Q3.12 Geostationary orbits (S)
A satellite is in geostationary orbit above a point on the equator. What is the radius of its orbit?
Could one engineer a geostationary orbit above Edinburgh?
Hint
Reveal

Hint
- If the orbit is geostationary what is its period?
- Sketch an orbit geostationary above Edinburgh. Where is its centre? So what is the direction of its acceleration? What is the problem with that?
Solution
Reveal

Solution
If the orbit is geostationary it also has to be circular. Denote the radius by R. Newton’s 2nd law applied to the satellite motion implies
![\[ \frac{GM_Em}{R^2} = \frac{mv^2}{R} \]](mastermathpng-0.png)
Now the satellite speed v is related to the radius of the orbit R and the period of the orbit (the time taken to complete one orbit) T by
![\[ v= \frac{2\pi R}{T} \]](mastermathpng-1.png)
![\[ \frac{GM_E}{R^2} = \frac{4 \pi^2 R}{T^2} \]](mastermathpng-2.png)
![\[ R=\left[\frac{GM_E T^2}{4 \pi^2}\right] ^{1/3} \]](mastermathpng-3.png)

To be stationary above Edinburgh a satellite would have to move in a circle whose plane is parallel to the equatorial plane, but which has Edinburgh lying on it. The centre of this circle lies on the earth’s axis, but not at the earth’s centre. |
The gravitational force (which acts towards the earth’s centre) is then no longer in the same direction as the centripetal acceleration. Thus such an orbit is not possible, unless one can create another force on the satellite, so that the total force (the sum of this one and the gravitational force) does act in the required direction. To do this the satellite would need to have its own propulsion system, and a suitable (large!) supply of fuel.