RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

  • question_answer
    Acceleration due to gravity at earths surface is \[g\,rn{{s}^{-2}}.\] Find the effective value of gravity at a height of 32 km from se. a level: \[({{R}_{e}}=6400\,km)\]

    A)  \[0.5\,g\,m{{s}^{-2}}\]                                 

    B)  \[0.99\,g\,m{{s}^{-2}}\]

    C)  \[1.01\,g\,m{{s}^{-2}}\]               

    D)         \[0.90\,g\,m{{s}^{-2}}\]

    Correct Answer: A

    Solution :

    For height \[h\] above the earths surface                 \[g=g\left( 1-\frac{2h}{{{R}_{e}}} \right)=g\left( 1-\frac{64}{6400} \right)\]                                                 \[(\because \,\,{{R}_{e}}=6400\,\,km)\]                 \[=g(1-0.01)\]                                 \[=0.99\,\,m{{s}^{-2}}\]


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