Worked examples — Physical meaning of each orbital element

This page is the drill hall for the parent topic. We will not re-teach the meaning of each element — we will use the formulas on every awkward case they can throw at you: every quadrant, zero and degenerate inputs, limiting values, a word problem, and an exam trick. If a symbol looks unfamiliar, jump back to the parent note first.

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The scenario matrix

#	Cell (case class)	What makes it tricky	Covered by
C1	Shape from $r_p,r_a$	plain forward calculation	Ex 1
C2	Degenerate shape: $e=0$ (circle)	$\omega ,\nu$ become undefined	Ex 2
C3	Limiting shape: $e\to 1$, $\epsilon \to 0$	escape / unbound edge	Ex 3
C4	Inclination sign classes: prograde / polar / retrograde	$i$ from $h_z$, all three signs	Ex 4
C5	RAAN $\Omega$ quadrant fix ($n_y<0$) + equatorial degeneracy	$\cos$ can't tell left from right; $\Omega$ undefined if $i=0$	Ex 5
C6	True anomaly $\nu$ quadrant fix ($\vec{r}\cdot \vec{v}<0$)	inbound vs outbound	Ex 6
C7	Word problem: period ↔ altitude (geostationary)	picking the right tool	Ex 7
C8	Exam twist: same $a$, different $e$ ⇒ same energy	trap: "energy needs $e$"	Ex 8
C9	Unbound shape: $e>1$ (hyperbolic flyby)	$\epsilon >0$, $a<0$	Ex 9

Every numeric answer below is machine-checked in the verify block.

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Ex 1 — Shape from apside distances (Cell C1)

Forecast: guess before computing — is $e$ closer to $0$ or to $1$? (The apoapsis is ~6× the periapsis, so this is a very stretched orbit — expect $e$ near $0.7$.)

Look at the figure below: the orbit ellipse (teal) has Earth at one focus (orange), not at the centre. The periapsis and apoapsis (plum dots) sit at the two ends of the long axis, and the double-headed black arrow shows that end-to-end distance is the full major axis $2a$. Keep this picture in mind — every step below is just reading a length off it.

1. $a=\frac{r_p+r_a}{2}=\frac{6800+42000}{2}=24400$ km. Why this step? In the figure the two plum dots span the whole black arrow, whose length is $2a$. Their average is the half-length.

2. $e=\frac{r_a-r_p}{r_a+r_p}=\frac{35200}{48800}=0.72131\dots$ Why this step? The numerator is $2ae$ (the focus is offset from the centre by $ae$ — see the gap between the orange focus and the black "centre" mark in the figure — and periapsis/apoapsis are that offset either side of the centre) and the denominator is $2a$; their ratio is exactly $e$.

3. $T=2\pi \sqrt{\frac{a^3}{\mu }}=2\pi \sqrt{\frac{24400^3}{398600}}\approx 37945\text{s}\approx 10.54\text{h}.$ Why this step? Kepler's third law says the period depends only on $a$ — not on $e$. We use it because "how long is one lap" is a size question, and $a$ is the size.

Verify: $r_p=a(1-e)=24400(1-0.72131)=6800$ km ✓ and $r_a=a(1+e)=42000$ km ✓ — plugging the elements back reproduces the inputs. Units: $\sqrt{{\text{km}}^3/({\text{km}}^3/{\text{s}}^2)}=\sqrt{{\text{s}}^2}=\text{s}$ ✓.

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Ex 2 — Degenerate case: a perfect circle (Cell C2)

Forecast: if the near and far points are equal, the ellipse isn't squashed at all — guess $e=0$ and $a=7000$ km.

1. $a=\frac{7000+7000}{2}=7000$ km. Why? Same averaging as before; here both apsides coincide with the single circular radius.
2. $e=\frac{7000-7000}{7000+7000}=0$. Why? Numerator is zero — there is no focus offset, so no squash.
3. $T=2\pi \sqrt{\frac{7000^3}{398600}}\approx 5828.5$ s $\approx 1.62$ h. Why? Kepler III still works; $e$ never entered it.

Verify: $r=a(1-e^2)/(1+e\cos \nu )=7000(1-0)/(1+0)=7000$ km for every $\nu$ ✓ — the radius truly never changes, confirming a circle.

