Maneuvers

2015-07-11

The six words you never say at NASA: “And besides — it works in Kerbal Space Program.” – xkcd, cartoonist and NASA roboticist

In this chapter, we will see how to change from a circular orbit to another.

Pro-/retro- grade burn

The vis-viva equation (*) tells us:

$v^{2} = G M (\frac{2}{r} - \frac{2}{r_{a} + r_{p}})$

For example, if we are at an apsis (apo- or peri-) and want to rise the opposite point, we need to speed up (burn prograde), and slow down to decrease it; the formula above tells us how much so.

A satellite on the blue orbit can switch to the red one by burning prograde (speeding up) at

B

; conversely it can switch from the red orbit to the blue one by burning retro grade at this same point.

When searching for good trajectories, we are interested in saving propellant. According to (*), this is the same as saving for $Δ v$ (althgouh proportionally). If $r$ is the apsis where the burn is performed, $r_{0}$ the opposite apsis before the burn and $r_{1}$ after:

v = - v_1 - v_0 -
1 v = - - -

Hohmann transfer

Now, assume we are in a circular orbit of radius $r_{0}$ and want to do a simple transfer to a circular orbit of radius $r_{1}$ . During a Hohmann transfer, we first raise our apoapsis to $r_{1}$ and then the periapsis (from the new apopsis).

We first switch from the blue orbit to the green one by burning at

B 1

and then from the green one to the red one by burning at

B 2

$\frac{1}{\sqrt{G M}} Δ v = | \sqrt{\frac{2}{r_{0}} - \frac{2}{r_{0} + r_{1}}} - \frac{1}{\sqrt{r_{0}}} | + | \frac{1}{\sqrt{r_{1}}} - \sqrt{\frac{2}{r_{1}} - \frac{2}{r_{0} + r_{1}}} |$

We can multiply both hands by $\sqrt{r_{0}}$ and set $x = \frac{r_{1}}{r_{0}}$ to get a simpler expression:

$\underset{α}{\underset{⏟}{\sqrt{\frac{r_{0}}{G M}}}} Δ v = | \sqrt{2 - \frac{2}{1 + x}} - 1 | + | \frac{1}{\sqrt{x}} - \sqrt{\frac{2}{x} - \frac{2}{1 + x}} |$

Note that decreasing an orbit by half costs about as much as the converse (doubling it), but dividing it by four costs twice as much as the converse.

Bi-elliptical transfer

The idea is to use three burns instead of two.

During a bi-elliptical transfer, we use two intermediate orbits (green, then violet); the idea is that it will be easier to raise the green periapsis from a higher apoapsis

1 v = - - 1 -
+ - - -
+ - 1 - -

Again, we set $x = \frac{r_{2}}{r_{0}}$ and $y = \frac{r_{1}}{r_{0}}$ and:

_ v = - - 1 -
+ - - -
+ - 1 - -

Inclination change

Remember, we are only considering circular orbits. The formulas and derivations below only make sense for circular orbits. We advise you to set your inclination in a circular orbit before any subsequent maneuver.

(Anti-)normal burn

Consider the orbital plane in which a satellite is moving. We are interested in the effect of an acceleration orthogonal to the plane (normal or antinormal). For this, we study the evolution of the velocity.

The satellite is heading towards

\vec{v}

and an acceleration is applied to it so that during a time

d t

, its velocity is changed by

\vec{d v}

As seen in figure (?), we can easily find the change in inclination:

= v
t = a t
= a t = v

Straight burn

Doing a $180^{\circ}$ inclination chage using a constant radial burn like exposed above yields a $Δ v$ proportional to the current orbital velocity $v$ : $Δ v = π v \approx 3 v$ . However, simply going retrograde until the speed is reverse only yields $Δ v = 2 v$ for the same result.

You can find another derivation of the cost in [1] [2].

A rotation of angle

θ

from velocity vector

\vec{v_{0}}

\vec{v_{1}}

is done in a straight change in speed

Δ v

We need to compute $Δ v$ for given $v = | \vec{v_{0}} | = | \vec{v_{1}} |$ and $θ$ . Because the triangle is isosceles, the altitude and the median from $P$ are one so:

$Δ v = | 2 v \sin \frac{θ}{2} |$

For $θ = π$ , we get $Δ v = 2 v$ which is the expected result.

An inclination change of about

30^{\circ}

already costs half the orbital speed;

60^{\circ}

costs as much as the orbital speed.

Bi-elliptical inclination change

Whatever the method used for the inclination change, the cost is proportional to the current orbital speed. Thus, it is more efficient to do such a maneuver at low speed (e.g. at apoapsis). /u/ObsessedWithKSP demonstrated a maneuver similar to the bi-elliptical transfer for a more efficient plane change [3] [4].

A formal derivation of the optimal inclination change has been published by /u/listens_to_galaxies [1] [2].

Starting in the blue plane on the circular orbit, the spacecraft first burns prograde in

B 1

to raise its apoapsis to

B 2

; once there, its speed is lower and it can proceed to the inclination change to the red plane effectively; finaly, it burns retrograde back in

B 1

to return to a circular orbit.

Radial in/out burn

TODO

Arbitrary burn

TODO

“[D]erivation of the [optimal] inclination change transfer orbit”,
2014, Reddit submission, /u/listens_to_galaxies,
https://pay.reddit.com/r/KerbalAcademy/comments/23v5wo
“Bi-elliptic inclination change transfer orbit”,
2014, imgur album, /u/listens_to_galaxies,
https://imgur.com/a/uM7PK
“How to do a bi-elliptic inclination change transfer orbit in one picture.”, 2014, Reddit submission, /u/ObsessedWithKSP,
https://pay.reddit.com/r/KerbalSpaceProgram/comments/23ri7a
“How to do a bi-elliptic inclination change transfer orbit in one picture.”, 2014, Reddit submission, /u/ObsessedWithKSP,
https://pay.reddit.com/r/KerbalSpaceProgram/comments/23ri7a