# Optimal time and coordinates of intersection of a spacecraft with an object on orbit

Spacecraft and an object are rotating on orbit. The spacecraft is controlled via the velocity vector change, may change its velocity to intersect another object.

Needed to find the time and coordinates of meeting, as both of them are dynamic objects. The optimal solution- minimum velocity change.

Ideas?

Another question: What is the maximum value of the delta V? It depend on the fuel, according to Cialcovski equation. Obviously, its value should be restricted by something else. Isp? And how?

• What is an 'SC'? Have you tried to do any calculations yourself? Anything? It's better to post some of your own work first, then it would be easier for someone to see where you need help. Right now it looks like you just want someone to do a bunch of work for you. Welcome to stackexchange! Why not take a moment and take the tour? – uhoh Apr 15 '17 at 3:18
• SC-spacecraft. No, I need answer to second question and discuss ideas for first. I don't want someone solve it for me) – Sonya Seyrios Apr 15 '17 at 8:18
• OK so if it is a general problem - the two orbits can be arbitrarily different orbital parameters, you may be asking for quite a lot. What do you mean "meet"? Can they just pass by each other with a high velocity, or do you want them to attain the same orbit so they could dock? Can you at least list some of the reading you've done so far on the subject of orbital maneuvers? Even looked at other questions and answers here in SXSE for reference? There are many good answers here already that may give you an idea of how big of a question this really is. – uhoh Apr 15 '17 at 10:00
• Meet- just pass each other. I don want you to solve the problem. I'm just collecting Ideas. If you have good references, please share – Sonya Seyrios Apr 15 '17 at 10:04
• Oh, its better than 'Google your question' Thanks! – Sonya Seyrios Apr 15 '17 at 10:15

As far as I know, no analytical solution to the problem has been found, and it's normally obtained through numerical optimization - like the Simplex method.

First, let's take care of the Tsiolkovsky's rocket equation: delta-V depends on specific impulse, dry mass of the craft (without fuel), and amount of fuel (or wet mass of the craft.) Since the first two don't change during the flight, only the third is actually a variable, not a constant. Still, delta-V may convert to different change of trajectory, and different change of energy, depending on what maneuver and where is performed - in particular, look up Oberth Effect.

Now for your problem: you're asking for minimal burn that creates a fly-by encounter between two arbitrary initial orbits; encounter speed doesn't matter, the crafts don't need to dock to each other.

You will need orbital motion equations of both crafts - for one central body you can use the conic section formulas / orbit equation, for a more complex system you'll use hamiltonian mechanics to derive equations of motion in the non-uniform gravitational field. You'll also need a third orbit, a generalized parametric equation for one, which will be your transfer orbit.

Then you need to bind the three: write a set of equations where:

• the position of the target is the same as position of the craft in the transfer orbit, at a certain time (the encounter)
• the position of the spacecraft in the initial orbit is the same as in the transfer orbit (transfer burn).

Add an equation of delta-V of the departure burn: difference of velocities at the moment of transfer burn between the initial orbit and the transfer orbit.

Your set of equations is under-constrained: your free variables are time of departure, and departure delta-V (3 variables, a vector). This set of equations then is the subject to numerical optimization: finding such parameters that the magnitude of the departure burn is least - and where TIME IS REASONABLY CONSTRAINED!

This last part will be a big headache. That departure burn must occur before encounter, this is obvious. But the encounter largely depends on relative position of the two craft, and if their orbits aren't resonant (orbital period ratio isn't a simple fraction) one can always find a more optimal departure burn - if it's to happen enough millions of years away.

Simply, the best, most optimal transfer exists, which requires the craft to be in certain alignment to each other. Least delta-V to reach nearest point of the target orbit, and the target just happens to be there at the point of time when you arrive. Usually it won't - it will be in an entirely different place of its orbit. But if you wait long enough, you will be aligned to encounter it pretty close to that optimal point, and you will waste very little fuel vs the optimal for that adjustment. If you wait longer though, it will pass even closer, so you can do this more optimally. And so on, to infinity.

• Thanks. 1) The SC and object are on LEO. The orbits Keplerian elements are close to each other. 2) I can't wait too long. All this should be done in 20 days maximum. 3) About the delta V: The fuel mass depends on delta V, not vice versa. I need to know, what is the maximum delta V, that the SC may make? It can't be 10 km/s, even if fuel enough, yes?) Does it depend on Isp and how? – Sonya Seyrios Apr 15 '17 at 16:08
• @SonyaSeyrios: If the orbits are co-planar, that simplifies your calculation a lot. As for delta-V, the Rocket Equation, remembering $v_e = I_{sp} \cdot g_{0}$. With "enough" fuel delta-V can be anything, but the fuel requirements grow exponentially - your craft becomes unreasonably big and expensive fast. And the mass depends on delta-V the same way as delta-V depends on mass, it's an equation! – SF. Apr 15 '17 at 16:33
• With "enough" fuel delta-V can be anything? Ok, If the dry mass is 1000 and total mass is 5000 and Isp=340, the delta V may be even 5366 m/s. Is it possible? – Sonya Seyrios Apr 15 '17 at 16:37
• I'm talking about impulsive velocity change – Sonya Seyrios Apr 15 '17 at 16:39
• @SonyaSeyrios: check yourself. I'm getting 5362m/s. But take: Dry mass 1000kg, Isp = 70s, delta-V = 35 km/s. You're getting mass of fuel = 2.4 Earth mass. – SF. Apr 15 '17 at 16:46