Combined maneuvers — optimal split between plane change and velocity change: Exercises

This page tests the parent topic. Every problem has a collapsible worked solution — try it first, then reveal. We build up from recognizing the formula to synthesizing a whole mission plan.

Everything here rests on one formula from the parent note, so let us restate it in plain words before we use it once.

Figure — Combined maneuvers — optimal split between plane change and velocity change

Look at the amber side of that triangle — that amber length is the $Δ v$ your engine must supply. The two cyan sides are $v_{1}$ and $v_{2}$ ; the white arc is $θ$ . When $θ = 0$ the two sides fold flat onto each other and the amber side shrinks to the plain difference $∣ v_{2} - v_{1} ∣$ ; when the sides are equal it becomes the isoceles-triangle base $2 v sin (θ /2)$ ; when $θ = 18 0^{\circ}$ the sides point opposite ways and the amber side stretches to $v_{1} + v_{2}$ . Keep this one picture in your head — every problem below is just this triangle with different side lengths and a different arc.

Level 1 — Recognition

L1.1

Problem. A spacecraft is in a circular orbit at speed $v = 7.8$ km/s. It must tilt the orbit plane by $θ = 1 5^{\circ}$ with no change in speed. What single-burn $Δ v$ does this cost?

Recall Solution

Equal speeds means $v_{1} = v_{2} = v$ — in the triangle figure both cyan sides have the same length. Plug into the master formula: $Δ v = v^{2} + v^{2} - 2 v^{2} cos θ = v 2 - 2 cos θ = 2 v sin (θ /2)$ The middle step used the identity $1 - cos θ = 2 sin^{2} (θ /2)$ — why? Because it turns the ugly $2 - 2 cos θ$ into a clean $2 sin (θ /2)$ , which is exactly the base of the isoceles velocity triangle (both equal sides $v$ , apex angle $θ$ ) — drop a perpendicular from the apex and each half-base is $v sin (θ /2)$ . $Δ v = 2 (7.8) sin (7. 5^{\circ}) = 15.6 \times 0.13053 = 2.036 km/s$ Answer: $\approx 2.04$ km/s. That is over a quarter of orbital speed for a mere $1 5^{\circ}$ tilt — plane changes are brutal.

L1.2

Problem. In the same triangle, what is $Δ v$ if $θ = 0^{\circ}$ , $v_{1} = 3.07$ km/s, $v_{2} = 1.61$ km/s?

Recall Solution

With $θ = 0$ the two velocity arrows point the same way, so the triangle collapses flat and the burn is pure speed change (a straight subtraction): $Δ v = v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos 0 = (v_{1} - v_{2})^{2} = ∣ v_{1} - v_{2} ∣$ $= ∣3.07 - 1.61∣ = 1.46 km/s$ Answer: $1.46$ km/s. The $cos 0 = 1$ collapses the triangle to a flat line — no turning, just braking.

Level 2 — Application

L2.1

Problem. An angled burn changes speed from $v_{1} = 10.15$ km/s to $v_{2} = 7.73$ km/s while turning by $θ = 1 0^{\circ}$ . Find $Δ v$ .

Recall Solution

Straight into the master formula — the two cyan sides now have different lengths and the arc is non-zero, so we need the full cosine (no shortcut applies): $Δ v = 10.1 5^{2} + 7.7 3^{2} - 2 (10.15) (7.73) cos 1 0^{\circ}$ Compute piece by piece (WHAT: build each term; WHY: the two squared terms are the two sides' lengths, and the cross term is how much the small angle "pulls" the tips together — a nearly-flat triangle):

$10.1 5^{2} = 103.0225$ (side one squared)
$7.7 3^{2} = 59.7529$ (side two squared)
$2 (10.15) (7.73) = 156.919$ , and $cos 1 0^{\circ} = 0.98481$ (close to 1 because the turn is small), so the cross term $= 154.536$ $Δ v = 103.0225 + 59.7529 - 154.536 = 8.239 = 2.87 km/s$ Answer: $\approx 2.87$ km/s. Notice it is close to the pure speed change $∣10.15 - 7.73∣ = 2.42$ — because $1 0^{\circ}$ barely tilts the triangle.

