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

• $\Delta v$ = the magnitude of the required burn (a change in the velocity vector, measured in m/s or km/s).
• $v$ = the orbital speed, which is the same before and after (a pure plane change keeps speed fixed).
• $\Delta i$ = the angle you rotate the velocity vector through = the change in inclination.
• The half comes from geometry: drop a perpendicular from the triangle's apex (angle $\Delta i$) to the base $\Delta \vec{v}$. It bisects the apex into $\Delta i/2$ and cuts the base in half. Each half-base is $v\sin (\Delta i/2)$, so the whole base is $2v\sin (\Delta i/2)$.

Since $\Delta v=2v\sin (\Delta i/2)$ and $\Delta i$ is the same for both, $\Delta v$ is directly proportional to $v$. So satellite A (faster) pays more. The ratio is exactly the speed ratio: $\frac{\Delta v_A}{\Delta v_B}=\frac{v_A}{v_B}=\frac{7.7}{3.1}\approx \mathrm{2.48.}$ A pays about 2.5× as much. (This is the reason to do plane changes where you are slow.)

(a) $\Delta i=0°$: $\Delta v=2v\sin (0°)=0$. No rotation, no burn — obvious. (b) $\Delta i=180°$: $\Delta v=2v\sin (90°)=2v$. You must fully reverse direction: kill your speed $v$, then build $v$ the opposite way, total $2v$. The formula captures this exactly.

Half-angle: $\Delta i/2=5°$. Then $\sin 5°\approx 0.08716$. $\Delta v=2(7.7)(0.08716)\approx 1.342\text{ km/s}=1342\text{ m/s}.$ A mere $10°$ costs over $1.3$ km/s — nearly a fifth of the orbital speed itself.

Rearrange $\Delta v=2v\sin (\Delta i/2)$ for $\Delta i$. First isolate the sine: $\sin \left(\frac{\Delta i}{2}\right)=\frac{\Delta v}{2v}=\frac{0.200}{2(7.7)}=\frac{0.200}{15.4}=\mathrm{0.012987.}$ Why $\arcsin$? We know the sine of the half-angle and want the angle back — $\arcsin$ is the operation that undoes sine. Since $0.012987$ is tiny, the angle is tiny (and positive), so no quadrant ambiguity: $\frac{\Delta i}{2}=\arcsin (0.012987)=0.012987\text{ rad}\Rightarrow \Delta i=0.025974\text{ rad}.$ Convert to degrees: $0.025974\times \frac{180}{\pi }\approx 1.488°$. So the biggest change is about $1.49°$ — showing how stingy plane changes are with a small budget.

First, unify units. The formula mixes $\Delta v$ and $v$, so both must be in the same unit. Convert the budget $\Delta v=400$ m/s $=0.400$ km/s (divide by $1000$), and we'll get $v$ out in km/s. Rearrange for $v$: $v=\frac{\Delta v}{2\sin (\Delta i/2)}$. Half-angle $\Delta i/2=10°$, $\sin 10°\approx 0.17365$. $v=\frac{0.400\text{ km/s}}{2(0.17365)}=\frac{0.400}{0.34730}\approx 1.152\text{ km/s}.$ You must be crawling at about $1.15$ km/s — the kind of speed you only reach at apoapsis of a high ellipse. Use vis-viva to find where that speed occurs.

Set $\Delta v=v$: $2v\sin (\Delta i/2)=v\Rightarrow \sin (\Delta i/2)=0.5\Rightarrow \Delta i/2=30°\Rightarrow \Delta i=60°$. Set $\Delta v=2v$: $\sin (\Delta i/2)=1\Rightarrow \Delta i/2=90°\Rightarrow \Delta i=180°$. So the burn first equals the whole orbital speed at exactly $\Delta i=60°$, and reaches its maximum $2v$ only at the full reversal $\Delta i=180°$. Because $\sin (\Delta i/2)$ never exceeds $1$ on $0°$–$180°$, $\Delta v=2v\sin (\Delta i/2)$ never exceeds $2v$ — it is capped at $2v$. The function is monotonically increasing on $0°$ to $180°$ (the sine of the half-angle rises steadily from $0$ to $1$), so bigger tilt always costs more, up to that hard ceiling of $2v$. The lesson: once $\Delta i>60°$ a single plane change costs more than an entire orbit's worth of speed, so engineers split large changes across slow orbit points instead.

