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Q.
At what height above the surface of earth the value of $''g''$ decreases by $2 \%$ ? [radius of the earth is $6400\, km$ ]
Gravitation
Solution:
Acceleration due to gravity above the surface of earth at a height $h$ is given $g^{\prime}=g\left(1-\frac{2 h}{R_{\theta}}\right)$
here, $g^{\prime}=0.98 g$
$\Rightarrow 0.98 =1-\frac{2 h}{R_{e}} $
$\Rightarrow \frac{2 h}{R_{e}} =0.02 $
$h =0.01 R_{e} $
$=0.01 \times 6400\, km $
$=64\, km$