Combined maneuvers — optimal split between plane change and velocity change

WHAT are we combining?

Two velocity vectors that differ in both magnitude and direction:

• $v_1$ = speed before the burn
• $v_2$ = speed after the burn
• $\theta$ = angle between the two velocity vectors (the plane-change angle)

The maneuver we must supply is the vector $\vec{\Delta v} = \vec v_2 - \vec v_1$.

HOW to derive the combined $\Delta v$ (law of cosines)

Place $\vec v_1$ and $\vec v_2$ tail-to-tail with angle $\theta$ between them.

$\vec{\Delta v} = \vec v_2 - \vec v_1$ $|\Delta v|^2 = \vec{\Delta v}\cdot\vec{\Delta v} = v_2^2 - 2\,\vec v_1\!\cdot\!\vec v_2 + v_1^2$

Since $\vec v_1\cdot\vec v_2 = v_1 v_2\cos\theta$:

Sanity checks (Forecast-then-Verify):

• $\theta = 0$ (no plane change): $\Delta v = \sqrt{(v_1-v_2)^2}=|v_2-v_1|$. ✅ pure speed change.
• $v_1=v_2=v$ (pure plane change): $\Delta v = v\sqrt{2-2\cos\theta}=2v\sin(\theta/2)$. ✅ the classic plane-change formula.

HOW to split the plane change optimally

Consider the standard problem: a Hohmann transfer between two circular orbits where the final orbit is also inclined by total angle $i$ relative to the initial. You have two burns:

• burn 1 (departure, at speed $v_p$ — high, near the smaller orbit),
• burn 2 (arrival, at speed $v_a$ — low, near the larger orbit).

Let the plane change $s$ be done at burn 1 and $(i-s)$ at burn 2. Each burn is a combined maneuver:

$\Delta v_1(s) = \sqrt{v_{1,i}^2 + v_{1,f}^2 - 2 v_{1,i}v_{1,f}\cos s}$ $\Delta v_2(s) = \sqrt{v_{2,i}^2 + v_{2,f}^2 - 2 v_{2,i}v_{2,f}\cos(i-s)}$

Total $\Delta v_{\text{tot}}(s) = \Delta v_1(s) + \Delta v_2(s)$. Minimize:

$\frac{d\,\Delta v_{\text{tot}}}{ds} = 0$

Worked Example 1 — LEO→GEO with inclination change

Given: transfer between $r_1$ (LEO, fast) and $r_2$ (GEO, slow), total inclination change $i=28.5^\circ$ (Cape Canaveral to equatorial GEO).

Approximate speeds (typical): perigee burn between $v_{1,i}=7.73$ and $v_{1,f}=10.15$ km/s; apogee burn between $v_{2,i}=1.61$ and $v_{2,f}=3.07$ km/s.

Step — do all plane change at apogee ($s=0$): $\Delta v_2 = \sqrt{1.61^2+3.07^2-2(1.61)(3.07)\cos 28.5^\circ}=1.83\text{ km/s}$ Why this step? Apogee has the smallest velocities, so $2v\sin$ costs least here.

Step — split optimally: numerically minimizing $\Delta v_{\text{tot}}(s)$ gives $s\approx 2.2^\circ$ at perigee, $26.3^\circ$ at apogee. Why this step? The equal-marginal-cost condition; a tiny share at perigee shaves a little more.

Step — compare with the naive "all at perigee": $\Delta v_1(\text{all }28.5^\circ)=\sqrt{7.73^2+10.15^2-2(7.73)(10.15)\cos28.5^\circ}=5.0\text{ km/s (!)}$ Why this step? Shows the penalty: doing plane change at the fast burn nearly doubles the whole mission cost.

Takeaway: save the plane change for apoapsis, keep only a whisker at perigee.

Worked Example 2 — Pure equal-speed plane change vs split

Suppose $v_1=v_2=v$ (same circular orbit, only tilt it by $\theta$).

Single burn: $\Delta v = 2v\sin(\theta/2)$.

Bi-elliptic trick (conceptual): raise apoapsis far out (cheap prograde burn), do the plane change there where $v$ is tiny (so $2v\sin(\theta/2)$ is tiny), then drop back. Why this step? Splitting velocity change (raise/lower) from where the plane change happens exploits low speed — for large $\theta$ this beats the single burn. Same principle as combined-maneuver splitting: put the turn where you're slow.

Flashcards

Connections

• Hohmann Transfer Orbit — the two-burn backbone this optimizes.
• Plane Change Maneuvers — the $2v\sin(\theta/2)$ special case.
• Bi-elliptic Transfer — extreme "turn where slow" strategy.
• Vis-viva Equation — supplies the speeds $v_1,v_2$ at each burn.
• Delta-v Budget — where these savings show up in mission planning.
• Law of Cosines — the geometric engine of the derivation.

Concept Map

Hinglish (regional understanding)