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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    The acceleration due to gravity becomes \[\left( \frac{g}{2} \right)\]g = acceleration due to gravity on the surface of the earth) at a height equal to

    A)  4R                                         

    B) R/4

    C) 2R                                          

    D)  R/2

    Correct Answer: B

    Solution :

    The acceleration due to gravity \[g=\frac{GM}{{{R}^{2}}}\] At a height h above the earth's surface, the acceleration due to gravity is \[g'=\frac{GM}{{{(R+h)}^{2}}}\] \[\therefore \]  \[\frac{g}{g'}={{\left( \frac{R+h}{R} \right)}^{2}}={{\left( 1+\frac{h}{R} \right)}^{2}}\]                 \[\frac{g'}{g}={{\left( 1+\frac{h}{R} \right)}^{-2}}\]                 \[=\left( 1-\frac{2h}{R} \right)\] But         \[g'=\frac{g}{2}\]                (given) \[\therefore \]  \[\frac{g/2}{g}=1-\frac{2h}{R}\]                 \[\frac{2h}{R}=\frac{1}{2}\]                 \[h=\frac{R}{4}\]


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