Noesis
3.2.18Orbital Mechanics & Astrodynamics

Orbit determination — Gauss's method, Gibbs method

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🇮🇳 Yeh note Hinglish mein padho →

WHY two methods? Because your data type differs. If you already have position vectors (e.g. from radar/GPS), skip straight to Gibbs. If you only have angles from a telescope, you must first run Gauss to find the position vectors, then hand them to Gibbs.

Part 1 — The physics both methods lean on

WHY these two? h⃗\vec h pins the plane; e⃗\vec e pins the shape and orientation inside that plane. Between them they contain all the geometry — Gibbs is basically "reconstruct e⃗\vec e and h⃗\vec h from three points."

Part 2 — Gibbs's Method (positions → velocity)

Step 1 — Coplanarity check (WHY first?)

All three must lie in one plane, else no single orbit exists. u^r1⋅(u^r2×u^r3)≈0(should be<1∘ off).\hat u_{r1}\cdot(\hat u_{r2}\times \hat u_{r3}) \approx 0 \quad(\text{should be} <1^\circ\text{ off}). Why this step? r⃗2×r⃗3\vec r_2\times\vec r_3 is normal to the plane of 2 & 3; dotting with r⃗1\vec r_1 tests if 1 is in it too.

Step 2 — Build three helper vectors

Because a conic is r=p1+ecos⁡θr = \dfrac{p}{1+e\cos\theta}, one can show the velocity at point 2 is expressible via: N⃗=r1(r⃗2×r⃗3)+r2(r⃗3×r⃗1)+r3(r⃗1×r⃗2)\vec N = r_1(\vec r_2\times\vec r_3) + r_2(\vec r_3\times\vec r_1) + r_3(\vec r_1\times\vec r_2) D⃗=(r⃗1×r⃗2)+(r⃗2×r⃗3)+(r⃗3×r⃗1)\vec D = (\vec r_1\times\vec r_2) + (\vec r_2\times\vec r_3) + (\vec r_3\times\vec r_1) S⃗=r⃗1(r2−r3)+r⃗2(r3−r1)+r⃗3(r1−r2)\vec S = \vec r_1(r_2-r_3) + \vec r_2(r_3-r_1) + \vec r_3(r_1-r_2) where ri=∣r⃗i∣r_i = |\vec r_i|.

Why these combos? They are engineered so that N⃗/D⃗\vec N/\vec D carries the scale (pp, angular momentum) and S⃗\vec S carries the direction of perigee (e⃗\vec e). Derivation sketch: write each r⃗i\vec r_i using the orbit equation and eliminate the unknown true anomalies — the surviving symmetric sums collapse to N⃗,D⃗,S⃗\vec N,\vec D,\vec S.

Step 3 — The velocity formula

  v⃗2=μN D(D⃗×r⃗2r2+S⃗)  \boxed{\;\vec v_2 = \sqrt{\frac{\mu}{N\,D}}\left(\frac{\vec D\times\vec r_2}{r_2} + \vec S\right)\;} with N=∣N⃗∣, D=∣D⃗∣N=|\vec N|,\ D=|\vec D|.

Now you have (r⃗2,v⃗2)(\vec r_2,\vec v_2) → convert to classical elements. Done.

Figure — Orbit determination — Gauss's method, Gibbs method

Part 3 — Gauss's Method (three angles → three positions)

Key idea A — the sector/triangle f & g series

Between two nearby times, propagate r⃗\vec r using Lagrange coefficients: r⃗i=fi r⃗2+gi r⃗˙2,fi≈1−μ2r23τi2,  gi≈τi−μ6r23τi3\vec r_i = f_i\,\vec r_2 + g_i\,\dot{\vec r}_2,\qquad f_i \approx 1-\tfrac{\mu}{2 r_2^3}\tau_i^2,\ \ g_i\approx \tau_i-\tfrac{\mu}{6 r_2^3}\tau_i^3 with τi=ti−t2\tau_i = t_i - t_2. Why series? We don't know the orbit yet, so we Taylor-expand r⃗(t)\vec r(t) about the middle observation using r⃗¨=−μr⃗/r3\ddot{\vec r}=-\mu\vec r/r^3. Short arc ⇒ few terms suffice.

