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NEET NEET SOLVED PAPER 2016 Phase-II

  • question_answer
    A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of\[{{g}_{0}}\],the value of acceleration due to gravity at the earth's surface, is

    A)  \[\frac{m{{g}_{o}}{{R}^{2}}}{2(R+h)}\]                   

    B)  \[\frac{m{{g}_{o}}{{R}^{2}}}{2(R+h)}\]

    C)  \[\frac{2{{m}_{o}}{{R}^{2}}}{R+h}\]                      

    D)  \[\frac{2{{m}_{o}}{{R}^{2}}}{R+h}\]

    Correct Answer: B

    Solution :

                \[Total=\frac{GMm}{2r}\]             Here, r= R + h and GM= \[{{g}_{o}}{{R}^{2}}\]             \[\Rightarrow E=-\frac{m{{g}_{o}}{{R}^{2}}}{2(R+h)}\]


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