[Orbital Mechanics] Kepler's Third Law - Launching Several ComSats in One Mission

[wmheric]

The context of this guide is based on the real-world use of communication satellites put in geosynchronous / geostationary orbits. In short, a geosynchronous orbit is an orbit with a period the same as the rotational period of the Earth; a geostationary orbit is a special case of a prograde synchronous orbit in which the orbit is directly above the Earth's equator. If our focus is solely on equatorial orbits in the same direction as Earth's rotation, the two terms basically refer to the same thing. Having satellites in a geostationary orbit is a great advantage for transmitting signals to and from the ground because the pointing of an antenna does not require further adjustments once it is set at the time of the installation. (Say the large dishes placed on the roof of skyscrapers; and if you haven't noticed, the pointing angle has something to do with the latitude at your location!)

In the current vanilla version of KSP, communication coverage is not a concern because you can control a probe, whether manned or not, from any location. The mod RemoteTech 2 (RT2) offers a more realistic situation where it is possible to control an unmanned probe only when there is a proper link to the KSC. Scott Manley has done

on how to deploy a network of three satellites in the configuration of an equilateral triangle in the Keostationary orbit; the trick is to put the probe carrier into a 4-hour eccentric orbit with its apoapsis at the Keostationary altitude and deploy a probe at each apoapsis pass. A crucial part of the maneuver is knowing the orbital period of your satellite. With the help of info panels provided by mods like Engineer Redux and MechJeb, this is rather straightforward. On the other hand, vanilla KSP does not currently provide a simple value readout.

The purpose of this guide is to introduce the Kepler's Third Law of Planetary Motion, which will be applied as a method to set up the 4-hour transfer orbit. We will then move on to talk about setting up a network in a regular polygon configuration in general. Although the Keostationary orbit is used as an example, the idea extends to the scope beyond circular and equatorial orbits. The same maneuver can be employed whenever you are trying to deploy a number of satellites sharing the same orbit with an equal time lag between them.

This guide will be divided into 4 parts. The first part will be an introduction to the law and some background information. Next, we will get to the practical example of getting communication satellites set up. In particular, we are using the 4-hour transfer orbit in Scott's demonstration. We will then move on to go through some more calculations and explore different cases of the application. It might get a bit mathematical at this point, but the goal is to try to understand it and not follow formulas blindly without knowing how they actually work (so this is not a step-by-step guide). Finally there will be some points to note when you are doing this in-game.

1) Background

The Kepler's laws of planetary motion are three empirical laws established in the early 17th century that describe the motion of planets around the Sun. These laws are empirical because they were published using data from astronomical observations, not being derived from first principles. Our focus here is the 3rd law, and let's go straight to the definition:

The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

All closed orbits are ellipses (a circle is just a special case of an ellipse!); the law says that if we measure the orbital periods P of all the planets, square the numbers, and at the same time, we measure the semi-major axes a of their orbits, cube the numbers, then the ratio of these squares and cubes between two planets will be the same! Mathematically, we'd write

where k is a constant (some fixed number), or, if we forget about the k,

Now, if we do a logarithmic trick (astronomers use it all the time, but I don't want to dive in the Math here) to the first equation we can get

which is the equation of a straight line with a slope of 3/2, or 1.5. (What line? If you have no idea what is going on here, don't worry, keep moving on.) Let's make this into a picture for the Solar System:

The data points are not trimmed by hand -- they are the actual values (from Wikipedia) for the planets, and it happens that all planets (and minor-planets too) (Pluto is not a planet, deal with it) lie so close to the red line that the discrepancy is virtually invisible! This shows that the Kepler's Third Law is a very good description of the reality. If you know the math about straight lines, you may have already noticed that the line passes through the origin (0, 0) while there is a y-intercept term (1/2 log k) in the equation. This is from the careful choice of units so that P is represented in years (~orbital period of the Earth) and a is represented in AU (~average Earth-Sun distance); in general, the line does not pass through the origin -- but its slope is always 3/2. In fact, if we consider the Kerbol System and use SI units, this is what we get:

I guess there is no need to explain how beautifully things fit together again.

