Question

At what height above the surface of earth acceleration due to gravity reduces by $1$%?

Answer 1

The new value of $gravity$ at a height $h$ is given as $g\prime =\frac{g{R}^2}{(R+h{)}^2}$

As visible from the formula that incresing $h$ results in decresing gravity.

So just put $g\prime =0.99g$ i.e. 1% less than $g$ where $g$ is gravity at surface. so we get $0.99g=\frac{g{R}^2}{(R+h{)}^2}$

Solving the equation we get $h=0.005R$