Moving satellites around in Realism Overhaul

[Dman979]

How do I calculate the degrees my satellite will change in its orbit?

For instance, let's say that I have a satellite at a perfect geostationary orbit. I want to set it up at 0^o 0' 0" N 150^o 0' 0" W. It is currently at 0^o 0' 0" N 95^o 0' 0" E.

I know that a lower orbit travels faster, and it will take less time to go to 150^o W if i get to a lower orbit than if i go to a higher one.

I lower my perigee so the orbital period is 20 minute less than synchronous. How many degrees will I shift each orbit, relative to my starting point, and what is the formula used to calculate this?

Edited January 18, 2016 by Dman979

[Snark]

Kerbin rotates once every six hours, i.e. 360 minutes, which is conveniently one degree per minute. Therefore, if you shorten your orbital period by 20 minutes, it will end up 20 degrees further east for each orbit.

Sometimes rocket science doesn't have to be complicated.

Edited January 18, 2016 by Snark

[Dman979]

@Snark

I'm glad to hear that it's not too complicated.

But I'm using Realism Overhaul, so it's Earth, not Kerbin. How does this affect the calculations?

[Snark]

1 hour ago, Dman979 said:

But I'm using Realism Overhaul, so it's Earth, not Kerbin. How does this affect the calculations?

Same principle, it's just that you don't have that convenient 1-minute-equals-1-degree setup, so you may actually have to pull out your calculator. 

• Let t_s be the sidereal orbital period of your satellite
• Let t_p be the sidereal rotation period of your planet
• let φ be the longitude that your satellite is over, in degrees

(remember, you want the sidereal day, not the solar day; Earth's sidereal day is a few minutes shorter than 24 hours)

In that case, the change dφ on each orbit will be given by:

dφ = 360 * (1 - t_p/t_s)

where the result will be positive for a net eastward change, and negative for a net westward change.

Edited January 18, 2016 by Snark