JIPMER Jipmer Medical Solved Paper-1998

  • question_answer
    The escape velocity from the earth is \[{{\upsilon }_{es\,.}}\] Then the escape velocity from a planet whose mass and radius are twice those of the earth is:

    A)  \[{{\upsilon }_{es\,}}\]            

    B)                         \[2{{\upsilon }_{es\,(\upsilon )}}\]

    C)  \[4{{\upsilon }_{es\,}}\]              

    D)         \[16{{\upsilon }_{es\,(\upsilon )}}\]

    Correct Answer: A

    Solution :

    Escape velocity from the earth is given by \[{{\upsilon }_{es}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] and        \[{{\upsilon }_{es(m)}}=\sqrt{\frac{2G{{M}_{p}}}{{{R}_{p}}}}\] hence,  \[\frac{{{\upsilon }_{es}}}{{{\upsilon }_{es(p)}}}=\sqrt{\frac{{{M}_{e}}}{{{R}_{e}}}\times \frac{{{R}_{p}}}{{{M}_{p}}}}\] \[=\sqrt{\frac{{{M}_{e}}}{{{R}_{e}}}\times \frac{2{{R}_{e}}}{2{{M}_{p}}}}=1\]     Hence,  \[{{\upsilon }_{es(p)}}={{\upsilon }_{es}}\]


You need to login to perform this action.
You will be redirected in 3 sec spinner