The statement of the law concerns the orbits of the planets in the Solar System, but it is not only limited to that. We can apply it to the satellites orbiting Kerbin. In fact, it can be applied everywhere, as long as the central body is the gravitationally dominant one (e.g., the Kerbol System, the Jool System, an artificial satellite system around Kerbin). (See Section 4 for more on this.) This means that we can apply the same period-semi-major-axis relation to the satellites we launch, as in the application below.

2) Application

Before you perform the transfer, there are four important points to know:

• the Keostationary orbit has a semi-major axis (radius) of 3468.75 km;
  
• the orbital period is the same as the rotation period of Kerbin, i.e., 6 h;
  
• the radius of Kerbin is (as in a perfect sphere, ignoring hills and mountains) 600 km;
  
• apsides are represented in-game as altitudes from the planet's surface (IMO there should be a way to switch between the two representations...), so the apsides will read 3468.75 km - 600 km = 2868.75 km.
  

For three satellites forming an equilateral triangle, this is what we are looking for:

Recall that our goal is to set up a 4-hour orbit with its apoapsis at an altitude of 2868.75 km. If you already have some experiences on Hohmann transfers, it is fairly easy to burn prograde at a low-Kerbin orbit to achieve such an altitude -- the challenge here is to get the period to be 4 hours at the same time. To do this, we have to figure out what the periapsis is:

The sum of the periapsis and the apoapsis is equal to two times the semi-major axis a; in other words,

To find a, we use the ratio in the second equation:

and we have a = 2647.1517 km. The periapsis is then at r_p = 2 * 2647.1517 km - 3468.75 km = 1825.553 km. So there you go, the altitude at periapsis is h = 1825.553 km - 600 km = 1225.553 km.

3) Further Calculations

The 4-hour transfer orbit works only for a 3-satellite configuration. What about a network in a regular polygon in general? In an equilateral triangle, the satellites are 120 degrees (1/3 of a full circle) apart, so we aim for a transfer orbit of a 4-hour period -- wait, shouldn't we have used a 2-hour orbit (1/3 of 6 hour)? Let's find out what altitude we need for that:

which gives a = 1667.6011 km, and r_p = 2 * 1667.6011 km - 3468.75 km = -133.548 km. Now you see why: a 2-hour orbit requires a negative periapsis and it is impossible! Do not get confused, it is fairly easy to get into a 2-hour orbit around Kerbin; for example, a circular orbit with radius 1667.6011 km (well beyond the atmosphere) will have a period of 2 hours. The 2-hour elliptical transfer orbit here is impossible because we require its apoapsis being at 3468.75 km. In other words, if you need a period of 2 hours, you just cannot have the apoapsis that high up. Or, in yet another words, if you need an apoapsis that high up, you cannot have a period as short as 2 hours.

In fact, there is a lower limit on how short the period can be if you fix the apoapsis at a certain altitude, if we do a little bit of math. Suppose we are transferring to a circular orbit of radius r_a, then the apoapsis of the transfer orbit must also be at r_a, and we have

To harvest some meaning out of this, consider two extreme cases:

1. r_p = r_a, which is the trivial case where your "transfer" orbit is just what you want, so what do you expect? f = 1;
   
2. r_p = 0, which is in principle a straight line (or an ellipse with a very, very small, tiny minor-axis) as the "orbit", and when the parent object is a point-mass, here f = 1/(2 * square root(2)), or about 0.35.
   

Therefore, no matter how hard you try, it is impossible, even in principle, to set up a transfer orbit with its period less than ~35% of that of the circular synchronous orbit. This shows how close the 2-hour orbit solution is -- 2/6 ~ 33%!

