Find the magnitude of acceleration due to gravity at a height of $10 \, km$ from the surface of the earth.

(N/A) The acceleration due to gravity at a height $h$ above the Earth's surface is given by the formula $g_{h} = g_{e} \left(1 - \frac{2h}{R_{e}}\right)$,where $h = 10 \, km = 10^{4} \, m$,$R_{e} = 6.4 \times 10^{6} \, m$,and $g_{e} = 9.8 \, m/s^2$.
Substituting the values:
$g_{h} = 9.8 \left[1 - \frac{2 \times 10^{4}}{6.4 \times 10^{6}}\right]$
$g_{h} = 9.8 [1 - 0.003125]$
$g_{h} = 9.8 [0.996875]$
$g_{h} = 9.769375 \, m/s^2$
Rounding to appropriate significant figures,we get $g_{h} \approx 9.77 \, m/s^2$.