Respuesta :

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)

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