Chapter 8: Kepler's Laws and the Two-Body Problem

"I first believed I was dreaming… But it is absolutely certain and exact that the ratio which exists between the periodic times of any two planets is precisely the ratio of the 3/2th powers of the mean distances." — Johannes Kepler, Harmonices Mundi (1619), on discovering his third law

Overview

For two thousand years, everyone knew that the heavens moved in circles. Circles were perfect, and the sky was perfect, so the planets must trace circles — and when they stubbornly refused to, astronomers added circles riding on circles (epicycles) until the models groaned under dozens of them and still missed. Then a nearsighted German mathematician named Johannes Kepler got hold of the most accurate planetary observations ever made, spent years failing to fit Mars to a circle, and finally did the unthinkable: he threw the circle away. Mars, he found, moves on an ellipse. That single act ended the ancient astronomy and began the modern one.

What Kepler could only describe, Newton would later explain. Kepler's three laws are the empirical shape of planetary motion — where a planet is, how fast it sweeps, how long it takes. Newton's law of gravity is the cause. And here is the moment this chapter is built around, the one that never stops being beautiful: Kepler's laws are not laws at all. They are theorems. Every one of them falls out of a single line — $\mathbf{F} = -\mu m\,\mathbf{r}/r^3$, the inverse-square gravity you met in Chapter 2 — when you turn the crank of calculus. The ellipse is not an assumption you feed in; it is a consequence you get out. Change the exponent in gravity from $2$ to anything else and the orbits stop closing into neat ellipses altogether. The specific, elegant shape of every orbit in the universe is the fingerprint of that exponent.

This is where the orbital-energy viewpoint of Chapter 6 becomes geometry. There we learned to compute an orbit's speed and energy; here we learn its shape, its orientation in space, and where the spacecraft actually is at a given clock time — the questions a navigator, a ground station, and a rendezvous pilot all need answered. By the end you will be able to pin down any orbit in the solar system with just six numbers, and predict where its occupant will be next Tuesday.

In this chapter, you will learn to:

• State Kepler's three laws and see them at work in real satellite orbits.
• Derive all three from Newton's gravity — the ellipse, the equal areas, and the period law — and, as a bonus, prove the energy relation $\varepsilon = -\mu/(2a)$ that Chapter 6 borrowed on credit.
• Read any orbit as a conic section set by its eccentricity, from a circle to an escaping hyperbola.
• Specify an orbit exactly with the six orbital elements — the "GPS coordinates" of a trajectory.
• Compute periods with Kepler's third law, and find a body's position in time by solving Kepler's equation.

Learning Paths

🚀 Space Enthusiast: Read 8.1 for the three laws and the story, then 8.3 (orbits as conic sections) and 8.4 (the six elements) — those two sections are what let you read any orbit you meet. You can skim the derivation in 8.2 and simply enjoy that the ellipse is forced by gravity.

📐 Engineering Student: Read everything and do the 8.2 derivation yourself before reading ours; it is the backbone of astrodynamics and it reappears in every later chapter of Part II. Section 8.6 (anomalies and Kepler's equation) is the one you will implement most often — master the iteration.

🎮 KSP Player: The map/orbit readout in your game is the six elements of 8.4, and the "time to apoapsis" counter is Kepler's equation (8.6) running live. Focus on 8.3–8.6 to understand what the numbers on your screen actually mean, then check your instincts against the worked examples.

🛰️ Industry Prep: The six orbital elements (8.4) and the anomaly relations (8.6) are the daily vocabulary of flight dynamics and the format of every two-line element set and ephemeris. Your Mission Design Checkpoint records your target orbit's complete element set — the backbone of your MDR.

8.1 Kepler's three laws

Before we explain planetary motion, we should say precisely what needs explaining. Working entirely by hand, without a telescope powerful enough to matter and certainly without calculus, Kepler distilled Tycho Brahe's mountain of naked-eye observations into three crisp statements. They remain exactly true for any two bodies bound by gravity, and every one of them will earn a physical meaning in the sections that follow.

Definition (Kepler's first law — the law of ellipses). The orbit of each planet is an ellipse with the Sun at one focus (not the center). For a satellite, read "Earth" for "Sun": a bound orbit is an ellipse with the primary body at one focus.

An ellipse is an oval with two special interior points, the foci. The Sun sits at one focus; the other is empty. The planet's distance from the Sun therefore changes over an orbit — closest at the point called perihelion (perigee, for Earth orbits), farthest at aphelion (apogee). A circle is the special case where the two foci merge at the center and the distance never changes. This is why we spent Chapter 6 watching a spacecraft speed up and slow down: on an ellipse, it must.

Definition (Kepler's second law — the law of equal areas). The line joining a planet to the Sun sweeps out equal areas in equal times. Near the Sun the planet moves fast; far from it, slow — in exactly the proportion that keeps the swept area's rate constant.

The second law is Kepler's most underappreciated, and it is really a statement about speed. Because the Sun–planet line must sweep area at a steady rate, a planet near perihelion (short line) has to move quickly to sweep the same area a slow planet sweeps with its long line at aphelion. It is the precise form of "low is fast, high is slow." We will see in §8.2 that it is nothing but the conservation of angular momentum in disguise.

Definition (Kepler's third law — the law of periods). The square of a planet's orbital period $T$ is proportional to the cube of its orbit's semi-major axis $a$: $T^2 \propto a^3$. For bodies orbiting the same primary, $T^2/a^3$ is the same constant for all of them.

The third law is the one that connects different orbits to each other. Double the size of an orbit and the period does not double — it grows by a factor of $2^{3/2} \approx 2.83$. Big orbits are slow not just in speed but in the total time to get around, and the third law says exactly how slow.

💡 Intuition: the second law is a planet's throttle you cannot touch. Picture the Sun–planet line as the hand of a clock whose length breathes in and out over the orbit. Kepler's second law says the area the hand sweeps per second is fixed. When the hand is short (perihelion), it must swing fast to keep the area rate up; when it is long (aphelion), it swings slowly. The planet has no choice — gravity enforces the schedule. This is why a comet spends decades crawling through the outer solar system and then whips around the Sun in a matter of weeks.

Let us make the third law concrete with two orbits you already know, both around Earth, using $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$ (from Chapter 2).

