s}}\]


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