L2.2

Problem. Compare doing a $28. 5^{\circ}$ plane change (with a speed change $1.61 \to 3.07$ km/s) as (a) two separate burns versus (b) one combined burn. Which wins and by how much?

Recall Solution

(a) Separate. First change speed ( $θ = 0$ ), then turn at the new speed $v_{2} = 3.07$ — this walks along two sides of a triangle instead of the direct third: $Δ v_{sep} = ∣3.07 - 1.61∣ + 2 (3.07) sin (14.2 5^{\circ}) = 1.46 + 6.14 (0.24615) = 1.46 + 1.511 = 2.97 km/s$ (b) Combined. One angled burn — the direct amber side of the triangle: $Δ v_{comb} = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 28. 5^{\circ}$ $= 2.5921 + 9.4249 - 9.8854 (0.87882) = 12.017 - 8.687 = 3.330 = 1.825 km/s$ Answer: combined $= 1.83$ km/s beats separate $= 2.97$ km/s — a saving of about 1.14 km/s. This is the Law of Cosines triangle inequality made concrete: the direct third side is shorter than going along the other two.

Level 3 — Analysis

L3.1

Problem. For the LEO→GEO transfer, the total plane-change budget is $i = 28. 5^{\circ}$ (recall: $i$ is the whole turn the mission must share between its two burns). Perigee burn: $v_{1, i} = 7.73$ , $v_{1, f} = 10.15$ km/s. Apogee burn: $v_{2, i} = 1.61$ , $v_{2, f} = 3.07$ km/s. Doing all $28. 5^{\circ}$ at perigee, find the perigee $Δ v$ , and compare to doing it all at apogee.

Recall Solution

"All at perigee" means the perigee share is $s = 28. 5^{\circ}$ (so its slice of the turn is the whole $i$ ), and the apogee share is $i - s = 0^{\circ}$ . Apply the master formula at each burn with its own speeds: $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 28. 5^{\circ}$ $= 59.7529 + 103.0225 - 156.919 (0.87882) = 162.775 - 137.905 = 24.871 = 4.987 km/s$ Apogee burn is then pure speed change ( $i - s = 0$ ): $∣3.07 - 1.61∣ = 1.46$ . Total $\approx 4.99 + 1.46 = 6.45$ km/s.

All at apogee ( $s = 0$ , apogee slice $= i$ ): perigee is pure speed change $∣10.15 - 7.73∣ = 2.42$ ; apogee combined $= 1.825$ (from L2.2). Total $\approx 2.42 + 1.825 = 4.25$ km/s.

Answer: perigee-turn total $\approx 6.45$ km/s vs apogee-turn total $\approx 4.25$ km/s. Turning at the fast perigee costs about 2.2 km/s more — because the products $v \cdot v$ in the cross term are enormous there. See Delta-v Budget.

L3.2

Problem. Verify the marginal-cost (optimal-split) condition is not satisfied when all plane change sits at apogee. Compute both sides at $s = 0$ (perigee share $s = 0$ , so the apogee share $i - s = 28. 5^{\circ}$ ).

Recall Solution

The condition (from the parent note) is $\frac{v _{1, i} v _{1, f} sin s}{Δ v _{1}} = \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2}}$ , where the left side is the marginal cost of one more degree of turn at perigee and the right side is the same at apogee.

Left (perigee) at $s = 0$ : $sin 0 = 0$ , so LHS $= 0$ .

Right (apogee) at $s = 0$ : numerator $= (1.61) (3.07) sin 28. 5^{\circ} = 4.9427 (0.47716) = 2.358$ ; denominator $Δ v_{2} = 1.825$ (from L2.2). RHS $= 2.358/1.825 = 1.292$ .