Worked Example: is $T^2/a^3$ really the same for LEO and GEO? Take a $400\ \text{km}$ low orbit ($a = 6{,}771\ \text{km}$, period $T \approx 5{,}545\ \text{s}$, which we will compute in §8.5) and a geostationary orbit ($a = 42{,}164\ \text{km}$, $T \approx 86{,}164\ \text{s}$). Kepler's third law claims the ratio $T^2/a^3$ is identical. Check it. $$\frac{T^2}{a^3}\bigg|_{\text{LEO}} = \frac{(5{,}545)^2}{(6{,}771)^3} = \frac{3.075\times10^{7}}{3.104\times10^{11}} = 9.90\times10^{-5}\ \text{s}^2/\text{km}^3.$$ $$\frac{T^2}{a^3}\bigg|_{\text{GEO}} = \frac{(86{,}164)^2}{(42{,}164)^3} = \frac{7.424\times10^{9}}{7.496\times10^{13}} = 9.90\times10^{-5}\ \text{s}^2/\text{km}^3.$$ The two orbits differ in size by a factor of six and in period by a factor of fifteen, yet $T^2/a^3$ agrees to three figures. That shared constant, we will prove, is exactly $4\pi^2/\mu_\oplus = 39.48/(3.986\times10^{5}) = 9.90\times10^{-5}\ \text{s}^2/\text{km}^3$ — the same number, out of pure gravity. The units check: $\text{s}^2/\text{km}^3 = (\text{km}^3/\text{s}^2)^{-1} = 1/\mu$.

📜 From History: eight minutes of arc. Kepler inherited the observations of Tycho Brahe, whose naked-eye instruments measured planetary positions to about one arcminute — the practical limit before telescopes. Kepler's best circular model of Mars fit almost everything, but it missed the data by up to eight arcminutes. A lesser scientist would have blamed the observations. Kepler, who trusted Tycho's precision, wrote that those eight minutes "led the way to the reformation of all of astronomy." He refused to round away a discrepancy smaller than a sixth of the width of a full Moon, and it cost him years and won him the ellipse. It is the founding parable of this book's obsession with sanity-checking numbers: the mistakes hide in the last digits you were tempted to ignore.

8.2 Deriving Kepler's laws from Newton's gravity

Kepler found what. Newton found why, and in doing so turned three empirical rules into three theorems. This section is the heart of the chapter and the most beautiful thing in Part II: watch three separate "laws of the heavens" tumble out of one equation. If you read only one derivation in this book slowly, make it this one.

The two-body problem. Two point masses attract each other by gravity and nothing else acts. Find their motion. When one mass hugely exceeds the other — Earth and a satellite, the Sun and a planet — the big body is essentially fixed and we track the small one's position vector $\mathbf{r}$ from the big one's center. This is the two-body problem, and it is the only orbital problem with an exact, closed-form solution. Everything in Part II is a refinement of it.

From Chapter 2, Newton's gravity on the small body gives it an acceleration toward the primary of magnitude $\mu/r^2$, directed along $-\hat{\mathbf{r}}$ (inward). So the entire dynamics is the single vector equation

$$ \ddot{\mathbf{r}} = -\frac{\mu}{r^{2}}\,\hat{\mathbf{r}} = -\frac{\mu}{r^{3}}\,\mathbf{r}, $$

where $\hat{\mathbf{r}} = \mathbf{r}/r$ is the outward unit vector, which we used to rewrite the force in terms of the full position vector $\mathbf{r}$ in the second form. (Strictly, $\mu = G(M+m)$; for $m \ll M$ it is just $GM$.) That is the whole physics. The rest is bookkeeping — but the bookkeeping is glorious.

Kepler's second law: angular momentum is conserved

Strategy first. We will show the spacecraft's specific angular momentum — angular momentum per unit mass, $\mathbf{h} = \mathbf{r}\times\dot{\mathbf{r}}$ — never changes. Two facts fall out immediately: the orbit lies in a fixed plane, and the Sun–planet line sweeps area at a constant rate. The second is Kepler's second law exactly.

Take the time derivative of $\mathbf{h} = \mathbf{r}\times\dot{\mathbf{r}}$:

$$ \dot{\mathbf{h}} = \dot{\mathbf{r}}\times\dot{\mathbf{r}} + \mathbf{r}\times\ddot{\mathbf{r}}. $$

The first term is a vector crossed with itself, which is zero. In the second, substitute the equation of motion $\ddot{\mathbf{r}} = -(\mu/r^{3})\mathbf{r}$:

$$ \mathbf{r}\times\ddot{\mathbf{r}} = \mathbf{r}\times\left(-\frac{\mu}{r^{3}}\mathbf{r}\right) = -\frac{\mu}{r^{3}}\,(\mathbf{r}\times\mathbf{r}) = \mathbf{0}. $$

So $\dot{\mathbf{h}} = \mathbf{0}$: specific angular momentum $\mathbf{h}$ is constant in both magnitude and direction. Because $\mathbf{h} = \mathbf{r}\times\dot{\mathbf{r}}$ is perpendicular to both $\mathbf{r}$ and $\mathbf{v}$, and it points in a fixed direction, the position and velocity are forever confined to the single plane perpendicular to $\mathbf{h}$. An orbit is flat. That is not obvious a priori — it is a gift of the central force.

Now the area. In a small time $dt$ the spacecraft moves by $d\mathbf{r} = \mathbf{v}\,dt$, and the thin triangle swept by the position vector has area $dA = \tfrac{1}{2}\lvert\mathbf{r}\times d\mathbf{r}\rvert = \tfrac{1}{2}\lvert\mathbf{r}\times\mathbf{v}\rvert\,dt = \tfrac{1}{2}h\,dt$. Therefore

$$ \frac{dA}{dt} = \frac{h}{2} = \text{constant}. $$

Equal areas in equal times — Kepler's second law, proven, and revealed to be nothing more than the conservation of angular momentum. $\blacksquare$

🔗 Connection: this is the $r_p v_p = r_a v_a$ we used in Chapter 6. At perigee and apogee the velocity is purely sideways (perpendicular to $\mathbf{r}$), so there $h = \lvert\mathbf{r}\times \mathbf{v}\rvert = r v$. Constant $h$ then forces $r_p v_p = r_a v_a$ — the exact relation Chapter 6 invoked to show the closest point is the fastest. We were quietly using Kepler's second law all along; now we own it.

Kepler's first law: the orbit is a conic

Strategy first. We want the shape of the path — $r$ as a function of the angle $\nu$. The trick, due to the same vector algebra, is to cross the equation of motion with the constant vector $\mathbf{h}$, integrate once, and then take a dot product with $\mathbf{r}$. Out comes $r(\nu)$ with no guessing. A constant vector of integration appears, and it turns out to be the eccentricity, pointing at perigee.

