Communications satellites are placed in a circular orbit where they stay directly over a fixed point on the equator as the earth rotates. these are called geosynchronous orbits. the radius of the earth is 6.37 * 106 m, and the altitude of a geosynchronous orbit is 3.58 * 107 m 1â22,000 miles2. what are (a) the speed and (b) the magnitude of the acceleration of a satellite in a geosynchronous orbit?

Refer to the diagram shown below.

Given:
R = 6.37 x 10⁶ m, the radius of the earth
h = 3.58 x 10⁷ m, the height of the satellite above the earth's surface.
Therefore
R + h = 4.217 x 10⁷ m

In geosynchronous orbit, the period of rotation is 1 day.
Therefore the period is
T = (24 h)*(60 min/h)*(60 s/min) = 86400 s

The angular velocity is
ω = (2π rad)/(86400 s) = 7.2722 x 10⁻⁵ rad/s

Part (a)
The tangential speed is
v = (R+h)*ω
   = (4.217 x 10⁷ m)*(7.2722 x 10⁻⁵ rad/s) 
   = 3066.7 m/s
   = 3.067 km/s

Part (b)
The centripetal acceleration is
a = v²/(R+h)
   = (3066.7 m/s)²/(4.217 x 10⁷ m)
   = 0.223 m/s²

Answers:
(a) The speed is 3.067 km/s
(b) The acceleration is 0.223 m/s²