Comet Intercept Tutorial

[theAstrogoth]

There are plenty of guides and tools for getting from one planet to another, but it’s a little more tricky to intercept a target in a highly eccentric orbit. Here, we’ll walk through the steps needed to calculate a transfer from an orbit around a starting planet (e.g. Kerbin) to a comet or asteroid in a solar orbit.

Part 1 – Doing the math

Step 1: Before we start…

In this tutorial, we’ll calculate a Hohman transfer from a planet (blue) with a circular orbit to a comet (green) with a highly eccentric orbit, as depicted in the figure below. We'll intercept the target at its apoapsis, since that's where it's traveling most slowly. This will cost us some extra Δv to get there, but requires much less Δv to match orbits upon arrival, saving us Δv overall.

It’s necessary to gather some information before we can perform some of the calculations below. We’ll need:

• The semi-major axes, a, of the starting orbit, the body you’re leaving, and your target
• The gravity parameters, µ, of all celestial bodies involved (e.g. Kerbin, the Sun)
• The size of the sphere of influence (SoI) of the planet you’re leaving

We’ll assume most orbits are circular to simplify things (that is, we assume that the apoapsis, periapsis, and semi-major axis are all equal). For situations where the target orbit is highly elliptic, such as that of a comet, you will also need:

• The apoapsis and periapsis altitudes of the target’s orbit
• The eccentricity, e, of the target’s orbit

You can calculate the semi-major axis and eccentricity of an elliptical orbit from the apoapsis and periapsis heights with the following formulas:

Note that all periapsis and apoapsis heights used here are distances from the center of a planet, NOT from the surface! You will need to add a body’s equatorial radius to the readout you see in-game.

For example, since Kerbin has an equatorial radius of 600km, an orbit with an altitude of 100km would have:

 

Step 2: Transfer phase angle

In this section, we’ll work out the timing of the trip: at what phase angle to depart and how long the transfer will take. If your target has a highly elliptical orbit, you can probably safely skip this section and assume a phase angle of close to 0°. However, if your target's orbit is more circular, it might be worthwhile to calculate a phase angle.

Since we’re using a Hohman transfer, we already know the periapsis and apoapsis of the transfer orbit, and we can calculate the orbital period, T:

Because the transfer orbit will start at the periapsis and end at the apoapsis, the total time elapsed will be half of the orbital period:

Now that we know the duration of the transfer, we can backtrack from arrival at the target to see what the phase angle should be at the beginning of the transfer. To do this, we’ll calculate the angular distance the target travels during the transfer. First, calculate the change in mean anomaly, which is the angular change of the orbit if it was circular, with the same semi-major axis (represented by the dashed green orbit in the above figure):

For circular orbits, the true anomaly, θ, is equal to the mean anomaly, so we have a phase angle:

For non-circular target orbits, we’ll need to do some more work to convert mean anomaly to true anomaly. We assume that our craft will arrive at the target’s apoapsis, so the target’s mean anomaly at departure is

Then, we have to solve Kepler’s Equation (which should be used with radians, not degrees),

to find the eccentric anomaly, E, at the start of the transfer. However, there is no analytical solution for E, so you have to use a numerical approximation. I came across a simple web tool that can do it for you, here.

Once you have the eccentric anomaly, you can then calculate the true anomaly, which will be equal to your phase angle.

 

It’s important to note that it’s very unlikely that your starting body will be 180° from the target’s apoapsis at the same time that it’s aligned with the target at this phase angle. You’ll want to get as close as possible to the calculated phase angle while still departing opposite the target’s apoapsis. A comet will be travelling very quickly at its periapsis, so big differences in phase angle don't necessarily correspond with big differences in departure time.

 

Step 3: Δv

Now, we’ll calculate the magnitude of the transfer burn.

We need to determine the excess velocity needed after escaping the SoI you’ll be leaving. First, let’s get the velocity, v, at the beginning of the transfer orbit. We can use the Vis-viva equation,

So, the velocity at the periapsis of the transfer orbit is

and since we assume its orbit is circular, the velocity of the starting body is

The difference between these is the hyperbolic excess velocity,

From the hyperbolic excess velocity, we can calculate the magnitude of the transfer burn using conservation of orbital energy,

To figure out what the Δv of the burn will be, we have to subtract the speed of the craft in the parking orbit.

Note that this is only the departure Δv. You can use a similar approach to calculate the Δv needed for arrival at the target.

 

 

Step 4: Ejection angle

With the ejection Δv calculated, we now need to determine at what ejection angle the burn will take place. We first need to calculate the eccentricity of the ejection hyperbola.

The angle from retrograde, β, is given by

and the ejection angle (from prograde) is then

 

Assumptions and pitfalls:

For the above plan, we assumed that both orbits are in the same plane. Since it’s likely that there will be some relative inclination between your starting body and the target, you will need to perform a plane-change maneuver during your transfer. These are most often performed either at the ascending/descending node or 90° before arrival at the target.

When we calculated the ejection angle, we simplified things by assuming that the SoI was large enough that the hyperbolic orbit was already traveling along its asymptote. This may introduce substantial error if the SoI is relatively small (e.g. Moho). To avoid this, one can calculate the true anomaly and flight path angle at the SoI.

One thing that was ignored above was the fact that the ejection from the starting body is not instantaneous. If a large amount of time passes in the starting SoI (between v₁ and v₂), it leads to inaccuracy of the ejection and phase angles.

 

Part 2 – Example mission to Billy-Bobbur-Bilgee-Jan 1

Here are some example calculations for planning an intercept with a comet in my KSP save with a wonderful name: “Billy-Bobbur-Bilgee-Jan 1.”

Given:

• Target apoapsis = 75,691,786,599 m
• Target periapsis = 6,877,938,363 m
• Parking orbit altitude = 100 km
• Kerbin’s equatorial radius = 600 km
• Kerbin’s semi-major axis = 13,599,840,256 m
• Kerbin’s gravity parameter = 3.5316000×10¹² m³/s²
• Kerbin’s SoI = 84,159,286 m
• Sun’s gravity parameter = 1.1723328×10¹⁸ m³/s²

Step 1:

          

Step 2:

          

Step 3:

          

Step 4:

          

 

Part 3 – Using flight planning tools

If you’re playing on PC, there are many mods and utilities available that can perform these calculations for you, which likely address some of the pitfalls discussed above.

Helpful mods/tools:

• Astrogator
• MechJeb
• Kerbal Engineer Redux
• KSP Trajectory Optimization Tool
• KSP Transfer Illustrator

Let’s compare the results of our calculations with those of my own web app, the KSP Transfer Illustrator.

Here are the orbital elements of Billy-Bobbur-Bilgee-Jan 1, taken from my savefile:

Semi-major Axis (m) 41284862481.571045 Eccentricity 0.8334029157804823 Inclination (°) 8.809621810913086 Argument of the Periapsis (°) 0 Longitude of the Ascending Node (°) 277.10321044921875 Mean Anomaly at Epoch (radians)  -0.2563119011155347  Epoch (s) 10654909.808515284

And here are some interactive 3D plots generated by the app (download and open in a web browser):

Interactive Ejection Plot

Interactive Transfer Plot

Results:

• Phase Angle: -87.95°
• Departure Burn: 1984.45 m/s prograde
• Ejection Angle: 115.51° from prograde

Not bad! The discrepancy in the Phase Angle result comes from the issue I mentioned earlier, and -87.95° was the closest we can get to -65.15° while still being lined up with the comet’s apoapsis.

Edited January 25, 2021 by theAstrogoth