Cross the equation of motion with $\mathbf{h}$ (a constant, so it slides in and out of the derivative):

$$ \ddot{\mathbf{r}}\times\mathbf{h} = -\frac{\mu}{r^{3}}\,\mathbf{r}\times\mathbf{h}. $$

The left side is $\dfrac{d}{dt}(\dot{\mathbf{r}}\times\mathbf{h})$. The right side, using $\mathbf{h} = \mathbf{r}\times\dot{\mathbf{r}}$ and the vector identity $\mathbf{r}\times(\mathbf{r}\times\dot{\mathbf{r}}) = \mathbf{r}(\mathbf{r}\cdot\dot{\mathbf{r}}) - \dot{\mathbf{r}}(\mathbf{r}\cdot\mathbf{r})$, works out (with $\mathbf{r}\cdot\dot{\mathbf{r}} = r\dot r$ and $\mathbf{r}\cdot\mathbf{r} = r^2$) to exactly $\mu\,\dfrac{d}{dt}\!\left(\dfrac{\mathbf{r}}{r}\right)$. So both sides are time derivatives, and we integrate once:

$$ \dot{\mathbf{r}}\times\mathbf{h} = \mu\,\frac{\mathbf{r}}{r} + \mathbf{B}, $$

where $\mathbf{B}$ is a constant vector of integration lying in the orbital plane. Now dot both sides with $\mathbf{r}$. On the left, the scalar triple product $\mathbf{r}\cdot(\dot{\mathbf{r}}\times\mathbf{h}) = (\mathbf{r}\times\dot{\mathbf{r}})\cdot\mathbf{h} = \mathbf{h}\cdot\mathbf{h} = h^2$. On the right, $\mu\,\mathbf{r}\cdot\mathbf{r}/r + \mathbf{r}\cdot \mathbf{B} = \mu r + rB\cos\nu$, where $\nu$ is the angle between $\mathbf{r}$ and the fixed vector $\mathbf{B}$. Thus

$$ h^{2} = \mu r + rB\cos\nu \quad\Longrightarrow\quad r = \frac{h^{2}/\mu}{1 + (B/\mu)\cos\nu}. $$

Define $e \equiv B/\mu$ and $p \equiv h^2/\mu$, and we have arrived:

$$ \boxed{\ r = \frac{p}{1 + e\cos\nu}\ } $$

This is the polar equation of a conic section with the focus at the origin — the primary body. It is Kepler's first law: bound orbits ($e < 1$) are ellipses with the primary at a focus. The constant of integration $\mathbf{B}$ points toward perigee (where $\cos\nu = 1$ makes $r$ smallest), and its size sets the eccentricity $e$. The ellipse was never assumed; it was forced by the $1/r^2$ in gravity. $\blacksquare$

🚪 Threshold Concept: the shape of every orbit is the signature of the inverse square. The single most reorganizing idea in orbital mechanics is that the elegant closed ellipse is not a coincidence and not an axiom — it is a theorem, and it depends delicately on gravity going as $1/r^2$ exactly. Only two force laws in all of physics produce orbits that close into simple, non-precessing curves: the inverse-square force (gravity) and the linear spring force. For any other exponent, an orbit would trace a slowly rotating rosette that never repeats. That our sky is full of clean, repeating ellipses is direct evidence that gravity is inverse-square — the orbits are the exponent, written across the heavens. Once you see Kepler's laws as consequences rather than rules, every orbit becomes a readout of the force that made it, and the whole subject snaps into a single beautiful piece. This is the chapter's assigned theme — orbital mechanics is beautiful — in its purest form.

Kepler's third law: the period follows from the area

We now have both the area rate ($dA/dt = h/2$) and the shape (an ellipse of semi-major axis $a$ and semi-minor axis $b$). Over one full period $T$, the position vector sweeps the entire area of the ellipse, which is $\pi a b$. Since area accrues at the steady rate $h/2$,

$$ \pi a b = \frac{h}{2}\,T \quad\Longrightarrow\quad T = \frac{2\pi a b}{h}. $$

Two geometric facts finish it. For an ellipse, $b = a\sqrt{1-e^2}$, and the semi-latus rectum satisfies $p = a(1-e^2) = h^2/\mu$, so $h = \sqrt{\mu\,a(1-e^2)}$. Substitute:

$$ T = \frac{2\pi a\cdot a\sqrt{1-e^2}}{\sqrt{\mu\,a(1-e^2)}} = \frac{2\pi a^{2}\sqrt{1-e^2}}{\sqrt{\mu a}\,\sqrt{1-e^2}} = \frac{2\pi a^{2}}{\sqrt{\mu a}} = 2\pi\sqrt{\frac{a^{3}}{\mu}}. $$

Squaring, $T^2 = \dfrac{4\pi^2}{\mu}\,a^3$ — Kepler's third law, with the proportionality constant $4\pi^2/\mu$ now explicit and identical for every orbit around the same primary, exactly as the LEO/GEO check in §8.1 found. Notice that $e$ has completely vanished: the period depends only on $a$, not on the orbit's shape. A nearly circular orbit and a wildly elongated one with the same semi-major axis take exactly the same time to go around. $\blacksquare$

🧩 Productive Struggle. Before reading on: two spacecraft leave the same point at the same instant, one on a circular orbit of radius $a$, the other on a thin, cigar-shaped ellipse of the same semi-major axis $a$ (it dives close to Earth, then swings far out). The third law says they return to the start together, with identical periods. How can that be, when the ellipse's path is so much longer? (Hint: think about where each spends its time, and use the second law.)

A bonus: the energy relation Chapter 6 borrowed

Chapter 6 used the relation $\varepsilon = -\mu/(2a)$ for a general ellipse but only proved it for a circle, promising the full proof here. We can now pay that debt in four lines, using the tools just built. Specific orbital energy $\varepsilon = v^2/2 - \mu/r$ is conserved (Chapter 6), so evaluate it at perigee, where the algebra is cleanest. There the velocity is purely tangential, so $v_p = h/r_p$; and $r_p = a(1-e)$ with $h^2 = \mu a(1-e^2)$. Then

$$ \varepsilon = \frac{v_p^{2}}{2} - \frac{\mu}{r_p} = \frac{h^{2}}{2r_p^{2}} - \frac{\mu}{r_p} = \frac{\mu a(1-e^2)}{2a^{2}(1-e)^{2}} - \frac{\mu}{a(1-e)}. $$

Using $1-e^2 = (1-e)(1+e)$ in the first term, it becomes $\dfrac{\mu(1+e)}{2a(1-e)}$, and

$$ \varepsilon = \frac{\mu(1+e)}{2a(1-e)} - \frac{\mu}{a(1-e)} = \frac{\mu}{a(1-e)}\left(\frac{1+e}{2} - 1\right) = \frac{\mu}{a(1-e)}\cdot\frac{e-1}{2} = -\frac{\mu}{2a}. $$

So $\varepsilon = -\mu/(2a)$ for any ellipse, not just a circle — and with it, the vis-viva equation $v^2 = \mu(2/r - 1/a)$ of Chapter 6 stands fully justified for every orbit. $\blacksquare$ The energy picture and the geometry picture are two views of one object.