Now back to the transfer problem. As we cannot use a 2-hour orbit, we resort to the closest possible one -- a 4-hour orbit. Strictly speaking, this puts the satellites with an angle of not 120, but 240 degrees apart. The second satellite being deployed is 240 deg "behind" the first one, and the third one is 240 deg behind the second one, so the third is 480 deg behind the first one; but then 480 deg is one full circle (360 deg) plus 120 deg, so in another perspective, the third is only 120 deg behind the first one (but it doesn't really matter -- we have an equilateral triangle anyway). Let's consider another configuration, where you are trying to set up 6 satellites in a regular hexagon.

If you need an apoapsis that high up, you cannot have a period as short as 2 hours ...

let alone 1 hour. The next choice is 5 hours. It should become kind of automatic for you to plug the numbers in at this point:

and get a = 3071.7474 km, or r_p = 2674.745 km. Since the orbital period of the transfer orbit is 5 hours, the carrier probe will return to the apoapsis and the previous satellite deployed will have travelled a bit short of a full circle (300 deg) so the next satellite will be 300 deg behind, or more conveniently this time, 60 deg ahead. Of course there are choices (actually, infinitely many choices in principle) greater than 6 hours; you just have to wait for the carrier probe to come back at the same position to deploy another satellite. Our choice here is to try to minimize the overall mission time, as well as the fuel usage by avoiding going into a higher orbit.

To extend this into a general context: it doesn't matter which planet/moon/object you're orbiting, it doesn't matter what the synchronous period of the object is; As long as you are comparing two orbits in the same SoI (i.e., same source of gravity), what you need are only the required semi-major axis a_sync and the fraction f depending on how you want the satellites to be separated, you can derive the altitude of the periapsis for your transfer orbit:

What's more, it doesn't matter how eccentric the orbit is (provided you don't go inside the atmosphere or even crash onto the surface ...) -- perhaps you want to set up a Molniya orbit with three satellites. The same holds:

Just bear in mind that apsides are shown in KSP as the altitude from the surface:

or if you want a single formula to work with:

4) Points to Note

Knowing how to deal with the maneuvers, there are some important points to note when doing it practically:

• Instead of setting up the transfer orbit from a low-Kerbin orbit, you can burn straight into the required apoapsis from launch to deploy the first satellite, circularize it and switch back to the carrier probe and bring up its periapsis, so that you can save some time, but you need to be quick.
  
• In RT2, Communotron 32 has a range of 5 Mm (5000 km), which is the highest among the omni-directional antennas. Yet in a equilateral triangle configuration, the separation between two satellites is 6008 km, which means that the satellites can talk to each other only by dish antennas, so be sure that you bring enough onboard. On the other hand, you can save some by pointing those on two of the satellites towards Kerbin, but you have to shorten one of the sides of the triangle so that the two lie within the cones.
  
• Your satellite will depend solely on its battery (well, the radioisotope generator is quite late in the tech tree if you're doing Carrer Mode, and who needs it when you can just bring enough battery) in an eclipse when Kerbin is blocking the Sun, so make sure you have enough to survive the eclipse. Also, bring slightly more for the occasions where you happen to find Mun in the way when you just come out of Kerbin's shadow, feeling relieved.
  

[EtherDragon]

Very nice tutorial!

I am working on a full YouTube video lesson on Kepler's Laws of Planetary Motion - and in doing so derived a very useful equation:

r = 1 / (2*sqrt(d^3 / t^3))

r is the portion of the target's orbit that will be completed during a Hohmann Transfer Orbit given that:

d is the Semi-Major Axis of the Destination

t is the Semi-Major Axis of the Hohmann Transfer Orbit

[wmheric]

Very nice tutorial!

I am working on a full YouTube video lesson on Kepler's Laws of Planetary Motion - and in doing so derived a very useful equation:

r = 1 / (2*sqrt(d^3 / t^3))

r is the portion of the target's orbit that will be completed during a Hohmann Transfer Orbit given that:

d is the Semi-Major Axis of the Destination

t is the Semi-Major Axis of the Hohmann Transfer Orbit

Thanks a lot! Yes, the equation you derived is very useful for insertion into orbit at the right position and at the right time. It merits a post for discussion (got so many ideas on what to do next! )

Looking forward to your video

[WafflesToo]

Five hours if you want a six-satellite network