MGIMS WARDHA MGIMS WARDHA Solved Paper-2003

  • question_answer
    Mass of a planet is 10 times that of earth and radius is 2 times that of earth. If at the earth, the escape velocity of a satellite is \[{{\upsilon }_{es(e),}}\] then escape velocity at the planet is :

    A) \[\sqrt{5}\,\,{{\upsilon }_{es(e)}}\]                        

    B) \[5\,\,{{\upsilon }_{es(e)}}\]

    C) \[2\sqrt{5}\,\,{{\upsilon }_{es(e)}}\]                      

    D) \[10\,\,{{\upsilon }_{es(e)}}\]

    Correct Answer: A

    Solution :

    \[{{\upsilon }_{e{{s}_{(e)}}}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] \[{{\upsilon }_{e{{s}_{(p)}}}}=\sqrt{\frac{2G{{M}_{P}}}{{{R}_{P}}}}\]      (Here:\[{{M}_{p}}=10\text{ }{{M}_{e}},\text{ }{{R}_{p}}=2{{R}_{e}}\]) \[{{\upsilon }_{e{{s}_{(p)}}}}=\sqrt{\frac{2G\times 10{{M}_{e}}}{2{{R}_{e}}}}\]                  ?.(2) From Eqs. (1) and (2), we have \[\frac{{{\upsilon }_{es(e)}}}{{{\upsilon }_{es(p)}}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\times \sqrt{\frac{2{{R}_{e}}}{2G\times 10{{M}_{e}}}}\] \[{{\upsilon }_{es(p)}}=\sqrt{5}{{\upsilon }_{es(e)}}\]


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