🔄 Check Your Understanding 1. Which conserved quantity is Kepler's second law equivalent to, and why does it also force the orbit to lie in a plane? 2. In the third law $T^2 = (4\pi^2/\mu)\,a^3$, the eccentricity $e$ does not appear. State in words what that means for two orbits of the same $a$ but different shapes. 3. The orbit equation is $r = p/(1 + e\cos\nu)$. At what value of $\nu$ is $r$ smallest, and what is that smallest $r$ in terms of $a$ and $e$?

Answers

1. Angular momentum (specific angular momentum $\mathbf{h} = \mathbf{r}\times\mathbf{v}$). It is conserved because gravity is a central force ($\ddot{\mathbf{r}}\parallel\mathbf{r}$, so $\mathbf{r}\times\ddot{\mathbf{r}} = 0$). Its direction is fixed, and since $\mathbf{r}$ and $\mathbf{v}$ are always perpendicular to that fixed direction, they stay in one plane. 2. They have exactly the same period — the same time to complete one orbit — even though one may be a circle and the other a long thin ellipse. Period cares only about orbit size ($a$), never shape. 3. Smallest at $\nu = 0$ (perigee), where $\cos\nu = 1$: $r_{\min} = p/(1+e) = a(1-e)$.

8.3 The orbit as a conic section

The equation we derived, $r = p/(1 + e\cos\nu)$, is worth dwelling on, because a single parameter in it — the eccentricity $e$ — decides the entire character of the trajectory. As $e$ runs from zero upward, the orbit morphs continuously from a circle, through ever more elongated ellipses, to the open escape curves. The same formula describes a communications satellite, a returning capsule, and an interstellar probe. That unity is the payoff of the two-body problem.

Definition (conic section). A conic section is any curve obtained by slicing a cone with a plane — a circle, ellipse, parabola, or hyperbola. Every solution of the two-body problem is a conic with the primary body at one focus, and the eccentricity $e$ selects which one:

Eccentricity	Conic	Energy $\varepsilon$	Bound?
$e = 0$	circle	$\varepsilon < 0$	yes
$0 < e < 1$	ellipse	$\varepsilon < 0$	yes
$e = 1$	parabola	$\varepsilon = 0$	marginal escape
$e > 1$	hyperbola	$\varepsilon > 0$	no (escapes)

The Greeks studied these four curves as slices of a cone two thousand years before anyone dreamed they were the paths of the planets — one of the great "pure math becomes physics" surprises in the history of science. Notice that the eccentricity's verdict (bound or free) matches exactly the energy sign we read in Chapter 6: they are two languages for the same fact. There is even a clean bridge between them, $e = \sqrt{1 + 2\varepsilon h^2/\mu^2}$, which makes $e<1 \Leftrightarrow \varepsilon<0$ visible at a glance. (You need not memorize it; the table is what you will use.)

Definition (eccentricity and semi-major axis, geometric). For an elliptical orbit, the semi-major axis $a$ is half the length of the long axis — the average of the closest and farthest distances — and the eccentricity $e$ measures how stretched the ellipse is, from $0$ (a circle) toward $1$ (a parabola). They connect the extremes of the orbit to its size and shape through $$r_p = a(1-e), \qquad r_a = a(1+e),$$ and, inverting, $a = \tfrac{1}{2}(r_p + r_a)$ and $e = \dfrac{r_a - r_p}{r_a + r_p}$, where $r_p$ is the periapsis (closest) radius and $r_a$ the apoapsis (farthest) radius. This is the same $a$ whose energy meaning, $\varepsilon = -\mu/(2a)$, you met in Chapter 6 (🔗) — now given its geometric face.

Worked Example: reading the geostationary transfer orbit as a conic. A satellite bound for GEO rides the transfer ellipse of Chapter 6: perigee at $r_p = 6{,}771\ \text{km}$ (a $400\ \text{km}$ parking orbit) and apogee at $r_a = 42{,}164\ \text{km}$ (GEO altitude). Its geometry, straight from the definitions: $$a = \frac{r_p + r_a}{2} = \frac{6{,}771 + 42{,}164}{2} = 24{,}468\ \text{km},$$ $$e = \frac{r_a - r_p}{r_a + r_p} = \frac{42{,}164 - 6{,}771}{42{,}164 + 6{,}771} = \frac{35{,}393}{48{,}935} = 0.723.$$ The semi-latus rectum is $p = a(1-e^2) = 24{,}468\,(1 - 0.5227) = 11{,}678\ \text{km}$. With those, the orbit equation predicts the radius at any point. For instance, at a true anomaly of $\nu = 90^\circ$ (a quarter-turn past perigee, where $\cos\nu = 0$), $r = p/(1 + 0) = p = 11{,}678\ \text{km}$ — the satellite is $11{,}677\ \text{km}$ from Earth's center at that moment. Sanity check: this lies between perigee ($6{,}771$) and apogee ($42{,}164$), as it must, and $e = 0.723$ places the orbit firmly in the "elongated ellipse" band ($0 < e < 1$), so it is bound — correct for a satellite that has not yet escaped.

⚠️ Common Misconception: "orbits are egg-shaped, fat at one end." An ellipse is perfectly symmetric about both its axes — it is not an egg or a teardrop. What is not symmetric is where the primary body sits: at a focus, off to one side, not at the center. So the orbit looks balanced, but the planet is close to one end and far from the other, which is why perigee and apogee speeds differ so wildly. Many people also misplace the primary at the center of the ellipse; Kepler's whole revolution was moving it to the focus. The center of the ellipse is an empty point in space with no physical significance — except, as we will see in §8.6, as the vantage point for measuring the eccentric anomaly.