Answer: LHS $= 0 \neq = 1.29 =$ RHS. Marginal cost of turning is zero at perigee but large at apogee — so shifting a sliver of the turn to perigee lowers total cost. The optimum sits slightly off $s = 0$ (around $s \approx 2. 2^{\circ}$ at perigee), exactly where the two sides balance.

Level 4 — Synthesis

L4.1

Problem. For the L3.1 numbers, numerically confirm the optimal split $s \approx 2. 2^{\circ}$ (perigee) / $i - s \approx 26. 3^{\circ}$ (apogee) beats the all-at-apogee plan, and report the total $Δ v$ at the optimum.

Recall Solution

We evaluate each burn's cost with its own slice of the turn: the perigee slice is $s = 2. 2^{\circ}$ , the apogee slice is $i - s = 26. 3^{\circ}$ .

Perigee ( $θ = s = 2. 2^{\circ}$ ). Why so small an angle here? Perigee is fast, so we let it carry only a whisker of the turn: $Δ v_{1} = 7.7 3^{2} + 10.1 5^{2} - 2 (7.73) (10.15) cos 2. 2^{\circ}$ Here $cos 2. 2^{\circ} = 0.99926$ — almost 1, because the angle is tiny, so the cross term $= 156.919 (0.99926) = 156.803$ barely shrinks from its flat-triangle value. The triangle is nearly collapsed, so the cost is close to the pure speed change: $= 162.775 - 156.803 = 5.972 = 2.444 km/s$ Apogee ( $θ = i - s = 26. 3^{\circ}$ ). Why the big slice here? Apogee is slow, so the expensive turning is cheapest here: $Δ v_{2} = 1.6 1^{2} + 3.0 7^{2} - 2 (1.61) (3.07) cos 26. 3^{\circ}$ $cos 26. 3^{\circ} = 0.89661$ (well below 1 — a real tilt), cross term $= 9.8854 (0.89661) = 8.863$ , which is smaller than the flat value, so the triangle opens up and the amber side grows: $= 12.017 - 8.863 = 3.154 = 1.776 km/s$ Total $= 2.444 + 1.776 = 4.220$ km/s.

Compare with all-at-apogee total $4.245$ km/s (L3.1). Answer: the optimum $\approx 4.22$ km/s shaves about 0.025 km/s (25 m/s) — small but free, and confirms the split is real. Relate to Hohmann Transfer Orbit budgeting.

L4.2

Problem. A satellite in a circular orbit ( $v = 7.8$ km/s) needs a $4 0^{\circ}$ plane change. Option A: single burn. Option B (bi-elliptic-style): burn prograde to raise apoapsis where speed drops to $v_{high} = 1.5$ km/s, do the whole plane change there, then drop back. Ignore the raise/lower cost for the comparison of the turn itself — compare only the turning costs.

Recall Solution

Both options are pure equal-speed turns, so both use $2 v sin (θ /2)$ (the isoceles-triangle base) — only the speed $v$ differs.

Option A turn cost. Why plug in $7.8$ ? The turn happens at full orbital speed, so both cyan sides are long: $2 v sin (2 0^{\circ}) = 2 (7.8) (0.34202) = 5.336 km/s$ Option B turn cost. Why plug in $1.5$ ? We deliberately moved the turn out to apoapsis where the sides shrink to $1.5$ , so the same $4 0^{\circ}$ apex angle spans a much smaller base: $2 v_{high} sin (2 0^{\circ}) = 2 (1.5) (0.34202) = 1.026 km/s$ Answer: the turn alone drops from $5.34$ to $1.03$ km/s — a factor of $5.2$ cheaper. Why exactly $5.2$ ? Because the base of an isoceles triangle scales linearly with the side length, so the cost ratio is just the speed ratio $v / v_{high} = 7.8/1.5 = 5.2$ . This is why Bi-elliptic Transfer shines for large plane changes: turning where you crawl is dramatically cheaper. Principle: turn where slow.

Level 5 — Mastery

L5.1

Problem. Prove that the combined burn is always at least as cheap as any two-step "change speed, then turn" plan for the same $v_{1} \to v_{2}$ and turn $θ$ . Then state the single condition under which they are exactly equal.

