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Q. Acceleration due to gravity is $g$ on the surface of the earth. Then the value of the acceleration due to gravity at a height of $32 \,km$ above earths surface is (Assume radius of earth to be $6400\, km$)

WBJEEWBJEE 2007

Solution:

$ g=\frac{g}{{{\left( 1+\frac{h}{{{R}_{e}}} \right)}^{2}}}$
$=\frac{g}{{{\left( 1+\frac{32}{6400} \right)}^{2}}}$
$=0.99\,g $