🔄 Check Your Understanding 1. An orbit has $e = 0$. What conic is it, and what is special about the primary's location? 2. A comet is measured to have $e = 1.2$. Is it bound to the Sun? Will it return? 3. For an ellipse with $a = 10{,}000\ \text{km}$ and $e = 0.4$, find the perigee and apogee radii.

Answers

1. A circle. The two foci coincide at the center, so the primary sits at the center and the distance $r = a$ never changes. 2. No — $e > 1$ is a hyperbola with $\varepsilon > 0$; it is unbound and will leave the solar system, never to return. (Interstellar visitors like 1I/ʻOumuamua were identified exactly this way.) 3. $r_p = a(1-e) = 10{,}000(0.6) = 6{,}000\ \text{km}$; $r_a = a(1+e) = 10{,}000(1.4) = 14{,}000\ \text{km}$.

8.4 The six orbital elements

We can now describe the shape of an orbit ($a$, $e$) and locate a body within its plane ($\nu$). But an orbit also floats in three-dimensional space at some tilt and swivel, and to specify it completely — to tell a ground station exactly where to point, or a navigator exactly what trajectory a spacecraft is on — we need six numbers. Why six? Because the state of a body is its position and velocity, three components each: six numbers. The orbital elements repackage those same six degrees of freedom into a set that is far more meaningful, because most of them don't change as the body coasts.

Definition (orbital elements). The six classical (Keplerian) orbital elements specify an orbit and a body's place on it, relative to a chosen reference plane (for Earth, the equatorial plane) and reference direction (the vernal equinox, a fixed direction in the sky). Two fix the size and shape, two orient the orbital plane in space, one orients the ellipse within that plane, and one locates the body:

Element	Symbol	What it sets
Semi-major axis	$a$	the size of the orbit (and, via Kepler III, the period)
Eccentricity	$e$	the shape — how elongated, from circle to escape
Inclination	$i$	the tilt of the orbital plane from the reference plane
Right ascension of the ascending node	$\Omega$	the swivel of the plane about the pole
Argument of periapsis	$\omega$	the orientation of the ellipse within its plane
True anomaly	$\nu$	where the body is on the orbit right now

Let us give the four new ones (beyond $a$ and $e$) their physical meaning, because the names are forbidding and the ideas are simple.

• Inclination $i$ is the angle between the orbital plane and the reference (equatorial) plane. An $i = 0^\circ$ orbit hugs the equator; $i = 90^\circ$ is a polar orbit passing over both poles; $i = 51.6^\circ$ is the International Space Station's tilt, chosen so it launches from and overflies a wide band of the Earth. Inclination is the single most expensive element to change once you are in orbit — a fact that will dominate Chapter 10.

• Right ascension of the ascending node $\Omega$ (RAAN) fixes how the tilted plane is rotated about Earth's axis. The ascending node is the point where the orbit crosses the reference plane going south-to-north; $\Omega$ is the angle from the reference direction (the vernal equinox) around to that node. If $i$ tells you how much the plane is tipped, $\Omega$ tells you which way the tip faces.

• Argument of periapsis $\omega$ is the angle, measured in the orbital plane, from the ascending node to periapsis. It says which way the ellipse points within its plane — whether the low point sits over the equator, the north, or the south. For a Molniya communications orbit, $\omega$ is deliberately set to $270^\circ$ so that apogee (where the satellite lingers) hangs over the northern hemisphere.

• True anomaly $\nu$ is the angle, measured at the focus (the primary), from periapsis to the body's current position — the one element that is always changing as the body moves. It answers "where is it now?" The other five are (in the ideal two-body world) constant.

💡 Intuition: the elements are an orbit's mailing address. Think of specifying a seat in a stadium. $a$ and $e$ are the size and shape of the bowl; $i$ and $\Omega$ say how the bowl is oriented on the ground; $\omega$ says which way row 1 faces; and $\nu$ is your seat number along the row. Five of the six are properties of the stadium and never change; only $\nu$, your position, advances with time. That is why flight dynamicists love the elements: they turn a fast-moving six-dimensional state into five constants and one clock.

🔧 Engineering Reality: where the elements actually come from — the TLE. In practice, no one hands you clean orbital elements; you extract them from tracking data (Chapter 13 is devoted to this inverse problem). The most common public format is the two-line element set (TLE) published by U.S. Space Command for every tracked object. Buried in its cryptic two lines of digits are exactly the quantities of this section: inclination, RAAN, eccentricity, argument of perigee, mean anomaly (not true — see §8.6), and the mean motion (from which $a$ follows by Kepler III). Feed a TLE and a date into a propagator and it returns a position in the sky. Every satellite-tracking app on your phone is doing precisely this, six elements at a time.

⚠️ Common Misconception: "an orbit needs the primary's mass as a seventh element." It does not. The six elements describe the geometry and phase of the orbit; the primary's gravitational parameter $\mu$ is a property of the body you orbit, not of the orbit, and it enters only when you convert elements into times and speeds (period, mean motion, vis-viva). Give someone your six elements and tell them you orbit Earth, and $\mu_\oplus$ is understood. Six numbers plus a named primary is a complete specification.

🔄 Check Your Understanding 1. Of the six elements, which one changes continuously as the spacecraft coasts, and which five are constant in the ideal two-body problem? 2. Which element would you change to move a satellite from an equatorial orbit to one that passes over the poles? Roughly what value would it take? 3. Two satellites share identical $a$, $e$, $i$, $\Omega$, and $\omega$ but have different $\nu$. What is the same about them, and what is different?

Answers

1. True anomaly $\nu$ changes continuously (it is the body's moving position); $a$, $e$, $i$, $\Omega$, $\omega$ are constant. (In real orbits, perturbations slowly drift the "constants" — that is Chapter 12.) 2. Inclination $i$; a polar orbit has $i \approx 90^\circ$. 3. They are on the exact same orbit (same ellipse, same plane, same orientation); they differ only in where along it each currently sits — one is ahead of the other. This is the situation of two satellites in a train, or a chaser and its target before rendezvous.

8.5 Period from Kepler's third law

We derived $T = 2\pi\sqrt{a^3/\mu}$ in §8.2; now let us put it to work, because the period and its cousin the mean motion are the bridge between geometry and the clock. This is the equation you will reach for whenever a question involves time: how often a satellite passes overhead, how long a transfer takes, when a constellation repeats its ground track.

Definition (mean motion). The mean motion $n$ is the average angular rate of a body around its orbit — the angle it would advance per second if it moved uniformly: $$n = \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^{3}}}.$$ It has units of radians per second (or, in the TLE world, revolutions per day). Real orbital motion is not uniform (Kepler's second law), but $n$ is the steady rate that carries the mean anomaly of §8.6, and it makes many period calculations one step shorter.