Why a derivative? We want the rate $d(\Delta v)/d(\Delta i)$ — "how much does the cost change per unit angle" — and the derivative is exactly the tool that answers "rate of change of one quantity with respect to another." $\frac{d}{d(\Delta i)}[2v\sin (\Delta i/2)]=2v\cdot \frac{1}{2}\cos (\Delta i/2)=v\cos (\Delta i/2).$ At small $\Delta i$, $\cos (\Delta i/2)\approx 1$, so the slope $\approx v=7.7$ km/s per radian. Convert to per degree by multiplying by $\pi /180\approx 0.01745$: $7.7\times 0.01745\approx 0.1344\text{ km/s per degree}=134\text{ m/s per degree}.$ Every extra degree near zero costs about 134 m/s in LEO. (Matches the parent note's $2°$-GPS example scaled by speed.)

Look at the figure again: the red arrow $\Delta \vec{v}$ is the base of an isosceles triangle. The angle between ${\vec{v}}_1$ and $\Delta \vec{v}$ equals $90°+\Delta i/2$ measured on that triangle — i.e. the burn points almost exactly perpendicular to the velocity (it tips slightly backward by $\Delta i/2$). For an infinitesimally small change ($\Delta i\to 0$) the burn is purely sideways (normal to $\vec{v}$), which makes sense: a sideways nudge rotates a vector without changing its length to first order. The sign (left vs right / north vs south) sets whether inclination increases or decreases.

Combined (law of cosines, two different sides at angle $\Delta i$): $\Delta v_{\text{comb}}=\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \Delta i}.$ With $\cos 28°\approx 0.88295$: $\Delta v_{\text{comb}}=\sqrt{1.6^2+3.1^2-2(1.6)(3.1)(0.88295)}$ $=\sqrt{2.56+9.61-8.7589}=\sqrt{3.4111}\approx 1.847\text{ km/s}.$ Separate = speed change first, then a pure plane change at the new speed $v_2=3.1$: $\Delta v_{\text{sep}}=\underset{\text{speed}}{\underbrace{(3.1-1.6)}}+\underset{\text{plane}}{\underbrace{2(3.1)\sin 14°}}=1.5+6.2(0.24192)=1.5+1.500=3.000\text{ km/s}.$ Combined ($1.85$ km/s) beats separate ($3.00$ km/s) by over a km/s — the triangle inequality in action: one straight step is shorter than two legs of a bent path. This is why Hohmann transfers fold the plane change into the apogee burn.

One shot: $\Delta v=2(7.7)\sin 20°=15.4(0.34202)\approx 5.267$ km/s. Split into two $20°$ burns: each costs $2(7.7)\sin 10°=15.4(0.17365)\approx 2.674$ km/s, total $2\times 2.674=5.348$ km/s. Splitting at the same speed is slightly worse ($5.35>5.27$): each small chord is shorter, but two chords sum to more than the single larger chord (again the triangle inequality, now the other way). Splitting only helps if you do the sub-burns where $v$ is smaller — the real saving comes from lower speed, not from subdividing the angle.

Why Tsiolkovsky? It's the tool that turns a required $\Delta v$ into a mass ratio. See Tsiolkovsky rocket equation. $\frac{m_0}{m_f}=e^{\Delta v/v_e}=e^{1.342/3.0}=e^{0.44733}\approx \mathrm{1.5641.}$ Propellant fraction $=1-\frac{m_f}{m_0}=1-\frac{1}{1.5641}=1-0.6393=0.3607$. About 36% of the spacecraft's mass is burned as fuel — for a single $10°$ tilt. Plane changes are ruinous.

