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BVP Medical BVP Medical Solved Paper-2011

  • question_answer
    A satellite is moving in a circular orbit at acerta in height above the earths surface. It takes\[i\alpha \sqrt{\theta }\] s to complete one revolution with acentric petal acceleration equal to 9.32 m/s2, the height of the satellites orbit above the earths surface is \[\gamma \]

    A)  70km                                   

    B)  120km

    C)  170 km                                

    D)  220 km

    Correct Answer: C

    Solution :

                    Here  \[-\frac{1}{2}mg{{R}_{e}}\] Centripetal acceleration \[{{\alpha }_{1\,}}and\,{{\beta }_{2}}\] \[{{Y}_{1\,}}and\,{{Y}_{2}}\]        \[\frac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=\frac{2}{3}\]       \[\frac{{{Y}_{1}}}{{{Y}_{2}}}\] \[{{v}_{H}},{{v}_{N}}\,and\,{{v}_{o}}\]   \[{{v}_{H}}>{{v}_{N}}>{{v}_{o}}\]                 \[{{v}_{o}}>,{{v}_{N}}\,=\,{{v}_{H}}\] \[{{v}_{o}}>,{{v}_{H}}\,=\,{{v}_{N}}\]Height of satellites orbit above the earths surface                 \[{{v}_{N}}>,{{v}_{o}}\,=\,{{v}_{H}}\]                 \[=1.013\times {{10}^{5}}N/{{m}^{2}}\]                 \[\overset{0}{\mathop{A}}\,\]


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