Orbital Rendezvous Theory question

[Maltman]

So imagine two spacecraft orbiting kerbin have the same periapsis at the exact same point. One of the craft has a circular orbit. The second craft has an elliptical one.

Is there a way to calculate the change in angle between the two crafts relative positions at each passing using only the second crafts apoapsis in contrast to the first crafts(since their periapsis are the same)??

[bewing]

Using just the value for the Ap of the elliptical vessel? Definitely not. You can't even quite calculate the angular change using the "Ap"s and "Pe"s, because it's not a constant.

To get an approximate angular change, you need to calculate the difference in orbital times, as a fraction of one of the orbital times. Then convert to radians or degrees. But in reality the number of degrees per hour varies with average altitude. And the orbital time is calculated with the SMA (which is the average of the Ap and the Pe -- measured from the center of the celestial body).

 

[Spricigo]

To calculate angle phasing between both orbits we use Kepler Laws. 

Quote

1. The orbit of a planet is an ellipse with the Sun at one of the two foci.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.^[1]
3. The square of the orbital period of a planet is proportional to the cube of thesemi-major axis of its orbit.

 

 Was about to  say there is no shortcut... However more accurate to say that is the shortcut . 

[Maltman]

Thanks guys I appreciate the answers. 

 

Can you give me an example of the type of maths required for the problem?because I am having trouble understanding the application of them. Let me give you an example  

Two orbits of Kerbin.  Both periapsis at the exact same point at say 100km. Both craft have perfectly circular orbits that are exactly the same. However the craft are 10° apart with the second craft in front of the first. 

So what does the apoapsis of the second craft need to be in order for those 10° to be closed and for a perfect rendezvous on the next passing of the periapsos? 

Also is there a magical tutorial that explains all this somewhere? 

[OHara]

The type of maths needed are just the fractional powers in Kepler's third law cited above -- for figuring angles in the one orbit that is circular after a whole number of circuits of the other orbit, because angles in the circular orbit advance uniformly in time.

The periods of the orbits T1 and T2 follow T2 / T1 = [(Ap2 + Pe2) / (Ap1 + Pe1)] ^ (3/2)

You wanted to catch up 10° in one cycle of a modified orbit, so you want to adjust T2 = T1 + (10°/360°)T1
T2/T1 = 370°/360°;  (370/360)^(2/3) = 1.018;   Ap2 + Pe2 = 1.018×(Ap1 + Pe1)

So if Ap1=Ap2 = 700km (100km above sea level) we want
Pe2 + 700km = 1.018×(700km + 700km)  =>  Pe2 = 726km (126km above sealevel)

Edited July 29, 2016 by OHara
corrected 2/3 -> 3/2 as everyone else had correctly, and given bewings quoted radius of Kerbin, finished the calculatino

[ElWanderer]

The equation for orbital period is 2.π.√(a^3 / mu)

a is the semi-major axis. Bear in mind that is the average of the apoapsis and periapsis from the centre of the orbit I.e. if you're orbiting Kerbin add 600km to the average of the displayed Ap and Pe.

mu is the gravitational parameter, formed of the planet's mass times Newton's gravitational constant (it's often written as GM instead of mu). You can find this value on the wiki page for each planet.

You can rearrange the equation to get the semi-major axis you will need for a given orbital period a = ((mu.period^2)/(4.π^2))^1/3

You need to know the time until the other craft gets to the intercept point to use this, though.

I have made extensive use of OhioBob's page: http://www.braeunig.us/space/orbmech.htm

I needed to calculate this for my rendezvous program (kOS script), but in-game it's easier just to plot a manoeuvre node at the intercept and watch the intercept markers move as you add/remove delta-v to the burn.

[bewing]

OK, so the first thing to keep in mind is that the Kerbin Ap is measured from the surface, and that's at 600000m -- so you need to add that back in.

With the orbital period formula that Spricigo linked to, you should get a number of about 32.64 minutes for a circular orbit with a 100km altitude. (You can get the gravitational parameter from the wiki: http://wiki.kerbalspaceprogram.com/wiki/Kerbin)

Since one of the ships is 10 degrees ahead, that means it's 1/36th of an orbit ahead -- or .91 of a minute ahead.

So now we want to figure out an orbit with a period of 33.55 minutes -- because that's how long it needs to take both ships to get back to where ship#1 is right now.

So, inverting the formula, SMA = cube root of (GravParam * t * t / (4 * pi * pi) ) = 713019 meters. So that's the average of the Pe and the Ap. So since we want the Pe to be 100000 (+600000), that means we need to set the Ap of the ship in front to 126038 (+600000) meters.

 

[Spricigo]

 

5 hours ago, Maltman said:

Can you give me an example of the type of maths required for the problem?because I am having trouble understanding the application of them. Let me give you an example . 

Also is there a magical tutorial that explains all this somewhere? 

Actually you don't need number crunching to figure out rendezvous maneuvers.  All you need to know is higher orbits have longer periods. 

If we start with craft A 10° ahead of craft B in the same orbit, we need to raise craft A orbit toet craft B catch up.  ( lowering craft B orbit do the same)  Once we reduce the angle enough we bring back craft A for the same orbit. 

 

Anyway,  Scott Manley  can explain it much better.