# Phasing maneuvers; methods to travel between satellites with very similar circular orbits

If you imagine two satellites orbiting earth, each in a circular orbit in the same plane and at the same altitude, but varying in their angular position, the simplest way in my mind to travel from a rendezvous with satellite A to a rendezvous with satellite B is to either speed up or slow down such that after one orbital period, you arrive back at the same altitude, but at the angular position of the target. I believe this is also the most dV efficient method, but please let me know if I'm mistaken.

However, if you instead imagine traveling between, say, jupiter and an artificial satellite 5 degrees ahead of jupiter in the same orbit, this method would work but take something like 12 years despite the fact that the separation is not that great.

My question is what other options exist in this case. Surely you could accelerate greatly and just travel in a roughly straight line to the target, but this would be outrageously expensive. It seems to me that there must be a set of maneuvers that minimize both time and dV requirements given some arbitrary valuation for each (e.g. 1 year = 10pts, 1000m/s = 40pts, find maneuvers that minimize total score).

Edit:

I made an image that shows one kind of phasing maneuver I'm aware of that takes less than one period and works (at least for small phase changes).

• Do you ask for travel between satellites within the same orbital plane as well as at the same orbital radius?
– Uwe
Sep 26 '20 at 20:28
• Yes. I'll add that Sep 26 '20 at 20:37
• I think you are asking about a phasing maneuver, and the tradeoffs between how quickly it happens and how much total delta-v is required for the two impulses, and I think that that should be a pretty simple relationship to calculate. For example, if I orbit at 7700 m/s and I do a 7.7 m/s burn, I will have changed my speed by 1 part per thousand. If I want to change my position by 36 degrees or 100 parts per thousand of a circle, I'll have to wait for about 100 periods or 6.4 days before doing a second 7.7 m/s burn pointed in the other direction. Is that the kind of thing you are looking for?
– uhoh
Sep 26 '20 at 21:05
• Well... you can either have a quick phasing, which will cost lots of delta-v, or you can have a low delta-v one, which will take a long time. You can't magically have both. The calculation is rather straight-forward utilizing the vis-viva equation (as long as the orbits are co-planar, of course). Sep 26 '20 at 22:45
• Yes, that term "phasing" was unknown to me with respect to orbital mechanics. That seems exactly like what I'm looking for. Sep 27 '20 at 14:57

These manoeuvres typically come in two flavours:

1. Manoeuvres adjusting the orbital period, meeting up with the target at the origin after some number of revolutions.
2. Manoeuvres entering a transfer orbit, encountering the target at some different part of the orbit ("point and thrust" in its extreme case)

This is a three-variable solution space, with phase angle, delta-v and transfer time as its axis.

For any given phase angle, there are always two extreme solutions, one requiring zero delta-v but infinite transfer time, and one requiring zero transfer time but infinite delta-v. This motivates finding a compromise, since both time and delta-v is often only available in a quantity less than infinite.

Definitions:

To keep things simple, I'm assuming a two circular orbits with equal radius. These have unit radius, unit velocity. Scale to your actual use case.

Ahead in the orbit is the positive direction of the phasing angle.

These rely on the neat fact that if you do some impulsive manoeuvre, you will eventually reach the exact same location again, and will be able to "undo" the manoeuvre to get back the original orbit. Phasing using this method relies on the intermediate orbit having a different orbital period.

As the rendezvous point is fixed, the transfer time is quantised, and is on the form:

$$t = 2\pi n - \theta$$

Where $$n$$ is a whole number and $$\theta$$ is the phasing angle.

## increased orbital period

This helps increase the phasing angle. Useful when it's already close to $$2\pi$$, and you want it to roll around to 0. That is, a "trailing" target.

This can be done by raising the apoapsis, in fact, the optimal way of increasing orbital period this is by raising the apoapsis.

Adds the constraint $$n \geq 2$$.

We will be in the transfer orbit for $$\lfloor{t\rfloor}$$ orbits, meaning the orbital period of it is $$\frac{t}{\lfloor{t\rfloor}}$$. Clearly, this approaches 1 when $$t$$ is very large.

