CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2007

  • question_answer
    The escape velocity of a body on the surface of earth is 11.2 km/s. If the mass of the earth is doubled and its radius halved, the escape velocity becomes

    A)  5.6 km/s                             

    B)  11.2 km/s

    C)  22.4 km/s          

    D)         44.8 km/s

    E)  67.2 km/s

    Correct Answer: C

    Solution :

    Escape velocity from earth \[({{v}_{e}})=\sqrt{\frac{2GM}{R}}\] \[\frac{v_{e}^{}}{{{v}_{e}}}=\sqrt{\frac{2G(2M)}{R/2}\times \frac{R}{2GM}}\] Or           \[\frac{v_{e}^{}}{11.2}=2\] or \[\text{v}_{e}^{}=22.4\text{ }km/s\]


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