J & K CET Medical J & K - CET Medical Solved Paper-2014

  • question_answer
    The escape velocity from the surface of the earth is\[11.2\text{ }km{{s}^{-1}}\]. What is the escape velocity in a planet whose radius is three times that of the earth and on which the acceleration due to gravity is three times of that on the earth?

    A)  \[11.2\text{ }km{{s}^{-1}}\]       

    B)  \[22.4\text{ }km{{s}^{-1}}\]

    C)  \[33.6\text{ }km{{s}^{-1}}\]      

    D)  \[5.6\text{ }km{{s}^{-1}}\]

    Correct Answer: C

    Solution :

     Escape speed on the earth surface \[{{v}_{e}}=\sqrt{2{{g}_{e}}{{R}_{e}}}\] \[11.2=\sqrt{2{{g}_{e}}{{R}_{e}}}\]                ...(i) Escape speed on the planet surface \[{{v}_{p}}=\sqrt{2{{g}_{p}}{{R}_{p}}}\] \[=\sqrt{2(3{{g}_{e}})(3{{R}_{e}})}\] \[\left[ \begin{align}   & \because {{g}_{p}}=3{{g}_{e}} \\  & {{R}_{p}}=3{{R}_{e}} \\ \end{align} \right]\] \[{{V}_{p}}=3{{V}_{e}}\] \[=33.6m/s\]


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