# Tutorial: How to meet objects in orbit

This tutorial describes the process of meeting an object that is in orbit around a larger body. The object can be a space station, a piece of debris, a stranded kerbonaut, or even a planet orbiting the sun or a moon orbiting a planet. I will describe the case of rendezvous with another ship in the orbit of Kerbal, but the instructions should apply to all those other cases (except that rendezvous with a massive object such as a moon or planet is easier, because unlike a space ship they will capture you in their gravity well once you get reasonably close).

It isn't really a step-by-step tutorial, it is more of an explanation for how to solve the problem of meeting objects in space. The idea is that you can read this, and then you'll be able to work out the steps yourself.

# Basics/Set-up

Let's say there is an object in stable orbit (everything becomes much messier if the target orbit intersects the atmosphere) around something. You want to meet this object.

First you need to be orbiting the same body as the target. There are ways of achieving this, they are described elsewhere so I won't go into them.

Second, you need to be on the same orbital plane as the target. It will be very difficult to rendezvous a target on equatorial orbit when you are on a polar orbit! Again, the process of aligning orbits is described elsewhere, so I will leave that as an exercise to the reader.

## The phase angle problem

Now, let's say you've matched orbits with your target. You have the exact same orbital trajectories. But, there is a problem - when you are over the western hemisphere the target is over the eastern hemisphere, and you are literally a world apart at all times! Of course, the problem is that you are at a different phase than your target.

Naturally, if you want to be at the same place at the same time with a target, it is necessary to have the same orbit. However, this is not enough: You must also be at the same part of the orbit at the same time. It is relatively simple to alter the shape of the orbit: You fire thrusters for so long at such point, and it immediately bends the rest of your orbit. However, there is no direct way to make you "advance" along your orbit without changing the orbit itself - if you try to thrust forward you will change the radius of your orbit.

In physics, periodic movement is described such that there is a period (how long it takes to complete one iteration), and there is also a parameter called "phase" which describes what part of the current iteration is complete. The phase is usually 0, or a number larger than 0 and smaller than the period - you can think of it as measuring how much of one period is complete at an arbitrary starting point.

### Phase angle of orbits

In the case of orbits, the natural quantity to measure is angle of rays connecting the orbiting object and the center point of the orbit. In this case, we are not particularly interested in the individual phases of our orbiting objects, since the orbital phase only matters to us as a relative term. Therefore, we simply look at the angle between the target and our own ship: this is the phase angle.

The phase angle will be something between 0 and 360 degrees, or 180 and -180 degrees depending on how you measure it. I prefer 180 and -180, because the procedure for meeting targets that are ahead of you (positive angle) is slightly different than meeting those behind you (negative angle).

### Phase angle and orbital period

We now have a way of quantifying orbital phase geometrically, using the phase angle. But recall that I said: In physics, the phase shows how far along a period you are - therefore, the phase and period must have the same units. So your phase is actually measured in days, hours, minutes and seconds, just like your orbital period, but your phase angle is an angle (usually in degrees).

Observe that phase (in seconds) and phase angle are closely related. If you measure the phase angle $\phi$ to be between 0 and 360 degrees (or if you use -/+180 like me, just add 180 to it), and the phase $t_{\phi }$ is in seconds, then ${\frac {\phi }{360}}={\frac {t_{\phi }}{T}}$ (where $T$ is orbital period in seconds) will always hold. For this reason, I won't take too much care about which term I use when.

# Synchronizing orbital phase

Having established the concepts involved, we go back to the initial problem: You are in the same orbit as your target, but have a large phase angle. How to reduce the phase angle to zero? (when it is zero, you would right on top of your target)

It is actually not necessary to be in the same orbit first. In fact, manipulating the phase is easier and more efficient if you do it while you are still doing your transfer - for example, if you are on Hohmann transfer orbit which is tangent to the target orbit, you're already halfway there. So, if you are in the same orbit, the first thing to do is get out of the orbit into another one. That said, I will still describe the problem as if you are already in the same orbit, and now want to synchronize, since that is simpler to explain.

## Choosing a transfer orbit

The basic principle is this: As orbits get bigger, their periods increase. Higher is slower - orbit radius is inversely proportional to orbital velocity. So if you are "ahead" of your target and want it to catch up, you should go into a higher orbit (one with a longer period) and hang out there for a while, and come back. If you need to catch up your target, you go into a lower orbit (with a shorter period). This is why I said earlier -/+180 is better for measuring the phase angle: If you measure from you to target (your angle minus target's angle), positive means you go up and negative means you go down.

I call this new, temporary orbit a transfer orbit, and this is because the Hohmann transfer orbit happens to be convenient for this purpose. However, keep in mind this is a bit different from what is normally meant by "transfer orbit" (orbit for transferring from one altitude to another). In this tutorial, I will say "transfer orbit" when I mean an orbit that you stay in while waiting for your phase to sync up with your target.

There are orbits besides a Hohmann elliptical that can serve this function, possibly better than the Hohmann. But generally, whatever orbit you choose must have two properties:

• Have an orbital period different from your target's orbit
• Be easy to transfer back to your target's orbit from

Besides having a longer/shorter period, a key property of your transfer orbit is by how much it is longer. A larger period difference requires using more fuel, while a very small period difference will make you wait for a long time for the sync.

If your current period is $T$ and your current phase difference is $t_{\phi }$ . The quickest thing to do would be to thrust and make a Hohmann elliptical with a period of $T+t_{\phi }$ . After one revolution in this orbit, you will arrive back at your starting point at almost the same time as your target. You must now quickly burn retrograde to pull your orbit back to match the target's orbit again. You are now back to being in the same orbit as your target, but your phase is much smaller. You can now repeat the whole process to get as close as you like.

If $t_{\phi }$ is large, you will find that you need a lot of fuel to do the above. You can save your $\Delta v$ by instead using a smaller orbit. For example, you can use a transfer orbit with a period of $T+t_{\phi }/3$ . Every time you complete this orbit, you phase with your target will decrease by a third. So you will now wait for 3 revolutions, and then burn retrograde. Since this orbit is much smaller, you can save a lot of fuel in this way.

Let's go back to an orbit with period $T$ and your phase difference is $t_{\phi }$ . You are considering a transfer orbit with period $T+\Delta t$ . This means you will wait $(T+\Delta t){\frac {t_{\phi }}{\Delta _{t}}}=t_{\phi }+T{\frac {t_{\phi }}{\Delta t}}$ . $t_{\phi }$ is constant and smaller than $T$ , so the second term dominates: The ratio $\rho ={\frac {t_{\phi }}{\Delta t}}$ determines how long you will wait.
Now, thrusting prograde/retrograde basically injects however much $\Delta v$ you spent into your orbital velocity, which determines your period. So, for circular-ish orbits, we can say $T\approx {\frac {2\pi r}{v_{o}}}$ (where $r$ is orbit radius and $v_{o}$ is orbital velocity). If $\omega ={\frac {v}{r}}$ , then after burning some $\Delta v$ your $\omega$ (angular frequency) will be ${\frac {v+\Delta v}{r}}$ and since $T={\frac {2\pi }{\omega }}$ , we get $T+\Delta t\approx {\frac {2\pi r}{v+\Delta v}}$ . So there is a roughly linear relation between how much $\Delta v$ you burn and how many seconds of $\Delta t$ you shave off. If you burn twice the fuel, you can get done in half the time - so it's a question of how much you value your $\Delta v$ and how much you value your time.