Question

You want to put a satellite in a geosynchronous orbit around the earth. (This means the asteroid takes 1 day to complete a turn around the earth.)
(a) At what height above the surface do you need to put it?
(b) How fast is it moving?
(c) How does the answer to (b) compare to the escape speed from the earth, for an object at this height?

Answer 1

A geosynchronous orbit is an orbit of a satellite around Earth in which the rotation period of satellite exactly matches with rotation period of Earth i.e. one day (23 hours, 56 minutes, and 4 seconds).

(a) Earth's geosynchronous orbit is 35786 km from the surface. So, we need to put a satellite 35786 km above from the surface.

(b) The speed of satellite can be calculated by this equation

  

Where, G = 6.67 x 10^-11 Nm²/kg²

M = 6 x 10²⁴ kg (Mass of Earth)

R = 6400 km (Radius of Earth)

h = 35786 km

R + h = 42186 km = 4.2186 x 10⁴ km = 4.2186 x 10⁷ m

So,

or,   

or,   

or,   

(c) Escape speed from Earth is 11.2 km/s which is much greater than speed calculated in (b) i.e. 3.08 km/s.

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