Celestia Forums

How do I determine the geosynchronous orbit for a satellite around a given planet?

I tried using astrosynthesis, but it doesn't work very well at computing that when making planets from scratch.

PlutonianEmpire wrote:How do I determine the geosynchronous orbit for a satellite around a given planet?

I tried using astrosynthesis, but it doesn't work very well at computing that when making planets from scratch.

To compute a "geosynchronous" (Clarke) orbit for a given planet, one needs to know: 1) the mass of the planet, 2) the planet's period of rotation, and 3) the equatorial radius of the planet. With this data in hand, the first thing to calculate is the planet's rate of angular motion using the formula

w = 2 pi radians/length of day (in seconds)

Next, w is plugged into the equation

r = ((GM)/w^2)^(1/3)

where G is the Gravitational Constant and M is the mass of the planet in kilograms.

Finally, subtract the equatorial radius of the planet (in kilometers) from r to get the radius of the equatorial Clarke orbit.

Hope this helps,

Radtech

radtech wrote:

PlutonianEmpire wrote:How do I determine the geosynchronous orbit for a satellite around a given planet?

I tried using astrosynthesis, but it doesn't work very well at computing that when making planets from scratch.

To compute a "geosynchronous" (Clarke) orbit for a given planet, one needs to know: 1) the mass of the planet, 2) the planet's period of rotation, and 3) the equatorial radius of the planet. With this data in hand, the first thing to calculate is the planet's rate of angular motion using the formula

w = 2 pi radians/length of day (in seconds)

Next, w is plugged into the equation

r = ((GM)/w^2)^(1/3)

where G is the Gravitational Constant and M is the mass of the planet in kilograms.

Finally, subtract the equatorial radius of the planet (in kilometers) from r to get the radius of the equatorial Clarke orbit.

Hope this helps,

Radtech

That's right with subtracting the equatorial radius of the planet if you want the geosynchronus height above the surface.
But to plug the correct value into Celestia you need r as you calculated it, as Celestia takes the radius for any orbiting element as distance from the center of gravity.

Regards,

Guckytos

radtech wrote:w = 2 pi radians/length of day (in seconds)

Next, w is plugged into the equation

r = ((GM)/w^2)^(1/3)

where G is the Gravitational Constant and M is the mass of the planet in kilograms.

The correct formula for the semi-major axis a (in meters) is in fact:

[tex]a=\sqrt[3]{GM\left(\frac{P}{2\pi}\right)^2}[/tex]

where P = the planet's (sidereal) rotation period in seconds.
... For the Earth this is approx. 86,164 s (23h 56m 04 s).

Again, subtract the planetary radius to get the altitude.

Chuft-Captain wrote:

radtech wrote:r = ((GM)/w^2)^(1/3)

The correct formula for the semi-major axis a (in meters) is in fact:

[tex]a=\sqrt[3]{GM\left(\frac{P}{2\pi}\right)^2}[/tex]

Your formula is essentially the same as radtech's, and they both give the same result.

Hi friends!
The formulas ARE exactly the same!!
Ciao