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3.2.22 · D5Orbital Mechanics & Astrodynamics

Question bank — Plane change maneuvers — Δv = 2v·sin(Δi - 2)

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The one governing fact behind almost every trap: Here is the orbital speed (a magnitude, in km/s), is the angle you rotate the orbital plane by, and is the ==magnitude of the velocity-vector change== the burn must supply.

True or false — justify

A satellite's speed is the same before and after a pure plane change, so no fuel is needed.
False. Speed (a magnitude) is unchanged, but is a vector and its direction rotates; . Rotating a vector always costs thrust.
Doubling the orbital speed doubles the plane-change cost for the same .
True. is directly proportional to ; the sine factor depends only on , so at twice the speed you pay twice the .
Doubling the angle doubles .
False. depends on linearly but on through , which is not linear. E.g. going from to raises from to , a factor of , not .
For small , cost grows roughly proportionally to .
True. For tiny angles (in radians), so — nearly linear. This is why the small-angle GPS nudge in the parent note scaled cleanly.
A plane change costs because .
False. You plug in the half angle: . A reversal is the most expensive case — you kill and rebuild it the other way.
Combining a speed change and a plane change into one burn always costs less than doing them separately.
True. By the triangle inequality, one vector side of the velocity triangle is never longer than two sequential sides; the single combined burn beats the sum of two burns.
A plane change never alters the orbit's size or shape.
True (by definition). A pure plane change keeps and the radius at the burn point unchanged, so via Orbital velocity — vis-viva equation the semi-major axis is unchanged; only the plane's tilt moves.
You should perform plane changes at periapsis to get them over with quickly.
False. Since and speed is maximum at periapsis, that is the most expensive spot. Do them at apoapsis, where is smallest.

Spot the error

"Since , we have ."
The error is subtracting magnitudes instead of vectors. Correct is : subtract the vectors first, then take the magnitude of the difference.
"For , ."
The angle inside the sine must be halved: . The half-angle is forced by the isosceles-triangle geometry (the perpendicular bisector splits into two equal halves).
"The velocity triangle has sides , with included angle minus ."
The included angle between and is exactly itself, not its complement — you rotated by , so that's the angle the two vectors span.
"A change is cheap, so cost per degree is small no matter the angle."
Cost per degree rises as the angle grows, because curves upward toward . The near-linear cheapness only holds for small angles.
"Plane-change is small, so the propellant mass is small too."
Tsiolkovsky makes mass ratio grow exponentially with . A modest-looking near the exhaust speed can still demand more propellant than payload.
"To reverse orbit direction (), just fire once with ."
You need , not . You must remove your entire velocity and then build up in the opposite direction — two full speeds' worth.

Why questions

Why does the half-angle appear instead of the full ?
Dropping a perpendicular from the apex of the isosceles velocity triangle to the base bisects both the apex angle and the base; each half-base is , so the full base is .
Why is the velocity triangle isosceles rather than a general triangle?
Because a pure plane change leaves the speed unchanged, — two equal sides. If speeds differed (a combined burn), the triangle would be scalene and you'd use the full law of cosines.
Why do we reach for the law of cosines here rather than simple subtraction?
We know two side lengths (, ) and the angle between them (), and want the third side (). The law of cosines is exactly the tool that turns "two sides + included angle" into the opposite side.
Why does make apoapsis the smart burn point?
The cost scales directly with your current speed, and by energy conservation speed is lowest at apoapsis; rotating a slow vector is cheap, rotating a fast one is dear.
Why is a plane change described as "as expensive as reaching orbit"?
, so the burn equals the entire orbital speed (~ km/s in LEO) — comparable to the whole cost of getting to orbit in the first place.
Why do engineers fold plane changes into the Hohmann apogee burn?
At high apogee the speed is small, so the cost is small; and combining the plane rotation with the circularization burn (one vector step) beats two separate burns by the triangle inequality.
Why does get rewritten as in the derivation?
The half-angle identity converts the awkward into a perfect square, which lets you take a clean square root and land on instead of a messy radical.

Edge cases

What is when ?
Zero: . No rotation means no vector change, hence no burn — the trivially correct base case.
What is the maximum possible plane-change , and at what angle?
The maximum is , at where ; you fully reverse direction. No plane change can cost more than twice your speed.
If the orbital speed (hypothetically at rest), what is ?
Zero, since vanishes when . With no velocity vector to rotate, "changing the plane" of motion is meaningless — there is no motion.
For the combined-burn formula, what happens as with ?
gives — a pure speed change with no rotation, exactly a scalar difference.
For the combined-burn formula, what happens when ?
It collapses to , recovering the pure plane-change formula — a good consistency check.
Does a plane change alter the orbital period?
No. Period depends on the semi-major axis, which depends on speed and radius at the burn point (via vis-viva); a pure plane change touches neither, so the period is preserved.
If you split one into two equal plane changes of done at the same speed, do you save fuel?
No — you lose. Two burns of each sum to , which exceeds the single for any positive angle (the sine curve is concave). Splitting only helps if a slower point is available.

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