Manipal Engineering Manipal Engineering Solved Paper-2010

  • question_answer
    At what height h above earth, the value of g becomes g/27 (.R = Radius of earth)

    A)  3R                                         

    B)  \[\sqrt{2}R\]

    C)  \[(\sqrt{2}-1)R\]             

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

    Correct Answer: C

    Solution :

    The value of acceleration due to gravity at height\[h\](when\[h\]is not negligible as compared to\[R\])                 \[g=g\frac{{{R}^{2}}}{{{(R+h)}^{2}}}\] Here,     \[g=\frac{g}{2}\] \[\therefore \]  \[\frac{g}{2}=g\frac{{{R}^{2}}}{{{(R+h)}^{2}}}\] or            \[\frac{1}{2}=\frac{{{R}^{2}}}{{{(R+h)}^{2}}}\] or            \[\sqrt{\frac{1}{2}}=\frac{R}{R+h}\] or            \[R+h=\sqrt{2}R\] \[\therefore \]  \[h=(\sqrt{2}-1)R\]


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