Question

What height above the surface of Earth acceleration due to gravity by one percentage?

Answer 1

Dear student,
$$\begin{gathered} g\text{'}=g-0.01g=0.99g \\ g\text{'}=g\left(1-\frac{2h}{R}\right) \\ \frac{0.99g}{g}=1-\frac{2h}{R} \\ -\frac{2h}{R}=-0.01 \\ h=\frac{0.01\times 6400}{2}km \\ h=32km \end{gathered}$$
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