# How to calculate the change in inclination due to perturbations?

I would like to know the how much the inclination for a generic elliptical orbit changes over an unspecified period. The variation of inclination fluctuates as per a thread listed at the bottom.

I'd like a more analytical solution than having to run a GMAT/STK simulation for an extended period of time and plotting the inclination to see the variation.

In The New SMAD (page 211) it refers to the disturbing potential but doesn't list it, and then I've become a bit stuck in working out how I'd use that to get an answer.

The aim here is to calculate the inclination change required if the secular variation in inclination was to be minimized. So I'm looking to find how large a plane change is needed and how often to effectively maintain the current orbit within a specified tolerance.

This all fits in with trying to come up with the generic stationkeeping requirements for any elliptical orbit as part of an optimization study so the aim is more trying to find a workable solution at relatively low computation requirement (as it will have to be run for a large number of scenarios) then coming up with a super high fidelity solution.

• Let me double check on what you are asking. Would an answer be a list of expressions for the rate of change of Keplerian orbital elements? For example like the rate of nodal precession for an elliptical orbit shown in this question but of course a complete list? There must be a nice listing of these somewhere, at least for $J_2$. It might also be a good idea to explain which perturbations you are interested in.
– uhoh
Mar 28, 2018 at 0:44
• Yes, a list for rate of change of the Keplerian elements as shown would be ideal. The list given in The New SMAD would be ideal if it were in terms of J2 components. Although I'm not sure that would work for inclination, I don't think the orbit simply has a reduction in inclination over time but more fluctuates, although I may be mistaken on that. Mar 28, 2018 at 8:51
• Okay thanks! I am not sure if I can do this myself, but I am sure someone here will be able to.
– uhoh
Mar 28, 2018 at 8:59