Question

at what distance above the surface of the earth is the acceleration due to gravity 0.980m/s² if the acceleration due to gravity at the surface has magnitude 9.80m/s²?

Answer 1

30635

Answer 2

$$
\large{
F =mg= G\frac{m\times m_{earth}}{r^2}\\
g=G\frac{m_{earth}}{r^2}\
}
$$
So @ 10% of g
$$
\large{
0.980=G\frac{m_{earth}}{r^2}\\
\implies r=\sqrt{\cfrac{G m_{earth}}{0.980}}
}
$$
Distance from the surface is
$$
\large{
r-R_{earth}
}
$$
Where \(R_{earth}\) is the radius of the Earth.

This turns out to be about 1800 kilometers or 1118 miles.