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.

#### Solution

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 \[g' = \frac{Gm}{\left( R + h \right)^2}\]

At h = 36000 km, we have :

\[g' = \frac{Gm}{\left( 36000 + 6400 \right)^2}\]

At the surface, we have :

\[g = \frac{Gm}{\left( 6400 \right)^2}\]

\[ \therefore \frac{g'}{g} = \frac{6400 \times 6400}{42400 \times 42400}\]

\[ = \frac{256}{106 \times 106} = 0 . 0228\]

\[ \Rightarrow g' = 0 . 0227 \times 9 . 8 = 0 . 223 \left[ \text { Taking }\ g = 9 . 8 \ \text{ m/ s}^2 \text {at the surface of the earth} \right]\]

For a 120 kg equipment placed in a geostationary satellite, its weight will be mg' = 120 × 0.233

\[\Rightarrow\] 26.76 ≈ 27 N