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

This page assumes you have seen nothing. We build each symbol brick by brick, always with a picture, and only once a brick is placed do we allow ourselves to stack the next on top.

1 — A vector: an arrow with a length and a direction

Look at figure s01. The Earth's centre sits at a dot we call the origin. The arrow reaches out to where the satellite is right now. That arrow is the satellite's position.

Why do we need an arrow and not just a plain number? Because "where" in space is not one number — it needs three (left–right, forward–back, up–down). An arrow bundles all three into one object. We write those three numbers as a triple: $\vec{r}=(r_x,r_y,r_z).$ Each of $r_x,r_y,r_z$ is a component — how far the arrow reaches along one of three fixed directions. Those three fixed directions are the axes of a coordinate frame; for orbits around Earth we use one anchored to the sky (the ECI frame).

2 — Adding and subtracting arrows: vector addition (tip-to-tail)

Look at figure s02. This "tip-to-tail" hop is the workhorse of the whole topic: every satellite position is built by chaining arrows — known hops plus unknown hops — so that you can write down where the satellite is even when one leg of the journey is still unknown.

Why mention subtraction now? Because later orbit geometry constantly needs the arrow between two points — e.g. how the position changed from one observation to the next — and that "between" arrow is exactly a subtraction ${\vec{r}}_i-{\vec{r}}_j$.

3 — Scaling an arrow: scalar multiplication

So "travel a distance $\rho$ along direction $\hat{\rho }$" is written $\rho \hat{\rho }$: take the length-1 direction arrow and stretch it to length $\rho$. We will use exactly this in §8 to lay down the satellite's position. Scaling and adding together are all you need to build any arrow out of a few reference arrows — that fact underlies the Lagrange blend ${\vec{r}}_i=f_i{\vec{r}}_2+g_i{\dot{\vec{r}}}_2$ in §10.

4 — Length of a vector: $r=|\vec{r}|$

For a triple $(r_x,r_y,r_z)$ the length comes from Pythagoras in 3D: $r=|\vec{r}|=\sqrt{r_x^2+r_y^2+r_z^2}.$

In the worked example the parent computes $r_1=7360.5$ km straight from this rule (plug the three numbers into the formula above and you land on it). The little subscript ($1$, $2$, $3$) just labels which observation; the arrow and its length work the same way for each.

5 — Velocity $\vec{v}$ and the dots on top ($\dot{\vec{r}}$, $\ddot{\vec{r}}$)

Look at figure s03. The satellite sits on its curved path. The green arrow $\vec{r}$ reaches it from Earth's centre; the blue arrow $\vec{v}$ leaves the satellite along the path (tangent to it). The pink arrow $\ddot{\vec{r}}$ points back toward the Earth — gravity always pulls inward.

Why is acceleration the one that points at Earth? Because gravity is the only force out here, and gravity pulls toward the mass. Newton's law of gravity says the pull's strength is $\mu /r^2$ (weaker the farther out you go — the famous inverse-square), and its direction is straight back toward Earth, i.e. along $-\hat{r}$ where $\hat{r}=\vec{r}/r$ is the unit arrow pointing outward. Multiply strength by direction: $\ddot{\vec{r}}=-\frac{\mu }{r^2}\hat{r}=-\frac{\mu }{r^2}\cdot \frac{\vec{r}}{r}=-\frac{\mu }{r^3}\vec{r}.$

6 — The gravity number $\mu =GM$

Why bundle $G$ and $M$ into one symbol? Because in every orbit formula they only ever appear together as a product. Nature never lets you feel $G$ and $M$ separately from an orbit — only their combined pull. Naming that product $\mu$ saves ink and reflects the physics. It is the single knob that sets the pace of everything: bigger $\mu$ ⇒ faster orbits, tighter curves.

7 — The cross product $\vec{a}\times \vec{b}$

Look at figure s04. Two arrows $\vec{a}$ and $\vec{b}$ lie flat on the board; their cross product $\vec{a}\times \vec{b}$ rises straight up out of the board (right-hand rule), at a right angle to the sheet. The shaded parallelogram they span has an area, and that area is exactly the length $|\vec{a}\times \vec{b}|$: stretch the two arrows apart and the parallelogram (and the cross product) grows; line them up and the parallelogram flattens to nothing.

