Uttarakhand PMT Uttarakhand PMT Solved Paper-2004

  • question_answer
    The orbital velocity of an artificial satellite in a circular orbit just above the earths surface is \[\upsilon \] The orbital velocity of a satellite orbiting at an altitude of half of the radius, is :

    A)  \[\frac{3}{2}{{\upsilon }_{0}}\]

    B)  \[\frac{2}{3}{{\upsilon }_{0}}\]

    C)  \[\frac{\sqrt{2}}{3}{{\upsilon }_{0}}\]

    D)  \[\frac{\sqrt{3}}{2}{{\upsilon }_{0}}\]

    Correct Answer: C

    Solution :

     Given: \[{{R}_{1}}={{R}_{e}}\] \[{{R}_{2}}={{R}_{e}}+\frac{{{R}_{e}}}{2}=\frac{3}{2}{{R}_{e}}\] The orbital velocity of satellite is \[{{\upsilon }_{0}}=\sqrt{\frac{G{{M}_{e}}}{R}}\] \[\Rightarrow \] \[{{\upsilon }_{0}}\propto \sqrt{\frac{1}{R}}\] Hence,   \[\frac{{{\upsilon }_{1}}}{{{\upsilon }_{2}}}=\sqrt{\frac{{{R}_{2}}}{{{R}_{1}}}}\] \[=\sqrt{\frac{3{{R}_{e}}}{2{{R}_{e}}}}=\sqrt{\frac{3}{2}}\] \[{{\upsilon }_{2}}=\sqrt{\frac{2}{3}}{{\upsilon }_{1}}\] \[=\sqrt{\frac{2}{3}}{{\upsilon }_{0}}\] \[(\because {{\upsilon }_{1}}={{\upsilon }_{0}})\]


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