Maneuvers
20150711
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 visviva equation (*) tells us:
$${v}^{2}=GM(\frac{2}{r}\frac{2}{{r}_{a}+{r}_{p}})$$
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.
When searching for good trajectories, we are interested in saving propellant. According to (*), this is the same as saving for $\mathit{\Delta}v$ (althgouh proportionally). If $r$ is the apsis where the burn is performed, ${r}_{0}$ the opposite apsis before the burn and ${r}_{1}$ after:
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).
$$\frac{1}{\sqrt{GM}}\mathit{\Delta}v=\sqrt{\frac{2}{{r}_{0}}\frac{2}{{r}_{0}+{r}_{1}}}\frac{1}{\sqrt{{r}_{0}}}+\frac{1}{\sqrt{{r}_{1}}}\sqrt{\frac{2}{{r}_{1}}\frac{2}{{r}_{0}+{r}_{1}}}$$
We can multiply both hands by $\sqrt{{r}_{0}}$ and set $x=\frac{{r}_{1}}{{r}_{0}}$ to get a simpler expression:
$$\underset{\mathit{\alpha}}{\underset{\u23df}{\sqrt{\frac{{r}_{0}}{GM}}}}\mathit{\Delta}v=\sqrt{2\frac{2}{1+x}}1+\frac{1}{\sqrt{x}}\sqrt{\frac{2}{x}\frac{2}{1+x}}$$
Bielliptical transfer
The idea is to use three burns instead of two.
1 v
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Again, we set $x=\frac{{r}_{2}}{{r}_{0}}$ and $y=\frac{{r}_{1}}{{r}_{0}}$ and:
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Inclination change
(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.
As seen in figure (?), we can easily find the change in inclination:
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a t
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a t
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v
Straight burn
Doing a ${180}^{\circ}$ inclination chage using a constant radial burn like exposed above yields a $\mathit{\Delta}v$ proportional to the current orbital velocity $v$: $\mathit{\Delta}v=\mathit{\pi}v\approx 3v$. However, simply going retrograde until the speed is reverse only yields $\mathit{\Delta}v=2v$ for the same result.
You can find another derivation of the cost in [1] [2].
We need to compute $\mathit{\Delta}v$ for given $v=\overrightarrow{{v}_{0}}=\overrightarrow{{v}_{1}}$ and $\mathit{\theta}$. Because the triangle is isosceles, the altitude and the median from $P$ are one so:
$$\mathit{\Delta}v=\left2v\mathrm{sin}\frac{\mathit{\theta}}{2}\right$$
For $\mathit{\theta}=\mathit{\pi}$, we get $\mathit{\Delta}v=2v$ which is the expected result.
Bielliptical 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 bielliptical 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].
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“Bielliptic inclination change transfer orbit”,
2014, imgur album, /u/listens_to_galaxies,
https://imgur.com/a/uM7PK“How to do a bielliptic inclination change transfer orbit in one picture.”, 2014, Reddit submission, /u/ObsessedWithKSP,
https://pay.reddit.com/r/KerbalSpaceProgram/comments/23ri7a“How to do a bielliptic inclination change transfer orbit in one picture.”, 2014, Reddit submission, /u/ObsessedWithKSP,
https://pay.reddit.com/r/KerbalSpaceProgram/comments/23ri7a