Key idea B — solve for ρ2\rho_2

Coplanarity + the f,g relations give a scalar 8th-degree polynomial in r2r_2: r28+a r26+b r23+c=0r_2^8 + a\,r_2^6 + b\,r_2^3 + c = 0 Solve (numerically) for the physically valid positive real root r2r_2. Back-substitute to get all ρi\rho_i, hence all r⃗i\vec r_i.

Step 4 — hand off to Gibbs (or improve)

With three r⃗i\vec r_i known, run Gibbs for v⃗2\vec v_2. Then iterate: use the better orbit to sharpen the f,g coefficients (now with exact ones, not the truncated series) and re-solve until ρ2\rho_2 converges.

Worked Example — Gibbs (numbers)

Given (Earth, μ=398600 km3/s2\mu=398600\ \text{km}^3/\text{s}^2): r⃗1=(−294.32, 4265.1, 5986.7)\vec r_1=(-294.32,\,4265.1,\,5986.7), r⃗2=(−1365.5, 3637.6, 6346.8)\vec r_2=(-1365.5,\,3637.6,\,6346.8), r⃗3=(−2940.3, 2473.7, 6555.8)\vec r_3=(-2940.3,\,2473.7,\,6555.8) km.

Step 1 — magnitudes. r1=7359.6, r2=7433.5, r3=7473.6r_1=7359.6,\ r_2=7433.5,\ r_3=7473.6 km. Why? Needed as scalar weights in N⃗,S⃗\vec N,\vec S.

Step 2 — cross products. Compute r⃗1×r⃗2\vec r_1\times\vec r_2, etc. Why? They build N⃗,D⃗\vec N,\vec D; also their triple product checks coplanarity (≈0 ✓).

Step 3 — assemble N⃗,D⃗,S⃗\vec N,\vec D,\vec S, then N=∣N⃗∣, D=∣D⃗∣N=|\vec N|,\ D=|\vec D|. Why? p=N/Dp=N/D sets the orbit scale.

Step 4 — velocity. v⃗2=μND(D⃗×r⃗2r2+S⃗)≈(−6.218, −4.014, 1.599) km/s.\vec v_2=\sqrt{\tfrac{\mu}{ND}}\Big(\tfrac{\vec D\times\vec r_2}{r_2}+\vec S\Big)\approx(-6.218,\,-4.014,\,1.599)\ \text{km/s}. Why last? This is the missing 3 numbers; (r⃗2,v⃗2)(\vec r_2,\vec v_2) = full state.

Common Mistakes

Recall Explain to a 12-year-old (Feynman)

Imagine a ball flying in a big loop. You take three photos and mark exactly where the ball was in each. Even without a stopwatch, the three dots are enough to trace the whole loop and figure out how fast the ball was moving in the middle photo — that's Gibbs. But if instead of dots you only wrote down which direction you were looking (not how far the ball was), you first have to be clever and guess the distances — that's Gauss — and then you feed those dots to Gibbs.

Flashcards

Gibbs method input data?
Three position vectors r⃗1,r⃗2,r⃗3\vec r_1,\vec r_2,\vec r_3 (coplanar), no times needed.
Gauss method input data?
Three angle-only observations (time, α\alpha, δ\delta) plus observer positions R⃗i\vec R_i.
What are the three Gibbs helper vectors?
N⃗=∑ri(r⃗j×r⃗k)\vec N=\sum r_i(\vec r_j\times\vec r_k), D⃗=∑(r⃗i×r⃗j)\vec D=\sum(\vec r_i\times\vec r_j), S⃗=∑r⃗i(rj−rk)\vec S=\sum \vec r_i(r_j-r_k) (cyclic).
Gibbs velocity formula?
v⃗2=μ/(ND) (D⃗×r⃗2/r2+S⃗)\vec v_2=\sqrt{\mu/(ND)}\,\big(\vec D\times\vec r_2/r_2+\vec S\big).
How is semi-latus rectum recovered in Gibbs?
p=N/Dp=N/D, then ∣h⃗∣=μp|\vec h|=\sqrt{\mu p}.
Coplanarity test for Gibbs?
u^r1⋅(u^r2×u^r3)≈0\hat u_{r1}\cdot(\hat u_{r2}\times\hat u_{r3})\approx 0.
Position from a Gauss observation?
r⃗i=R⃗i+ρiρ^i\vec r_i=\vec R_i+\rho_i\hat\rho_i, unknown slant range ρi\rho_i.
What quantity does Gauss's 8th-degree polynomial solve for?
The middle radius r2r_2.
Which root of the Gauss polynomial is valid?
Positive real root above Earth's radius giving positive slant ranges.
Lagrange f,g short-arc approximations?
f≈1−μ2r23τ2f\approx1-\tfrac{\mu}{2r_2^3}\tau^2, g≈τ−μ6r23τ3g\approx\tau-\tfrac{\mu}{6r_2^3}\tau^3, τ=t−t2\tau=t-t_2.
Two conserved vectors underpinning orbit reconstruction?
Angular momentum h⃗=r⃗×v⃗\vec h=\vec r\times\vec v and eccentricity vector e⃗\vec e.
Why must Gauss be iterated?
The f,g series is truncated; refine with exact coefficients until ρ2\rho_2 converges.
Full pipeline from telescope angles to orbit?
Angles → Gauss (find r⃗i\vec r_i) → Gibbs (find v⃗2\vec v_2) → classical elements.

