CMC Medical CMC-Medical VELLORE Solved Paper-2008

  • question_answer
    What is the escape velocity for a body on it surface of a planet on which the accelerant due to gravity is \[{{(3.1)}^{2}}m{{s}^{-2}}\]and whose radius is 8100 km?

    A)  \[2790\,km\text{-}{{s}^{-1}}\]                  

    B)  \[27.9\,km\text{-}{{s}^{-1}}\]

    C)  \[\frac{27.9}{\sqrt{5}}\,km\text{-}{{s}^{-1}}\]                   

    D)  \[27.9\sqrt{5}\,km\text{-}{{s}^{-1}}\]

    E)  \[\frac{2.79}{\sqrt{5}}km\text{-}{{s}^{-1}}\]

    Correct Answer: C

    Solution :

                    Escape velocity, \[{{v}_{e}}=\sqrt{2gR}\] Given, \[g={{(3.1)}^{2}}m{{s}^{-2}}\]                 \[R=8100\,km=8100\times {{10}^{3}}m.\] \[\therefore \]  \[{{v}_{e}}=\sqrt{2\times {{(3.1)}^{2}}\times 8100\times {{10}^{3}}}\]                 \[{{v}_{e}}=12.5\times {{10}^{3}}m\text{/}s\] \[\Rightarrow \]               \[{{v}_{e}}=12.5\,km\text{/}s=\frac{27.9}{\sqrt{5}}km{{s}^{-1}}\]


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