Recall Solution

A note on notation first. Until now $v_{1}, v_{2}$ have been plain speeds (lengths of the cyan sides). To talk about subtracting velocities we need their arrow form — a velocity has both a size and a direction. We write an arrow on top, $v_{1}$ and $v_{2}$ , to mean "the full velocity, size and pointing"; their plain-letter versions $v_{1} = ∣ v_{1} ∣$ and $v_{2} = ∣ v_{2} ∣$ are just the lengths. The burn is literally the arrow that carries you from the tip of $v_{1}$ to the tip of $v_{2}$ .

Setup. The combined burn is the vector $Δ v = v_{2} - v_{1}$ , whose length is the master formula (the amber side of the triangle in the figure). A two-step plan traverses two sides of a triangle to get from the tip of $v_{1}$ to the tip of $v_{2}$ : one leg for the speed change, one leg for the turn. The theorem. By the triangle inequality (the straight route is never longer than a detour): $∣ v_{2} - v_{1} ∣ \leq ∣ leg_{1} ∣ + ∣ leg_{2} ∣ = Δ v_{sep}$ Concretely, with the turn at speed $v_{2}$ : $v_{1}^{2} + v_{2}^{2} - 2 v_{1} v_{2} cos θ \leq ∣ v_{2} - v_{1} ∣ + 2 v_{2} sin (θ /2)$ . Equality condition. A triangle degenerates to a straight line (equality) only when the two legs are collinear — i.e. when there is no turn, $θ = 0$ . Then both sides equal $∣ v_{2} - v_{1} ∣$ . Answer: combined $\leq$ separate always; equal iff $θ = 0$ . See Law of Cosines and Vis-viva Equation for where $v_{1}, v_{2}$ come from.

L5.2

Problem. A mission needs $v_{1} = 7.73 \to v_{2} = 10.15$ km/s with a $28. 5^{\circ}$ turn at perigee and $v_{1} = 1.61 \to v_{2} = 3.07$ km/s at apogee, but you may put any split $s$ of the $28. 5^{\circ}$ at perigee. Show, using the marginal condition, why the optimum $s$ is small and positive (not zero, not large). Give the qualitative sign argument only.

Recall Solution

Define $g (s) = \frac{d Δ v _{tot}}{d s} = \frac{v _{1, i} v _{1, f} sin s}{Δ v _{1} ( s )} - \frac{v _{2, i} v _{2, f} sin ( i - s )}{Δ v _{2} ( s )}$ . Here $s$ is the perigee share of the total turn $i = 28. 5^{\circ}$ , and $i - s$ the apogee share. We want $g (s) = 0$ .

At $s = 0$ : first term $= 0$ (since $sin 0 = 0$ ); second term $= 2.358/1.825 = 1.292 > 0$ , so $g (0) = - 1.292 < 0$ . Slope is negative — increasing $s$ still lowers total cost.
At $s = i = 28. 5^{\circ}$ : now the second term vanishes ( $sin (i - s) = sin 0 = 0$ ), and the first term is huge and positive (large perigee $v \cdot v$ products). So $g (i) > 0$ — pushing all the turn to perigee is far past optimum.
$g$ is continuous and goes from negative to positive, so a root sits between, and because the perigee products $v_{1, i} v_{1, f} = 78.4$ dwarf the apogee products $v_{2, i} v_{2, f} = 4.94$ (ratio $\approx 16$ ), the balance is struck at a small $s$ : even a tiny $sin s$ at perigee generates the marginal cost that a much larger $sin (i - s)$ at apogee produces. Answer: optimum is small and positive — the numbers give $s \approx 2. 2^{\circ}$ . The heavy perigee $v \cdot v$ term forces the turn almost entirely to apogee, but not quite all of it.

Recall One-line self-test

Why does most of the plane change end up at apogee? ::: Because the cross-term products $v \cdot v$ are tiny at the slow apogee, so each degree of turn is cheap there — marginal cost balances at a small perigee share.