Orbital period is calculated from the semi-major axis by $$2\pi\sqrt{a^3}$$, so from that we can find the needed apoapsis height:

$$r_A = 2\sqrt[3]{\left(\frac{t}{\lfloor{t\rfloor}}\right)^2} - 1$$

From the vis-viva equation, the total delta-v is:

$$\Delta v = 2\sqrt{2 - \frac{1}{\sqrt[3]{\left(\frac{t}{\lfloor{t\rfloor}}\right)^2}}} - 2$$

The worst case scenario for this is when the target is slightly leading, and one tries to reach it as fast as possible by changing into a transfer orbit with an orbital period close to 2.

$$\Delta v_{worst} = \frac{4}{\sqrt(3)} - 2 \approx 0.31$$

That would however be a very misguided use of resources, since the following method should have been used instead in this case:

## reduced orbital period

This helps reducing the phasing angle. Useful when it's already low, that is, a "leading" target.

This can be done by lowering the periapsis. Unfortunately, lowering the periapsis isn't always the optimal way of reducing orbital period. This follows as a corollary to the optimality of increasing apoapsis to increase orbital period.

Additionally, there's often the constraint that some lower limit exists for the periapsis, since there's often a planet in the way. These two complications makes analysis slightly more difficult.

Nevertheless, one may still use a similar strategy as in the case with the increased orbital period, this time lowering the periapsis into an elliptical transfer orbit. The orbital period is now $$\frac{t}{\lceil{t\rceil}}$$. Unlike the previous strategy, $$n$$ is not required to be 2 or greater, but as an additional constraint, the orbital period is not allowed to be lower than $$\frac{1}{2\sqrt{2}}$$.

$$\Delta v = 2 - 2\sqrt{2 - \frac{1}{\sqrt[3]{\left(\frac{t}{\lceil{t\rceil}}\right)^2}}}$$

You should generally check the delta-v for both the increased and reduced orbital period strategies, as they have different costs but similar transfer times (except in the $$n = 1$$ case where only the reduced period solution may exist). The best choice depends on the phasing angle.

## reduced orbital period, improved

The most efficient way of reducing orbital period is to first lower the periapsis, and then circularise. In other words, a Hohmann transfer (bi-elliptic transfers are not useful for this particular problem).

This increases the manoeuvre from two impulses to four:

1. Start Hohmann transfer to towards lower circular orbit
2. Break into lower orbit.
3. Start Hohmann transfer back.
4. Enter original orbit at target phasing angle.

Unlike the two previous strategies, transfer time is not quantised, since the transfer back into the original orbit can be done at an arbitrary phase angle, because the intermediate orbit is circular.

The free variable here is the periapsis radius ($$r_P$$). Once it's selected, the transfer time and delta-v cost follows. It's a trade-off between slow but cheap transfers at a higher periapsis radius, and faster but more expensive transfers at a lower periapsis radius.

Delta-v cost, per usual Hohmann calculations (notably, it's independent of phasing angle):

$$\Delta v = 2\left(1 - \sqrt{2 - \frac{2}{1 + r_P}} + \sqrt{\frac{2}{r_P} - \frac{2}{1 + r_P}} - \sqrt{\frac{1}{r_P}}\right)$$

Transfer time is a bit more involved. The phase angle is reduced both in the two Hohmann transfer legs, and during the time spent in the lower circular orbit:

1. Transfer legs. The orbital period of the full elliptical orbit is $$2\pi\sqrt{\frac{(r_P + 1)^3}{8}}$$. This then reduces the phasing angle by $$2\pi - 2\pi\sqrt{\frac{(r_P + 1)^3}{8}}$$.
2. Circular orbit. While the angular velocity of the original orbit is 1, it's higher in the lower orbit. This leads to a reduction of the phasing angle of $$\frac{1}{\sqrt{r_P^3}} - 1$$ per unit time.

Total transfer time is therefore:

$$t = 2\pi\sqrt{\frac{(r_P + 1)^3}{8}} + \frac{\theta - 2\pi + 2\pi\sqrt{\frac{(r_P + 1)^3}{8}}}{\frac{1}{\sqrt{r_P^3}} - 1}$$

At very cheap, long transfer time uses of this (that is, the time spent in the inner circular orbit is much higher than the Hohmann transfers), the equations can be simplified:

$$\Delta v \approx 2\sqrt{\frac{1}{r_P}} - 2$$

$$t \approx \frac{\theta}{\frac{1}{\sqrt{r_P^3}} - 1}$$

# Meeting up at a different location

## Sensible direct trajectories

TODO

(also quite an important TODO for fast phasing at reasonable cost. It's what's illustrated in the image in your question)

(outline: Pick some target $$r_P < 1$$ and $$r_A > 1$$, so the spacecraft goes through a slower of faster arc before reaching $$r=1$$ again.

The transfer time is not straight forward, and I don't quite have the equations for picking an optimal $$r_P$$ and $$r_A$$ at your delta-v budget.

