Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    Mass of moon is \[1/81\] times that of earth and its radius is \[1/4\] the radius of earth. If escape velocity on the surface of the earth is\[11.2\,\,km/s\]. Then the value of escape velocity at surface of the moon will be

    A) \[5\,\,km/s\]                    

    B) \[2.5\,\,km/s\]

    C) \[0.5\,\,km/\sec \]                         

    D) \[0.14\,\,km/\sec \]

    Correct Answer: B

    Solution :

    Here: Mass of moon\[{{M}_{m}}=\frac{{{M}_{e}}}{81}\] Radius of moon\[{{R}_{m}}=\frac{{{R}_{e}}}{4}\] Escape velocity on the surface of earth \[{{V}_{es(s)}}=11.2\,\,km/s\] The relation for escape velocity is                 \[v=\sqrt{\frac{2\,\,GM}{R}}\propto \sqrt{\frac{M}{R}}\] Hence   \[\frac{{{v}_{es(m)}}}{{{v}_{es(s)}}}=\sqrt{\frac{{{M}_{m}}}{{{R}_{m}}}\times \frac{{{R}_{e}}}{{{M}_{e}}}}\]                 \[=\sqrt{\frac{{{M}_{e}}}{81}\times \frac{4}{{{R}_{e}}}\times \frac{{{R}_{e}}}{{{M}_{e}}}}=\frac{2}{9}\]      \[{{v}_{es(m)}}=\frac{2}{9}\times 11.2=2.5\,\,km/s\]


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