Question bank — Combined maneuvers — optimal split between plane change and velocity change

True. By the triangle inequality the third side $\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \theta }$ can never exceed the two-leg path $|v_2-v_1|+2v_2\sin (\theta /2)$; they are equal only in degenerate cases (e.g. $\theta =0$).

False. The equal-speed cost is $2v\sin (\theta /2)=2v\sin 30^{\circ }=v$ exactly — you spend your whole orbital speed again just to turn.

False for small angles and false in general — cost goes as $\sin (\theta /2)$, not $\theta$. For tiny angles $\sin (\theta /2)\approx \theta /2$ so it is roughly linear, but for large angles the sine flattens and eventually the cost saturates.

True. $2v\sin (90^{\circ })=2v$ — you must kill your entire velocity and rebuild it in the opposite sense, the most expensive turn possible.

False. What must be equal is the marginal cost per extra degree, not the degrees themselves. Since the burns run at very different speeds, the optimal split of $i$ is lopsided — most of $i$ ends up at the slow burn.

True. When $v_{1,i}{v}_{1,f}=v_{2,i}{v}_{2,f}=k$, the balance condition becomes $\frac{k\sin s}{\Delta v_1}=\frac{k\sin (i-s)}{\Delta v_2}$; with equal speed products the two $\Delta v$ functions have identical shape, so the equation is satisfied when $\sin s=\sin (i-s)$, i.e. $s=i-s\Rightarrow s=i/2$.

False. You pay extra to raise and re-lower apoapsis. It only wins when the plane-change angle is large enough that the savings on $2v\sin (\theta /2)$ beat the raise/lower cost — the logic behind Bi-elliptic Transfer.

True. Setting $v_1=v_2=v$ gives $\sqrt{2v^2-2v^2\cos \theta }=v\sqrt{2-2\cos \theta }=2v\sin (\theta /2)$ via the half-angle identity.

True. The speeds $v_{1,i},v_{1,f},v_{2,i},v_{2,f}$ that enter the condition come from the Vis-viva Equation evaluated on the actual transfer ellipse, so a different transfer (e.g. Bi-elliptic Transfer) shifts where the plane change wants to live.

Periapsis is where $v$ is largest, and turning cost scales with $v$. Doing all the turn there ($2v_p\sin (\theta /2)$) is the single most expensive choice; the turn belongs at apoapsis where $v$ is smallest.

That formula adds two separate burns (walking two sides of the velocity triangle) and overcounts. The single combined burn is the triangle's third side $\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \theta }$, which is $\le$ the sum by the triangle inequality.

No — it sets the two derivatives $\frac{d\Delta v}{ds}$ equal in magnitude (marginal costs), i.e. $\frac{v_{1,i}{v}_{1,f}\sin s}{\Delta v_1}=\frac{v_{2,i}{v}_{2,f}\sin (i-s)}{\Delta v_2}$. The total burns themselves are usually quite unequal.

At $\theta =0$, $\cos \theta =1$, giving $\sqrt{v_1^2-2v_1{v}_2+v_2^2}=|v_2-v_1|$, not $\sqrt{v_1^2+v_2^2}$. The cross term does not vanish — it is maximal.

Apoapsis has the smallest speed, not zero speed. The cost $2v_a\sin (\theta /2)$ shrinks with $v_a$ but is never zero for a real orbit ($v_a>0$ always).

$\cos \theta$ is defined for every angle; the formula holds for all $\theta$ from $0$ to $180^{\circ }$. Beyond $90^{\circ }$, $\cos \theta$ simply goes negative, making the $-2v_1{v}_2\cos \theta$ term add to the cost.

The total cost $\Delta v_1(s)+\Delta v_2(s)$ is minimized where its derivative vanishes: $\frac{d\Delta v_1}{ds}+\frac{d\Delta v_2}{ds}=0$. Since raising $s$ raises burn 1 but lowers burn 2, this says the marginal cost of one more degree of turn must be equal at both burns — if it weren't, you'd shift a degree to the cheaper burn and save, so you can't already be at the minimum.

Differentiate one burn's cost: only $\cos s$ carries $s$, so $\frac{d\Delta v_1}{ds}=\frac{v_{1,i}{v}_{1,f}\sin s}{\Delta v_1}$. The marginal cost of a degree of turn is thus proportional to the local speed products $v_i{v}_f$. The slow (apoapsis) burn has tiny products, so its marginal cost stays low until it has absorbed most of the turn.

Pythagoras needs a right angle between $v_1$ and $v_2$; here the included angle $\theta$ is arbitrary (look at the velocity triangle above). The Law of Cosines generalizes Pythagoras to any included angle, which is exactly the velocity-triangle situation.

The equal-speed velocity triangle is isosceles; dropping a perpendicular from the apex bisects both $\theta$ and the base $\Delta v$. In the resulting right half-triangle $\sin (\theta /2)=(\Delta v/2)/v$, giving $\Delta v=2v\sin (\theta /2)$ — the half-angle identity made geometric.

A speed change costs only $|v_2-v_1|$ — the difference of speeds. A plane change costs $2v\sin (\theta /2)$, which is scaled by the full speed $v$; you must redirect the entire velocity vector sideways.

Delta-v measures vector magnitude, not tasks. Two separate burns trace two sides of the velocity triangle; the combined burn is the direct third side, and a straight path is never longer than a detour (triangle inequality).

Either ordering traces the same two legs of the triangle in opposite order, and addition is commutative — so the total is identical. The $2v_2$ factor just fixes the plane change to whichever speed you are moving at when you turn; doing it after speeding up to $v_2$ is why the second leg uses $v_2$, not $v_1$.

The balance condition weights each burn by its speed products $v_i{v}_f$; those come from the Vis-viva Equation on the transfer orbit, so faster/slower endpoints reshuffle where the turn is cheapest.

Zero — no change in speed or direction means no burn. The formula gives $\sqrt{v^2+v^2-2v^2}=0$, the sanity floor.

The speed products at both burns approach equality, so the equal-marginal-cost condition pushes the split toward symmetric, $s\to i/2$.

$v_a\to 0$, so the turn cost $2v_a\sin (\theta /2)\to 0$ — asymptotically free out there. But the total is not free: you first pay a raise burn $\Delta v_{\text{up}}=|v_{\text{trans, peri}}-v_p|$ to fling apoapsis outward and a return burn $\Delta v_{\text{down}}$ to drop back, and these two burns stay finite (and grow with how far you go), which is why the full Bi-elliptic Transfer delta-v never actually reaches zero.

There is nothing to combine it with at that point, so you pay the pure $2v\sin (\theta /2)$. The lesson shifts to location: perform it where $v$ is smallest, i.e. at apoapsis.

$\Delta v=\sqrt{0+v_2^2-0}=v_2$ — you simply need to build the full final velocity from scratch; the plane-change angle no longer matters because there is no initial direction to turn from.

They tie. Combined gives $2v$; separating gives $|v-v|+2v\sin 90^{\circ }=2v$. This is the degenerate case where the "triangle" collapses to a straight line and the inequality becomes equality.