The cost is however given by:

$$v_{horizontal} = r_p \sqrt{\frac{2}{r_P} - \frac{2}{r_P + r_A}}$$

$$v_{vertical} = \sqrt{2 - \frac{2}{r_P + r_A} - v_{horizontal}^2}$$

$$\Delta v = 2\sqrt{v_{vertical}^2 + (v_{horizontal} - 1)^2}$$

:outline end)

### A class of such trajectories, nadir catch-up burns.

These are the direct phasing manoeuvres that meet up on the exact oppsite side of the orbit. These are easy to calculate, hence why they are included here before completing this section.

Given some delta-v spent burning directly towards the parent body, the orbit's semi-major axis is given by $$a = \frac{1}{2 - \left(1 + \left(\frac{\Delta v}{2}\right)^2\right)}$$

And the periapsis by $$r_P = a - \sqrt{a^2 - a}$$

The transfer time in the fast arc is then:

$$t = 2 \sqrt{a^3} \tan^{-1}\left(\frac{a}{(a - r_P) \sqrt{r_P (2a - r_P)}}\right)$$

From which we can find the phase angle:

$$\theta = \pi - t$$

## Torchships

In the extreme case, where available acceleration is arbitrarily high and delta-v is much higher than the orbital velocity, the problem approaches a simple relationship to distance:

$$t = \frac{2\sqrt{\sin(\theta)^2 + (1 - \cos(\theta))^2}}{\Delta v}$$

## Low-thrust phasing

Certain spacecraft, such as those propelled by ion engines, have very limited thrust. They can therefore not perform impulsive burns, and instead follow constant thrust spirals.

In such low-thrust spirals, the spacecraft is always in an approximately circular orbit. The delta-v cost between two such circular orbits is remarkably simple:

$$\Delta v = v_1 - v_2$$

That is, simply the difference between the orbital velocities.

A low-thrust spacecraft would spiral up (or down) until half the phasing angle has been reached, and then return in a similar spiral transfer to cover the other half.

The relationship between transfer time and phasing angle is here highly non-linear. The change in phasing angle at any instant is $$\theta' = \omega - 1$$, where $$\omega$$ is the current angular velocity, which itself is given by $$\omega = \frac{v}{r}$$, where $$v$$ and $$r$$ is the current velocity and orbital radius.

Integrating over time is then:

$$\Delta \theta = \int 1 - (1 - at)^3 dt$$

Where $$a$$ is acceleration. With the base orbit as the origin, it yields the result:

$$\Delta \theta = \frac{\frac{t^4}{4} - t^3 + \frac{3t^2}{2}}{a}$$

(positive $$a$$ is here spiralling outwards)

The practical result of this is that spiralling outwards is the optimal for phasing angles up to ~80 degrees in the trailing direction, while all other phasing angles are achieved faster by spiralling inwards:

• Fantastic answer so far, I'm looking forward to reading the completed fast phasing section. One thing I'd suggest is adding some definition of variables where it may not be apparent to some, such as "a" being semi-major axis in the equation for orbital period. Sep 30 '20 at 15:40
• Thanks for this excellent answer! I posted a related question about low-thrust phasing for general elliptical orbits a couple of days ago and I was wondering if you could please help answer that if possible. I've finally saved up enough rep to comment here :) Jul 12 at 16:47

I have never computed the energy efficiency but my experience with Kerbal Space Program is that for catching stuff in basically your orbit but a bit ahead of you by far the easiest answer is to burn towards the planet. Likewise, for something a bit behind you burn away from the planet. In both cases you will make a matching burn half an orbit later. Note that the required burns go up rapidly as the angular distance goes up, the relationship is very non-linear but I do not know the formula.

Otherwise, you'll have to use a staging orbit.

• The interesting question here is whether a nadir burn is the most efficient strategy, or if some angle between nadir and prograde/retrograde is more efficient. Oct 1 '20 at 13:35
• Towards the planet (radial) and not prograde/retrograde ? Are you sure ? Oct 1 '20 at 15:37
• @Antzi It's a perfectly valid strategy for fast phasing (less than one orbit). Oct 1 '20 at 21:11
• @Antzi Yes, I mean towards/away from the planet. You get intercept in I believe half an orbit. I've done it many, many a time rescuing Kerbals in low Minmus orbit. (Don't chase one too far ahead, though, low Minmus orbit is very low and it's easy to put your periapsis in the terrain!) Oct 2 '20 at 2:17