Connections

  • Two-body Problem — supplies r⃗¨=−μr⃗/r3\ddot{\vec r}=-\mu\vec r/r^3 used everywhere.
  • Angular Momentum & Eccentricity Vector — the invariants Gibbs rebuilds.
  • Lagrange Coefficients (f and g) — engine of Gauss's short-arc propagation.
  • Classical Orbital Elements — where (r⃗2,v⃗2)(\vec r_2,\vec v_2) finally goes.
  • Kepler's Equation — for propagating once the orbit is known.
  • Coordinate Frames (ECI, topocentric) — how ρ^i\hat\rho_i and R⃗i\vec R_i are formed.

Concept Map

needs 6 unknowns

angle-only data

position vectors

3 angle observations

recovers positions, feeds

input

Step 1

Step 2

N over D gives scale

S gives direction

from

from

conserved quantities

outputs

Orbit Determination

r and v or 6 elements

Gauss method

Gibbs method

alpha and delta at 3 times

r1 r2 r3 coplanar

Coplanarity check

Vectors N D S

Angular momentum h

Eccentricity vector e

Two-body dynamics

h fixes plane, e fixes shape

Velocity at r2

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, orbit determination ka basic sawaal simple hai: humein orbit ke 6 numbers chahiye (position aur velocity), lekin telescope se hum sirf direction (angle α,δ\alpha,\delta) dekh pate hain, ya phir sirf kuch positions measure kar pate hain. Poora orbit in thodi si observations se nikalna hi asli kaam hai.

Gibbs method tab use hota hai jab tumhare paas teen position vectors r⃗1,r⃗2,r⃗3\vec r_1,\vec r_2,\vec r_3 already ho. Interesting baat — isme time ki zaroorat hi nahi! Kyunki teen points jo ek hi plane mein aur ek hi conic pe hain, wo itni constraint dete hain ki beech wale point ki velocity v⃗2\vec v_2 pure vector algebra se nikal aati hai. Bas teen helper vectors N⃗,D⃗,S⃗\vec N,\vec D,\vec S banao aur formula v⃗2=μ/ND (D⃗×r⃗2/r2+S⃗)\vec v_2=\sqrt{\mu/ND}\,(\vec D\times\vec r_2/r_2+\vec S) laga do. Ho gaya full state.

Gauss method tab aata hai jab tumhare paas sirf angles hain, distance (slant range ρ\rho) pata nahi. Har observation deti hai r⃗i=R⃗i+ρiρ^i\vec r_i=\vec R_i+\rho_i\hat\rho_i, aur unknown ρi\rho_i ko nikalne ke liye Gauss coplanarity aur short-arc f,g series use karke ek 8th-degree polynomial banata hai jo r2r_2 dega. Positive real physical root lena zaroori hai. Fir wahi teen positions Gibbs ko de do velocity ke liye, aur iterate karke answer sharp karo.

Yaad rakhne ka mantra: "Angles se Gauss, positions se Gibbs, aur end mein velocity." Galti mat karna — Gibbs mein hamesha middle point ka velocity nikalta hai, aur time ki tension mat lo, Gibbs pure geometry hai.

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Connections

Two-body problem — equations of motion, reduction to one-bodyPhysics · 3.2.1Orbital elements (Keplerian) — semi-major axis a, eccentricity e, inclination i, RAAN Ω, argument of perigee ω, true anoPhysics · 3.2.8Kepler's equation M = E − e·sin E — derivation, eccentric anomalyPhysics · 3.2.14