Question

The geostationary orbit of the earth is at a distance of about 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.

Answer 1

The geostationary orbit of the Earth is at a distance of about 36000 km.
We know that the value acceleration due to gravity above the surface of the Earth is given by $g\text{'}=\frac{\mathrm{G}m}{{\left(R+h\right)}^2}$.
At h = 36000 km, we have:
$g\text{'}=\frac{\mathrm{G}m}{{\left(36000+6400\right)}^2}$
At the surface, we have:
$$\begin{gathered} g=\frac{\mathrm{G}m}{{\left(6400\right)}^2} \\ \therefore \frac{g\text{'}}{g}=\frac{6400\times 6400}{42400\times 42400} \\ =\frac{256}{106\times 106}=0.0228 \\ \Rightarrow g\text{'}=0.0227\times 9.8=0.223\left[\mathrm{Taking}g=9.8\mathrm{m}/{\mathrm{s}}^2\mathrm{at}\mathrm{the}\mathrm{surface}\mathrm{of}\mathrm{the}\mathrm{earth}\right] \end{gathered}$$

For a 120 kg equipment placed in a geostationary satellite, its weight will be
mg' = 120 × 0.233
$\Rightarrow$26.76 ≈ 27 N