Refer to the diagram shown below.
In geosynchronous orbit, the satellite makes one circular revolution at the equator in 1 day.
From tables, obtain
r = 42,164 km = 4.2164 x 10⁷ m, the earth's radius
g = 9.8 m/s², acceleration due to gravity.
Note that 1 day = 86400 s
The angular velocity of rotation is
ω = (2π)/86400 = 7.2722 x 10⁻⁵ rad/s
The centripetal acceleration on the satellite is
a = rω²
= (4.2164 x 10⁷ m) * (7.2722 X 10⁻⁵ rad/s)²
= 0.223 m/s²
The weight of the 2000 kg satellite is
W = m(g - a)
= (2000 kg) * (9.8-0.223 m/s²)
= 19154 N
= 19.154 kN
Answer: 19 kN (nearest integer)
