# Circular motion/Earth- I'm lost

1. Oct 25, 2008

### fluidistic

Recently I've been calculating that if there was a satellite in orbit around the Earth, to make it move in a cicular path, we'd have to hit it so that it acquires a speed of $$\sqrt{\frac{GM}{r}}$$. (You can check this out : https://www.physicsforums.com/showthread.php?t=266719). Everything was OK until I remembered my post https://www.physicsforums.com/showthread.php?t=255633&highlight=geostationary.
If I understood something about this last post, in order to put in orbit (circular motion) a satellite we have to do it at $$3.58 \cdot 10^5 \text{km}$$ from the surface of the Earth otherwise the satellite would deviate and then not describe a circular motion. (unless it has a reactor and correct its path).
But the first formula ($$\sqrt{\frac{GM}{r}}$$) seems to suggest (at least to me) that no matter how far from the center of the Earth the satellite is, it is always possible to apply an impulse on a satellite so that it will describe a circular path.
So my guess is that I'm wrong with the first formula but is it wrong? In fact I can't understand how the 2 formulas can work at the same time. One says the distance between the center of mass of the Earth and the satellite matters while the other say it doesn't at all. I also believe that the fact that one formula is for geostationary satellites doesn't matter since the other formula is applicable for all situations. I hope you understand where I'm clueless. What's happening? Thanks...

2. Oct 26, 2008

### Staff: Mentor

Look at the wikipedia Circular Orbit page. Given any arbitrary radius you can determine the velocity required to achieve a circular orbit at that radius using your formula above. There is another formula on the Wikipedia page that allows you to determine the orbital period. The geostationary orbit is the circular orbit with an orbital period of one day, that has a fixed radius. Any other radius can have a circular orbit but its period will generally be different from one day.

3. Oct 26, 2008

### fluidistic

Thanks a lot DaleSpam!! I get it now!