(a) Combined burn. Two sides $v_a=1.60$, $v_{\text{GEO}}=3.07$, included angle $27°$, $\cos 27°\approx 0.89101$: $\Delta v=\sqrt{1.60^2+3.07^2-2(1.60)(3.07)(0.89101)}$ $=\sqrt{2.5600+9.4249-8.7526}=\sqrt{3.2323}\approx 1.798\text{ km/s}.$

(b) Separate burns. Circularize: $|3.07-1.60|=1.47$ km/s. Then a pure plane change at the GEO speed $3.07$: $2(3.07)\sin 13.5°=6.14(0.23345)=1.433$ km/s. Total: $\Delta v_{\text{sep}}=1.47+1.433=2.903\text{ km/s}.$ Combining saves $2.903-1.798=1.105$ km/s — a huge margin.

(c) Propellant fraction (combined). Use Tsiolkovsky with $\Delta v=1.798$ km/s and $v_e=3.1$ km/s: $\frac{m_0}{m_f}=e^{1.798/3.1}=e^{0.58000}\approx 1.7860,\text{fraction}=1-\frac{1}{1.7860}=1-0.5599=\mathrm{0.4401.}$ About 44% of the mass is propellant with the combined burn. For comparison, the separate strategy ($\Delta v=2.903$ km/s) gives $\frac{m_0}{m_f}=e^{2.903/3.1}=e^{0.93645}\approx 2.551$, a fraction $1-1/2.551=0.6081$, i.e. about 61% — far worse.

(d) Design lesson. Do the plane change (i) at the slowest point — apogee, where $v$ is smallest — and (ii) combined with the circularization burn, exploiting the law-of-cosines triangle. Both tricks stack to slash the propellant bill from ~61% to ~44%. That is why real GEO missions fold the inclination removal into the apogee kick.

Separate cost, as a function of $\Delta i$: $\Delta v_{\text{sep}}(\Delta i)=|v_{\text{GEO}}-v_a|+2v_{\text{GEO}}\sin (\Delta i/2)=1.47+6.14\sin (\Delta i/2).$ Combined cost: $\Delta v_{\text{comb}}(\Delta i)=\sqrt{1.60^2+3.07^2-2(1.60)(3.07)\cos \Delta i}=\sqrt{11.9849-9.8248\cos \Delta i}.$ At $\Delta i=0$: combined $=|3.07-1.60|=1.47$, separate $=1.47$ — equal (as they must, no plane change). For all $\Delta i>0$ the combined value is strictly below separate (triangle inequality), so combining is always at least as cheap and strictly cheaper whenever $\Delta i>0$.

Numerical check at $\Delta i=27°$: combined $=\sqrt{11.9849-9.8248(0.89101)}=\sqrt{11.9849-8.7538}=\sqrt{3.2311}\approx 1.797$ km/s vs separate $1.47+6.14(0.23345)=2.903$ km/s. So there is no break-even where separate wins — combining dominates for every $\Delta i>0$, and the two are equal only at the trivial $\Delta i=0$.

Speed change from $2°$ tilt at $7.7$ km/s? ::: $\approx 269$ m/s (using $2\cdot 7.7\cdot \sin 1°$) Largest $\Delta i$ for a $200$ m/s budget at $7.7$ km/s? ::: $\approx 1.49°$ $\Delta v$ per degree near zero at $7.7$ km/s? ::: $\approx 134$ m/s per degree Combined burn formula? ::: $\Delta v=\sqrt{v_1^2+v_2^2-2v_1{v}_2\cos \Delta i}$ Why does combining always win for $\Delta i>0$? ::: triangle inequality — one vector leg is shorter than two sequential legs