Chapter 8 — Key Takeaways (Kepler's Laws and the Two-Body Problem)
A one-page reference. Reread this before an exam, or before you specify any orbit.
Kepler's three laws (and what each really is)
Law
Statement
Underlying principle
First (ellipses)
Orbits are ellipses with the primary at one focus
the conic solution $r = p/(1+e\cos\nu)$ of inverse-square gravity
Second (equal areas)
The primary–body line sweeps equal areas in equal times
conservation of angular momentum ($dA/dt = h/2$)
Third (periods)
$T^2 \propto a^3$
area of ellipse $\pi ab = (h/2)T$ ⇒ $T^2 = (4\pi^2/\mu)a^3$
The big idea: Kepler's laws are theorems, not axioms — all three fall out of
$\ddot{\mathbf{r}} = -\mu\mathbf{r}/r^3$. The ellipse is forced by the exponent $2$ in gravity.
specific energy from size (proved for any ellipse here; from Ch. 6)
Conic classification by eccentricity (= energy sign)
$e$
Conic
$\varepsilon$
$a$
Bound?
$0$
circle
$<0$
$=r$
yes
$0
ellipse
$<0$
$>0$
yes
$1$
parabola
$=0$
$\to\infty$
marginal escape
$>1$
hyperbola
$>0$
$<0$
no (escapes)
The six orbital elements
Element
Symbol
Sets
Constant as it coasts?
Semi-major axis
$a$
size (⇒ period, energy)
yes
Eccentricity
$e$
shape
yes
Inclination
$i$
tilt of plane from equator
yes
RAAN
$\Omega$
swivel of plane about the pole
yes
Argument of periapsis
$\omega$
orientation of ellipse in plane
yes
True anomaly
$\nu$
where the body is now
no — the only one that moves
Six numbers + a named primary (its $\mu$) = a complete orbit specification. Real "constants" drift under
perturbations — Chapter 12.
The three anomalies (find position in time)
True $\nu$ — physical angle at the focus (what you want). Eccentric $E$ — geometric middleman,
measured at the center. Mean $M$ — fictitious, grows uniformly with time ($M = n\,\Delta t$).
Ordering (perigee → apogee): $M \le E \le \nu$. All three equal at perigee ($0^\circ$) and apogee
($180^\circ$).
The near-circular shortcut: as $e \to 0$, the three anomalies converge — position advances almost
uniformly. The divergence between them is a direct readout of $e$.
Procedure — time → position: (1) $M = n(t - t_p)$; (2) solve $M = E - e\sin E$ for $E$ (Newton, start
at $E_0 = M + e\sin M$); (3) convert $E \to \nu$ and $r = a(1-e\cos E)$.
Decision aid — "which relation do I use?"
You know…
You want…
Use
$a$ (and $\mu$)
period / mean motion
$T = 2\pi\sqrt{a^3/\mu}$, $n = \sqrt{\mu/a^3}$
$T$ (and $\mu$)
semi-major axis
$a = (\mu T^2/4\pi^2)^{1/3}$
$n$ from a TLE
semi-major axis
$a = (\mu/n^2)^{1/3}$
$r_p, r_a$
$a, e$
$a=(r_p+r_a)/2$, $e=(r_a-r_p)/(r_a+r_p)$
$a, e, \nu$
radius
$r = a(1-e^2)/(1+e\cos\nu)$
time since perigee
true anomaly
Kepler's equation, then $E\to\nu$ (3-step procedure above)
$a$ (and $\mu$)
speed at radius $r$
vis-viva $v=\sqrt{\mu(2/r-1/a)}$ (🔗 Ch. 6)
Common pitfalls
Pitfall
Reality
Primary at the center of the ellipse
It sits at a focus, off to one side. The center is empty.
Confusing $M$ (or $E$) with the true anomaly $\nu$
They coincide only at perigee/apogee; for eccentric orbits $\nu$ runs well ahead of $M$.
Thinking period depends on eccentricity
$T$ depends on $a$ only. Same $a$, any shape ⇒ same period.
Reading $\nu$ straight off the clock
Only valid for near-circular orbits; solve Kepler's equation otherwise.
Using $24\ \text{h}$ for the GEO period
It is the sidereal day, $23\ \text{h}\ 56\ \text{m}$ ($86{,}164\ \text{s}$).
Treating Kepler's laws as postulates
They are consequences of inverse-square gravity (§8.2).
Doubling $a$ multiplies the period by $2^{3/2} = \mathbf{2.83}$.
Earth $\mu = 3.986\times10^{5}$; Moon $4.903\times10^{3}$; Mars $4.283\times10^{4}\ \text{km}^3/\text{s}^2$.
Mission / astrotools additions this chapter
MDR: recorded your target orbit's full six-element set $(a, e, i, \Omega, \omega, \nu)$ plus its
period and mean motion — the definitive orbit spec you carry to the capstone.
orbits.py: added period(mu, a) (Kepler III) and a sketch of elements_to_rv(...) (the
perifocal position/velocity; Chapter 9 adds the rotation to the
inertial frame). Energy functions circular_velocity, specific_energy, vis_viva come from Ch. 6.