Question:
The geostationary orbit of the earth is at a distance of about $36000 \mathrm{~km}$ from the earth's surface. Find the weight of $120 \mathrm{~kg}$ equipment placed in a geostationary satellite. The radius of the earth is $6400 \mathrm{~km}$.
Solution:
$g=\frac{G M}{R^{2}}$
$g^{\prime}=\frac{G M}{(R+h)^{2}}$
$\frac{g^{\prime}}{g}=\frac{G M /(R+h)^{2}}{G M / R^{2}}=\frac{R^{2}}{(R+h)^{2}}$
$g^{\prime}=\left[\frac{(6400)^{2}}{(36000+6400)^{2}}\right] \times g$
$g^{\prime}=0.0227 \times 9.8$
$g^{\prime}=0.0223$
$W=g^{\prime}=120 \times 0.223$
$W=27 \mathrm{~N}$