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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    Escape velocity at surface of earth is 11.2km/s. Escape velocity from a planet whose mass is the same as that of earth and        radius 1/4 that of earth, is :                                                                                          [BHU PMT-2004]

    A)  2.8km/s                              

    B)  15.6 km/s

    C)  22.4 km/s                          

    D)  44.8 km/s

    Correct Answer: C

    Solution :

                     At a certain velocity of projection the body will go out of the gravitational field of the earth and will never return of the earth, this velocity is known as escape velocity \[{{v}_{e}}=\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\] Given, \[{{M}_{e}}={{M}_{p}},\,{{R}_{p}}=\frac{{{R}_{e}}}{4}\] \[\therefore \]  \[\frac{{{v}_{p}}}{{{v}_{e}}}=\sqrt{\frac{{{M}_{e}}}{{{M}_{e}}}\times \frac{{{R}_{e}}}{{{R}_{e/4}}}}=\sqrt{4}=2\] \[\Rightarrow \]               \[{{v}_{p}}=2{{v}_{e}}=2\times 11.2\]                                                 \[=22.4\,\,km/s\]


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