Worked Example: the period of an ISS-like low orbit. Take a circular orbit at $400\ \text{km}$ altitude, $a = r = 6{,}371 + 400 = 6{,}771\ \text{km}$. Then $$T = 2\pi\sqrt{\frac{a^{3}}{\mu_\oplus}} = 2\pi\sqrt{\frac{(6{,}771)^{3}}{3.986\times10^{5}}} = 2\pi\sqrt{\frac{3.104\times10^{11}}{3.986\times10^{5}}} = 2\pi\sqrt{7.788\times10^{5}}.$$ Now $\sqrt{7.788\times10^{5}} = 882.5\ \text{s}$, so $T = 2\pi\,(882.5) = 5{,}545\ \text{s}$. Convert: $5{,}545/60 = 92.4\ \text{minutes}$. Sanity check: the real International Space Station, orbiting a bit higher near $420\ \text{km}$, circles Earth in about $93\ \text{minutes}$ — our number is right in the pocket, a little shorter because a slightly lower orbit is a slightly faster one. The units check too: $\sqrt{\text{km}^3/(\text{km}^3/\text{s}^2)} = \sqrt{\text{s}^2} = \text{s}$. This is why astronauts see roughly sixteen sunrises a day: $1{,}440\ \text{min} / 92.4\ \text{min} \approx 15.6$ orbits.

Worked Example: does geostationary orbit really take 23 h 56 m? A geostationary satellite must circle Earth exactly once per rotation so it hovers over one spot. Its defining radius is $a = 42{,}164\ \text{km}$. Kepler's third law gives $$T = 2\pi\sqrt{\frac{(42{,}164)^{3}}{3.986\times10^{5}}} = 2\pi\sqrt{\frac{7.496\times10^{13}}{3.986\times10^{5}}} = 2\pi\sqrt{1.8806\times10^{8}} = 2\pi\,(13{,}713\ \text{s}) = 86{,}164\ \text{s}.$$ Convert to hours, minutes, seconds: $86{,}164\ \text{s} = 86{,}164/3600 = 23.93\ \text{h} = 23\ \text{h} + 0.93\times60\ \text{min} = 23\ \text{h}\ 56\ \text{min}$. That is the famous answer — and it is not 24 hours. A geostationary satellite matches Earth's sidereal rotation period (one turn relative to the stars, $23\ \text{h}\ 56\ \text{min}\ 4\ \text{s} = 86{,}164\ \text{s}$), not the 24-hour solar day. Our computed $86{,}164\ \text{s}$ lands right on the sidereal day — the geostationary radius is defined to make it so. The four extra minutes matter: they are the reason the stars rise four minutes earlier each night, and the reason "geostationary" tracks the ground and not the Sun.

📜 From History: Newton's Principia and the "system of the world." Newton published the derivation you worked through in §8.2 in his 1687 Philosophiæ Naturalis Principia Mathematica. He proved that an inverse-square force yields Kepler's ellipses, equal areas, and the period law — and then, in a stroke of breathtaking unification, argued that the same force pulling an apple to the ground holds the Moon in its orbit. To do it he had to invent much of calculus. The general public could not follow the geometry, but they grasped the consequence: the heavens and the Earth obey one law. It is, arguably, the single most important calculation ever performed, and you have just retraced it.

🔄 Check Your Understanding 1. A satellite's semi-major axis is doubled. By what factor does its orbital period change? 2. Why is the geostationary period 23 h 56 m rather than 24 h 00 m? 3. What is the mean motion $n$ (in rad/s) of the $400\ \text{km}$ orbit above, and what does it represent?

Answers

1. By $2^{3/2} = 2\sqrt{2} \approx 2.83$. Period scales as $a^{3/2}$, so doubling $a$ multiplies $T$ by about $2.83$, not $2$. 2. Because a geostationary satellite matches Earth's rotation relative to the stars (the sidereal day, $23\ \text{h}\ 56\ \text{m}$), which is about four minutes shorter than the 24-hour solar day (the extra four minutes are Earth's daily progress around the Sun). 3. $n = 2\pi/T = 2\pi/5{,}545 = 1.133\times10^{-3}\ \text{rad/s}$ — the average angular speed of the satellite around Earth, about $0.065^\circ$ per second.

8.6 True, eccentric, and mean anomaly; Kepler's equation

We save the subtlest question for last, and it is the one mission operations lives on: given the time, where is the spacecraft? Kepler's second law is the obstacle. Because the body speeds up and slows down, its angular position (the true anomaly $\nu$) does not advance uniformly with time — you cannot just multiply an angular rate by the clock. Solving this cost Kepler himself enormous effort and produced one of the most famous equations in astronomy, an equation with no algebraic solution that we still crack by iteration today.

The strategy is to introduce two helper angles that bridge geometry and time. The true anomaly is the physical angle you actually want; the mean anomaly grows perfectly evenly with the clock; and the eccentric anomaly is the clever geometric middleman that connects them.

Definition (true anomaly $\nu$). The true anomaly is the actual angular position of the body, measured at the focus (the primary) from periapsis to the body. It is the physical "where is it" angle — the one in the orbit equation $r = p/(1 + e\cos\nu)$ — and it advances non-uniformly, fastest at perigee and slowest at apogee.

Definition (eccentric anomaly $E$). The eccentric anomaly is an auxiliary angle measured from the center of the ellipse, not the focus. Draw the circle of radius $a$ that circumscribes the orbit (the "auxiliary circle"); project the body straight up to that circle along the line perpendicular to the major axis; the angle from the center to that projected point is $E$. It is a geometric convenience with no direct physical speed, but it tames the algebra: it relates to position by the clean formulas $$r = a(1 - e\cos E), \qquad \cos\nu = \frac{\cos E - e}{1 - e\cos E}, \qquad \tan\frac{\nu}{2} = \sqrt{\frac{1+e}{1-e}}\,\tan\frac{E}{2}.$$

Definition (mean anomaly $M$). The mean anomaly is a fictitious angle that increases uniformly with time at the mean motion $n$: $M = n(t - t_p)$, where $t_p$ is the time of periapsis passage. It is the angle a phantom body moving at constant angular speed $n$ around the auxiliary circle would have. It has no direct geometric meaning on the real orbit, but it is the link to the clock: time in, angle out, with no eccentricity effects.

