# F=mv^2/r and Geosynchronouse orbits

1. Sep 23, 2005

### Elder

F=mv^2/r and Geosynchronouse orbits? Help Guys!

I’m having a mental block in understanding why geostationary satellites stay in the same place and don’t change orbits or simply fall out of the sky.

I may be using inaccurate information so please vet any and all suppositions I make here.

Using the formulas I found at this web site

I begin with determining the radius for a circular geosynchronous orbit the semi-major axis is given and it turns out to be 42168 Kilometers.

Now, I am assuming that if I calculate the Centrifugal Force on an object in geosynchronous orbit at this distance from the Center of the earth it should equate to the Gravitational force the earth exerts on it.

The formula I remember for Centrifugal force is F=mv^2/r

F= force
m= Mass
v=velocity

so an arbitrary 1 kilogram mass orbiting at r=42168 kilometer distance from the center of the earth traveling at 2(PI)r/24= 11039.56 kilometers /hour around the earth should have a gravitational force equal to the centrifugal force imposed on it by its orbit. Using the above formula F=mv^2/r, is equal to 2890.15 kilograms!

Does the earth impose a gravitational force of over 2890 kilograms on a 1 kilogram object at 42168 kilometers away? I must be doing something or many things wrong! Maybe the formulas or even the premise of the question is wrong.

Last edited: Sep 23, 2005
2. Sep 23, 2005

### Janus

Staff Emeritus
I think the biggest problem you are having is with units. First convert everything to meters, kilograms and seconds, and then do the math. You will find that you get an answer of .223 Newtons for the force on the satellite.

Do the same with the formula $F_g = \frac{G \mu m}{r^2}$, and you will get the same answer (within rounding error).

Last edited: Sep 23, 2005
3. Sep 24, 2005

### Janus

Staff Emeritus
You can also get the formula given in the web site by assuming that the gravitational force and centripetal force needed to hold the satellite in orbit are equal thusly.

For circular orbit centripetal force is:
$$F_c = \frac{mv^2}{r}$$

The force due to gravity is:
$$F_g = \frac{GMm}{r^2}$$

Now, $GM = \mu$ where M is the mass of the Earth, so:

$$F_g = \frac{\mu m}{r^2}$$

We are assuming that Fg=Fc, so

$$\frac{\mu m}{r^2} = \frac{mv^2}{r}$$

$$\frac{\mu }{r} = v^2$$

$$\sqrt{\frac{\mu }{r}} =v$$

The period is therefore:

$$T=\frac{2 \pi r}{\sqrt{\frac{\mu }{r}}}$$

$$T=\frac{2 \pi r \sqrt{r}}{\sqrt{\mu }}$$

$$T=\frac{2 \pi \sqrt{r^3}}{\sqrt{\mu }}$$

$$T=2 \pi \sqrt{\frac{r^3}{\mu }}$$

Last edited: Sep 24, 2005
4. Sep 25, 2005

### Elder

Thanks!

Thanks Janus

Very much!

As to your closing signature quote from Bertrand Russell, I thought Hum, interesting but something has been nawing at me since I read it.
Actually the problem is too many of us are arrogant enough to think we know the difference between fool, fanatic, and the wise.

Last edited: Sep 25, 2005