# Geosynchronous station keeping

1. ### BobG

2,338
I'm having problems figuring out geosynchronous station keeping requirements due to triaxiality.

So far, I've gotten as far as finding the equilibrium points at about 75 degrees and 255 degrees, but I can't get from there to the station keeping requirements for satellites located at other longitudes. The angular, or longitudinal, accleration should equal:

$$\ddot\lambda = -\left(\omega^2_E \left(\frac{R_E}{a}\right)^2\right) \left(-18J_{22} sin(2(\lambda - \lambda_{22}))\right)$$

$$\lambda$$ is longitude with $$\lambda_{22}$$ being a constant that goes along with $$J_{22}$$ for one of the Earth's spherical harmonics due to triaxiality.
$$\omega_E$$ is the rotation rate of the Earth, $$R_E$$ is the radius of the Earth, and a is the orbit's semi-major axis.

There's another variation of this I found in NASA's TM-2001-210854, Integrated Orbit, Attitude, and Structural Control Systems Design for Space Solar Power Satellites, but it yields the same results. They just merged the mean motion into the rest of the equation. Since the whole purpose of geosynchronous satellites is for the orbit's mean motion to match the Earth's rotation rate, I like the version where its separate, better.

I found the equilibrium points by finding where angular acceleration equaled zero. If I convert this to linear acceleration, I think I should get the acceleration necessary to stay in the same place for other longitudes. If projected over the course of one year, I get a maximum of around 5.203 meters/second/year, which is about 3 times too big.

Wertz and Larsen's Space Mission Analysis and Design just use the equation:

$$\deltaV = 1.715 sin(2(\lambda - \lambda_s))$$

Their equation does produce a realistic maximum. Unfortunately, 1.715 doesn't tell me anything.

Anyone know the missing link, here?

2. ### enigma

1,809
Staff Emeritus
Bob, I wish I could offer more help, but when this topic was covered in my space nav class, it went right over my head.

IIRC, Fundamentals of Astrodynamics and Applications by Vallado had a brief section on this. If they've got it in the library, it may be worth a look.

3. ### remcook

68
Something like this:

The potential of the '22' part of the gravitational potential is written out as:

$$U_{22}=\frac{GM}{r} \frac{Re^2}{r^2} P_{22} ( \sin \phi) J_{22} \cos 2 ( \lamda - \lambda_{22} )$$

with the associated Legendre polynomial:

$$P_{22} ( \sin \phi ) = 3 \cos^2 \phi$$
equals 3 for equatorial orbit.

Getting the acceleration in lambda-direction (alongtrack) means differentiating once wrt r and once wrt lambda:

$$a_{\lambda} = \frac{\partial^2 U}{\partial r \partial \lambda} = -3 \left( \frac{GM}{r} \frac{Re}{r^3} \right) \cdot 3 J_{22} \left( -2 \sin 2 ( \lambda - \lambda_{22} ) \right)$$

equating:

$$a_{\lambda} = r \ddot{\lambda}$$

filling in $$r=a$$

and noting that $$\omega_e = \sqrt{ \frac{GM}{a^3} }$$ by definition of the geostationary orbit (orbital rotational velocity equals Earth rotational velocity) gives the given equation.

edit - I'm not sure about this (the exact reason concerning the double differentiation and the definition of the acceleration). Have to look it up. But at least this gives the given equation :)

Last edited: May 25, 2004
4. ### BobG

2,338
This gets me to my original equation, but reading another perspective gives me a lot of help. I was kind of looking at it as if, by constantly accelerating (or decelerating) to compensate for the longitudinal drift, that I would hold the radius constant. True enough, but that doesn't mean that the acceleration doesn't have a radial component.

In other words, the change in velocity is actually:

$$\dot V=r\dot\omega + \dot{r} \omega$$

And $$\dot\omega$$ and $$\dot{r}$$ are always changing in opposite directions. I'll have to play with it a little and see if it works.

(I need to get a better reference for this, as well. The ones that offer a good explanation came up with same numbers I did and the ones with realistic numbers offer no explanation at all.)