Why does this topic lean on cross products so heavily? Because we constantly need to answer "which way does the orbit plane face?" A plane is named by the arrow poking out of it (its normal). The cross product is the machine that hands you that normal. Two facts we will reuse:

• $\vec{a}\times \vec{b}$ reverses if you swap the order: $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b})$.
• If $\vec{a}$ and $\vec{b}$ point the same way, the parallelogram is flat, area $=0$, so the cross product is the zero arrow.

This is the engine of the angular-momentum vector $\vec{h}=\vec{r}\times \vec{v}$ and of every ${\vec{r}}_i\times {\vec{r}}_j$ term in Gibbs's helper vectors $\vec{N},\vec{D},\vec{S}$.

8 — The dot product $\vec{a}\cdot \vec{b}$ and the coplanarity check

Look at figure s05. The dot product equals the length of $\vec{b}$ times the length of the shadow $\vec{a}$ casts along $\vec{b}$ (its projection). When $\vec{a}$ leans the same way as $\vec{b}$ the shadow is long and positive; swing $\vec{a}$ to a right angle and the shadow vanishes to zero; push past $90^{\circ }$ and the shadow falls on the far side, giving a negative number.

We need the dot product for one job in the parent note: the coplanarity check. Write ${\hat{r}}_1,{\hat{r}}_2,{\hat{r}}_3$ for the three unit position vectors (each ${\hat{r}}_i={\vec{r}}_i/r_i$). Three arrows lie in one flat plane only if the third has zero height above the plane of the other two. The recipe: ${\hat{r}}_1\cdot ({\hat{r}}_2\times {\hat{r}}_3)\approx 0.$ Read it inside-out. First ${\hat{r}}_2\times {\hat{r}}_3$ gives the arrow poking out of the plane of vectors 2 and 3 (§7). Then dotting with ${\hat{r}}_1$ asks "does vector 1 have any height along that poking-out direction?" If the answer is zero, vector 1 lies flat in the same plane — all three are coplanar, and a single orbit can thread them.

9 — The line of sight $\hat{\rho }$, the slant range $\rho$, the observer $\vec{R}$

Now that scalar multiplication (§3) and vector addition (§2) are in hand, we can build the satellite position honestly. Stretch the length-1 sight-line ${\hat{\rho }}_i$ to the true distance by scaling it: ${\rho }_i{\hat{\rho }}_i$. Then hop tip-to-tail from Earth's centre to the telescope and continue along that stretched sight-line: ${\vec{r}}_i={\vec{R}}_i+{\rho }_i{\hat{\rho }}_i.$ Look at figure s06: the known hop ${\vec{R}}_i$ (blue) to the observer, then the stretched sight-line ${\rho }_i{\hat{\rho }}_i$ (pink) out to the yellow satellite, giving the total arrow ${\vec{r}}_i$ (yellow). Gauss's entire labour is finding the missing distances ${\rho }_i$. Once found, each ${\vec{r}}_i$ is known and Gibbs takes over.

10 — Angles that name a spot on the loop: $\alpha ,\delta ,\theta ,e,p,h$

First, one word every conic needs.

Look at figure s07: the ellipse with Earth at one focus, perigee (closest point) marked, the true anomaly $\theta$ opening from perigee to the satellite, the semi-major axis $a$ (half the long way across), and $p$ drawn straight up from the focus.

These meet in the single equation that defines every orbit, the conic (orbit) equation: $r=\frac{p}{1+e\cos \theta }.$ Read it: the distance $r$ from the focus depends only on where you are on the loop ($\theta$), scaled by $p$ and squashed by $e$. When $\theta =0$ (perigee) the bottom is largest, so $r$ is smallest — closest approach, exactly as "perigee" should be.

The bridge from these numbers to a name-tag for the orbit is the set of classical orbital elements, and pinning time to position uses Kepler's equation.

11 — Lagrange coefficients $f$ and $g$, and the time-gaps ${\tau }_i$

Why does such a blend exist at all? Because the orbit lives in one flat plane and ${\vec{r}}_2$, ${\vec{v}}_2$ are two arrows spanning that plane — every other point in the plane is some scaled-and-added combination of those two (§2 addition and §3 scaling, exactly). So $f$ measures "how much of the starting position" and $g$ measures "how much of the starting velocity" you need.

12 — How every foundation feeds the topic

The two boxes below name the two methods the parent note builds on top of these foundations; this page supplies the arrows, lengths, products and series they consume.

Equipment checklist

Test yourself — say the answer out loud before revealing.