Geostaionary orbit equation?

[Carrot]

I am trying to set up a Geostaionary orbit network with 4 satellites. I know the parking / deploying orbit is 4 hours if I want to set up a 3 satellites network. So I tried to do some maths myself for 6*2/4=3 hours. But I found it is wrong. So may I ask for the equation and the explanation of it please? Thanks

[DrLicor]

Quite easy one. You know that kerbins day lasts 6hours. Which means if you have a parking orbit of 3hours, your 3rd sat will be in de same spot as the 1st. 

You get your second sat in the opposite of your first sat. That's not what you where aiming for.

What you need is a orbital time of 1,5 hours. But then your Pe will probably in kerbin. (not sure)

So what you need to do is to make your orbitaltime extent with 1,5 hours. Well, you know 1,5+1,5 is 3hour orbit. Doesn't work, so you need a parking orbit of 4,5 hours. 

Just now, DrLicor said:

Was on my mobile phone, didnt work out as I wanted

Edited January 9, 2017 by DrLicor

[GoSlash27]

Carrot,

 For any resonant orbit, the conversion between period and SMA is (n/d)^3/2 where n is the numerator period and d is the denominator period. If you wish to equally space 3 sats in GSO, your transfer orbit must have a period of either 2/3 or 4/3. If you're willing to spend more cycles in the transfer, you can tighten this up, but for the sake of simplicity we'll assume you don't want to do that.

For a 2/3 transfer orbit, the SMA is (2/3)^3/2= .544331 of the GSO. Assuming a 3,463 km SMA for the GSO, this means the SMA of the transfer would be 1,885 km. Since the Pe is as much below the SMA as the Ap is above it, this would set the Pe at 307 km, which is inside the planet's radius.

So instead, let's say we go 5/6 and inject one sat every 2 cycles. (5/6)^2/3 = .88555. .88555*3,463 km= 3,067 km. Pe is 2,671 km, or an altitude of 2,071 km.

The math works the same way for a larger transfer orbit (4/3, 7/6, etc.)

HTHs,
-Slashy

 

D'oh! I misread your OP and thought you wanted 3 sats instead of 4.

 Same rules apply though. 3/4, 7/8 (if you skip a cycle), 5/4, etc. The math works the same for all resonances and all gravity wells. Your Pe would be 435 km on a 3/4 cycle.

Best,
-Slashy

 

Edited January 9, 2017 by GoSlash27

[jebediah kerman reentering]

quick answer: 3463.334 km

[king of nowhere]

On 1/9/2017 at 12:02 AM, Carrot said:

I am trying to set up a Geostaionary orbit network with 4 satellites. I know the parking / deploying orbit is 4 hours if I want to set up a 3 satellites network. So I tried to do some maths myself for 6*2/4=3 hours. But I found it is wrong. So may I ask for the equation and the explanation of it please? Thanks

what kind of calculation is that? why are you doing *2/4, and what is 3 hours supposed to represent?

and what are you trying to achieve exactly?
if you want those satellites equally spaced, you have an easier time launching them at 2 hours interval, since that's how you equally divide in three a kerbin day.