The three anomalies all start together at periapsis ($\nu = E = M = 0$) and meet again at apoapsis ($\nu = E = M = 180^\circ$), but between those points they spread apart, with $M \le E \le \nu$ on the way out from perigee. The middleman $E$ is what makes the connection between the uniform-time angle $M$ and the physical angle $\nu$ solvable. That connection is the celebrated equation.

Definition (Kepler's equation). Kepler's equation relates the mean anomaly (time) to the eccentric anomaly (geometry): $$\boxed{\ M = E - e\sin E\ }$$ Given the time, you compute $M = n(t - t_p)$ directly, then solve this equation for $E$, then convert $E$ to the true anomaly $\nu$ and the radius $r$. The catch: Kepler's equation is transcendental — $E$ appears both alone and inside a sine — so it has no closed-form solution. You solve it numerically, by iteration.

🔧 Engineering Reality: an equation you cannot solve, solved a billion times a day. It is a genuine mathematical fact that $M = E - e\sin E$ cannot be rearranged into "$E = \ldots$" with elementary functions. For three centuries this drove astronomers to invent clever series and tables; today, every GPS receiver, every planetarium app, and every flight computer solves it on the fly by iteration, typically in two to four steps. That a 400-year-old "unsolvable" equation runs invisibly inside the phone in your pocket, locating satellites so it can locate you, is one of the quiet marvels of applied mathematics.

Solving Kepler's equation by iteration

The workhorse method is Newton's method. To solve $M = E - e\sin E$ for $E$, define $f(E) = E - e\sin E - M$ (we want $f = 0$). Its derivative is $f'(E) = 1 - e\cos E$. Newton's update is

$$ E_{k+1} = E_k - \frac{f(E_k)}{f'(E_k)} = E_k - \frac{E_k - e\sin E_k - M}{1 - e\cos E_k}. $$

A good starting guess accelerates convergence enormously; the standard choice is $E_0 = M + e\sin M$, a first-order correction that is already close. Let us grind one real case by hand.

Worked Example: where is a GTO satellite two hours after perigee? Take the geostationary transfer orbit of §8.3: $a = 24{,}468\ \text{km}$, $e = 0.723$. First its period and mean motion: $$T = 2\pi\sqrt{\frac{(24{,}468)^{3}}{3.986\times10^{5}}} = 2\pi\sqrt{3.675\times10^{7}} = 2\pi\,(6{,}062\ \text{s}) = 38{,}089\ \text{s} \ (= 10.58\ \text{h}),$$ so $n = 2\pi/T = 1.650\times10^{-4}\ \text{rad/s}$. Two hours after perigee, $t - t_p = 7{,}200\ \text{s}$, so the mean anomaly is $$M = n\,(t - t_p) = (1.650\times10^{-4})(7{,}200) = 1.1877\ \text{rad}\ (= 68.05^\circ).$$ Now solve $M = E - e\sin E$ for $E$ by Newton's method. Start at $E_0 = M + e\sin M = 1.1877 + 0.723\sin(1.1877) = 1.1877 + 0.723(0.9275) = 1.8583\ \text{rad}$.

$k$	$E_k$ (rad)	$f(E_k) = E_k - e\sin E_k - M$	$f'(E_k) = 1 - e\cos E_k$	$E_{k+1}$
0	$1.8583$	$-0.02271$	$1.2050$	$1.8772$
1	$1.8772$	$+0.00012$	$1.2181$	$1.8771$
2	$1.8771$	$\approx 0$	—	converged

So $E = 1.8771\ \text{rad}\ (= 107.55^\circ)$ — two iterations from a good guess. Convert to the true anomaly with $\cos\nu = (\cos E - e)/(1 - e\cos E)$: $$\cos\nu = \frac{\cos(1.8771) - 0.723}{1 - 0.723\cos(1.8771)} = \frac{-0.3015 - 0.723}{1 - 0.723(-0.3015)} = \frac{-1.0245}{1.2180} = -0.8412,$$ giving $\nu = 147.3^\circ\ (2.570\ \text{rad})$. And the radius, from $r = a(1 - e\cos E) = 24{,}468\,(1 - 0.723(-0.3015)) = 24{,}468\,(1.2180) = 29{,}802\ \text{km}$.

Sanity checks. The ordering $M < E < \nu$ holds: $68.05^\circ < 107.55^\circ < 147.3^\circ$, exactly as it must between perigee and apogee. The radius $29{,}802\ \text{km}$ lies between perigee ($6{,}771$) and apogee ($42{,}164$), and it is already past the halfway mark toward apogee — sensible, since the satellite covers most of its angle quickly near perigee (the second law again) and will spend the rest of the half-period crawling the last $33^\circ$ to apogee. Two hours of a $10.6$-hour orbit has carried it $147^\circ$ of the way around in angle but only partway in time.

🐛 Find the Error. A student needs the true anomaly of the GTO satellite three hours after perigee. They compute $M = n(t-t_p) = (1.650\times10^{-4})(10{,}800) = 1.782\ \text{rad}$, and then — reasoning that "the true anomaly is just the angle swept" — report $\nu = 1.782\ \text{rad} = 102^\circ$ directly, skipping Kepler's equation entirely. Why is this wrong, and will their answer be too small or too large?

Answer

They confused the mean anomaly with the true anomaly. $M$ is the fictitious uniform-time angle; it equals $\nu$ only at perigee and apogee. Between perigee and apogee the body runs ahead of the uniform schedule (it moved fast early), so the true anomaly is larger than the mean anomaly: $\nu > E > M$. Their reported $102^\circ$ is far too small — they must solve $M = E - e\sin E$ for $E$ (here $E \approx 2.31\ \text{rad}$), then convert to $\nu \approx 160^\circ$, well past halfway to apogee. Using $M$ in place of $\nu$ is the single most common error in position-in-orbit calculations, and it grows worse the more eccentric the orbit.

🔄 Check Your Understanding 1. Put the three anomalies in order of size for a body between perigee and apogee, and state where all three are equal. 2. Why can't Kepler's equation $M = E - e\sin E$ be solved for $E$ with algebra? 3. You are given the time since perigee and want the true anomaly. List the three steps in order.

Answers

1. $M \le E \le \nu$ on the arc from perigee to apogee (the body runs ahead of the uniform schedule). All three are equal at perigee ($0^\circ$) and at apogee ($180^\circ$). 2. Because $E$ appears both by itself and inside a sine — it is transcendental, with no rearrangement into elementary functions. You solve it numerically (e.g., Newton's method). 3. (i) Compute the mean anomaly $M = n(t - t_p)$ with $n = \sqrt{\mu/a^3}$; (ii) solve Kepler's equation $M = E - e\sin E$ for the eccentric anomaly $E$ by iteration; (iii) convert to the true anomaly via $\tan(\nu/2) = \sqrt{(1+e)/(1-e)}\,\tan(E/2)$ (and, if needed, $r = a(1-e\cos E)$).

