Manipal Medical Manipal Medical Solved Paper-2000

  • question_answer
    The escape velocity from the surface of the earth\[{{\upsilon }_{es}}\]. The escape velocity from the surface of a planet whose mass arid radius are three times those of earth; will be:

    A)  \[1/3{{\upsilon }_{e}}\]

    B)  \[9{{\upsilon }_{e}}\]

    C)  \[3{{\upsilon }_{e}}\]

    D)  \[{{\upsilon }_{e}}\]

    Correct Answer: D

    Solution :

     Here: Escape velocity on surface of the earth \[={{\upsilon }_{es}}\] Mass of the earth \[={{M}_{e}}\] Mass of the surface of the planet \[=3{{M}_{e}}\] Radius of planet \[=3{{R}_{e}}\] (where\[{{R}_{e}}\]is the radius of earth) As the escape velocity on surface of the earths is \[{{\upsilon }_{es}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}=\sqrt{\frac{{{M}_{e}}}{{{R}_{e}}}}\] Hence, \[\frac{{{\upsilon }_{es(e)}}}{{{r}_{es(p)}}}=\sqrt{\frac{{{\upsilon }_{e}}}{{{R}_{e}}}\times \frac{{{R}_{p}}}{{{M}_{p}}}}\] Or \[\frac{{{\upsilon }_{es(e)}}}{{{r}_{es(p)}}}=\sqrt{\frac{{{M}_{e}}}{{{R}_{e}}}\times \frac{3{{R}_{e}}}{3{{M}_{e}}}}=\frac{1}{1}\] Or \[{{\upsilon }_{es(p)}}={{\upsilon }_{es(e)}}\]


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