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Haryana PMT Haryana PMT Solved Paper-2008

  • question_answer
    The earth is assumed to be a sphere of radius    R.A platform is arranged at a height R from the surface of the earth. The escape velocity of a body from this platform is \[f{{v}_{e,}}\] where \[{{v}_{e}}\] is its escape velocity from the surface of the    earth. The value of\[f\] is

    A) \[\sqrt{2}\]                                        

    B) \[\frac{1}{\sqrt{2}}\]

    C) \[\frac{1}{3}\]                                   

    D) \[\frac{1}{2}\]

    Correct Answer: B

    Solution :

                    At a platform at a height h, escape energy = binding energy of sphere or            \[\frac{1}{2}mv_{e}^{2}=\frac{GMm}{R+h}\] or            \[v{{}_{e}}=\sqrt{\frac{2G\,\,M}{R+h}}=\sqrt{\frac{2G\,M}{2R}}\]  (\[\because \] \[h=R\]) But at surface of earth,                 \[{{v}_{e}}=\sqrt{\frac{2G\,\,M}{R}}\] As given, \[v{{}_{e}}=f\,\,{{v}_{e}}\] Hence,  \[\sqrt{\frac{2GM}{2R}}=f\sqrt{\frac{2G\,M}{R}}\] or            \[\frac{1}{2R}=\frac{{{f}^{2}}}{R}\] \[\therefore \]  \[f=\frac{1}{\sqrt{2}}\]


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