Mission Design Checkpoint: your orbit's full element set and orbits.py

Across this book you are designing one real mission — a communications satellite to GEO (Track A), a lunar lander (Track B), a Mars orbiter (Track C), or an asteroid rendezvous (Track D) — and, if you like, building the astrotools package to compute it. Chapter 6 started orbits.py with the energy functions and recorded your target orbit's energy; this chapter gives that orbit a complete geometric identity and adds the period.

The design. Open your Mission Design Review (MDR) and give your target orbit its full six-element set — the definitive specification you will carry to the capstone. Write down $a$, $e$, $i$, $\Omega$, $\omega$, and a reference $\nu$ (or, equivalently, an epoch and mean anomaly $M_0$). Then add the period from Kepler's third law, $T = 2\pi\sqrt{a^3/\mu}$, and the mean motion $n = 2\pi/T$. A Track-A GEO comsat, for instance, records $a = 42{,}164\ \text{km}$, $e \approx 0$, $i = 0^\circ$ (equatorial), $\Omega$ and $\omega$ undefined for a circular equatorial orbit (a real design fixes them once slightly non-circular), and $T = 86{,}164\ \text{s}$ (23 h 56 m). A Track-C Mars orbiter records the elements of its science orbit about Mars, using Mars's own $\mu = 4.283\times10^{4}\ \text{km}^3/\text{s}^2$. These six numbers plus the period are the backbone every later maneuver, ground-track, and coverage calculation hangs on.

The code. Extend astrotools/orbits.py with the period (a complete addition) and sketch the element-to-state converter that Chapter 9 will finish. Keep the signatures stable so the pieces compose.

import math

def period(mu, a):
    """Orbital period (s) from Kepler's third law: T = 2*pi*sqrt(a**3/mu).
    mu in m^3/s^2, a in m. (Ch. 8)"""
    return 2 * math.pi * math.sqrt(a**3 / mu)

MU_EARTH = 3.986e14        # m^3/s^2
R_LEO = 6.771e6            # m  (400 km altitude, ISS-like)
R_GEO = 42.164e6           # m  (geostationary, Track A)
print(round(period(MU_EARTH, R_LEO)))            # ISS-like period, seconds
print(round(period(MU_EARTH, R_GEO)))            # GEO period, seconds
print(round(period(MU_EARTH, R_GEO) / 3600, 2))  # GEO period, hours
# Expected output:
# 5545
# 86164
# 23.93

The first number is the $\approx 92$-minute low-orbit period, the second is the geostationary $86{,}164\ \text{s}$, and the third confirms it as $23.93\ \text{h} = 23\ \text{h}\ 56\ \text{m}$ — the sidereal day, produced by the module you will point at your mission's orbit. By the capstone, orbits.py will turn your six elements into positions and speeds at every waypoint of your trajectory; this chapter gave it the clock.

Summary

Kepler described the orbits; Newton proved them; the six elements pin them down; the anomalies place a body in time. Carry these forward:

Numbers worth memorizing: the Kepler constant $T^2/a^3 = 4\pi^2/\mu_\oplus = 9.90\times10^{-5}\ \text{s}^2/\text{km}^3$; a $400\ \text{km}$ orbit takes $\approx 92\ \text{min}$; GEO ($a = 42{,}164\ \text{km}$) takes $23\ \text{h}\ 56\ \text{m}$ (the sidereal day); doubling $a$ multiplies $T$ by $2.83$.

Spaced Review

Retrieval strengthens memory. Answer from memory before checking, then look back at the cited section. This chapter revisits Chapter 2 and Chapter 6.

1. (§8.2, Ch. 2) Chapter 2 gave Newton's gravity as an inverse-square force, $\mu/r^2$ inward. In §8.2 that single fact produced Kepler's second law almost immediately. What property of the force made the specific angular momentum $\mathbf{h}$ constant?
2. (§8.2, Ch. 6) Chapter 6 used $\varepsilon = -\mu/(2a)$ for any ellipse but only proved the circular case. Where in this chapter did the general proof come from, and at which point on the orbit was it easiest to evaluate?
3. (Ch. 6) Chapter 6 classified trajectories by the sign of $\varepsilon$. Match each energy sign to the eccentricity and conic type from §8.3.
4. (§8.5, Ch. 6) Chapter 6 showed a higher orbit is a slower orbit (smaller speed). Kepler's third law adds a second sense in which a higher orbit is "slower." What is it?

Answers

1. It is a central force — always directed along $\mathbf{r}$ (toward the primary). So the torque about the primary, $\mathbf{r}\times\ddot{\mathbf{r}}$, is zero, and $\mathbf{h} = \mathbf{r}\times\mathbf{v}$ cannot change. 2. From conservation of specific orbital energy: evaluate $\varepsilon = v^2/2 - \mu/r$ at perigee, using $v_p = h/r_p$, $r_p = a(1-e)$, and $h^2 = \mu a(1-e^2)$; the algebra collapses to $-\mu/(2a)$. Perigee was easiest because there the velocity is purely tangential. 3. $\varepsilon < 0 \leftrightarrow 0 \le e < 1 \leftrightarrow$ circle/ellipse (bound); $\varepsilon = 0 \leftrightarrow e = 1 \leftrightarrow$ parabola (escape); $\varepsilon > 0 \leftrightarrow e > 1 \leftrightarrow$ hyperbola (unbound). 4. A longer period: $T = 2\pi\sqrt{a^3/\mu}$ grows with $a$, so a higher orbit takes more time to go around — slower in both instantaneous speed and total orbital time.

What's Next

We can now describe any orbit exactly — its shape, its tilt, its period, and where its occupant will be at any moment. But we have treated all orbits as interchangeable mathematical objects, when in reality engineers reach for a particular orbit for a particular job: a weather satellite lives somewhere very different from a GPS satellite or a spy satellite or a communications relay, and each choice is a deliberate trade of altitude, inclination, and period against the mission's needs. In Chapter 9 we take the six elements we just built and go shopping with them — cataloging the orbits that missions actually use (LEO, MEO, GEO, Molniya, sun-synchronous), and why each altitude and inclination exists. Choosing an orbit, we will find, is the first real act of mission design — and the first big decision in your own.