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KVPY Sample Paper KVPY Stream-SX Model Paper-24

  • question_answer
    A satellite is moving with a constant speed u in circular orbit around the earth. An object of mass \['m'\] is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of the earth. At the time of ejection.  the kinetic energy of the object is:

    A) \[2m{{v}^{2}}\]

    B) \[m{{v}^{2}}\]

    C) \[\frac{1}{2}m{{v}^{2}}\]

    D) \[\frac{3}{2}m{{v}^{2}}\]

    Correct Answer: B

    Solution :

    Initially, kinetic energy\[=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m\frac{G{{M}_{e}}}{r}\]
    By conservation of M.E.,\[\frac{-G{{M}_{e}}m}{r}+KE=0\]
    \[KE=\frac{G{{M}_{e}}m}{r}=m{{v}^{2}}.\]


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