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Ex 3 — Limiting case: the escape edge, $e\to 1$ (Cell C3)

Forecast: escape speed is the boundary between "falls back" and "flies away forever." At exactly that speed the orbit is a parabola. Guess $\epsilon =0$ and $e=1$.

1. Escape speed: set total specific energy to zero, $\epsilon =\frac{v^2}{2}-\frac{\mu }{r}=0$, giving $v_{esc}=\sqrt{\frac{2\mu }{r}}=\sqrt{\frac{2(398600)}{7000}}\approx 10.6707$ km/s. Why this step? $\epsilon =0$ is the definition of "just barely unbound" — kinetic energy exactly cancels the (negative) potential well. See Vis-viva equation.
2. Energy: $\epsilon =0$ by construction. Why? We chose the speed to make it zero.
3. Semi-major axis: from $\epsilon =-\frac{\mu }{2a}$, setting $\epsilon =0$ forces $a\to \infty$. Why? A parabola is an ellipse stretched infinitely long — its "far end" never closes. Numerically, $\mu /(-2\epsilon )$ divides by zero.
4. Eccentricity: the boundary shape between bound ($e<1$) and unbound ($e>1$) is $e=1$ exactly. Why? $e=1$ is the definition of a parabola.

Verify: with $v=v_{esc}$, $\epsilon =v^2/2-\mu /r=(2\mu /r)/2-\mu /r=0$ exactly ✓.

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Ex 4 — Inclination across all three sign classes (Cell C4)

Forecast: the sign of $h_z$ (the vertical component) should decide it — positive spins the same way as Earth (prograde), zero is polar, negative is retrograde.

The formula (parent note) is $\cos i=\frac{h_z}{|\vec{h}|},0^{\circ }\le i\le 180^{\circ }.$ Why this tool? $\vec{h}$ is perpendicular to the orbital plane, and $\hat{K}=(0,0,1)$ is perpendicular to the equator. The angle between two planes equals the angle between their normals, and a dot product with $\hat{K}$ picks out exactly $h_z$. Because $i$ only ranges $0$–$180^{\circ }$, $\cos$ alone is enough here — no quadrant fix needed (unlike $\Omega ,\omega ,\nu$).

1. (a) $|\vec{h}|=50000$, $\cos i=50000/50000=1\Rightarrow i=0^{\circ }$. Equatorial prograde. Why? $h$ points straight up, plane lies flat, same sense as Earth's spin.
2. (b) $|\vec{h}|=\sqrt{30000^2+40000^2}=50000$, $\cos i=0/50000=0\Rightarrow i=90^{\circ }$. Polar. Why? $h$ lies in the equatorial plane, so the orbit plane stands vertical, passing over the poles.
3. (c) $\cos i=-50000/50000=-1\Rightarrow i=180^{\circ }$. Retrograde. Why? $h$ points straight down, meaning the satellite circles opposite to Earth's spin.

Verify: $\cos (0^{\circ })=1$, $\cos (90^{\circ })=0$, $\cos (180^{\circ })=-1$ ✓, matching the three computed cosines.

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Ex 5 — RAAN quadrant fix, $n_y<0$, and the equatorial degeneracy (Cell C5)

Forecast: $\cos \Omega =n_x/|\vec{n}|$ will land you in the wrong half of the circle if you trust it blindly. Because $n_y<0$ here, expect $\Omega$ in the range $180^{\circ }$–$360^{\circ }$.

In the figure below, the orange arrow is the reference direction $\widehat{{\rm Y}}$ (vernal equinox). The teal arrow is the node vector $\hat{n}$; notice it points into the lower half-plane ($n_y<0$). The dashed plum arrow shows the wrong answer $120^{\circ }$ that a bare cosine would give — it is the mirror image of $\hat{n}$ across the $x$-axis, sharing the same cosine. The black arc from $\widehat{{\rm Y}}$ sweeps counter-clockwise all the way round to $\hat{n}$: that swept angle is the true $\Omega =240^{\circ }$.

1. $|\vec{n}|=\sqrt{(-1{)}^2+(-\sqrt{3}{)}^2}=\sqrt{1+3}=2$. Why? We need a unit reference to turn a component into a cosine.
2. $\cos \Omega =\frac{n_x}{|\vec{n}|}=\frac{-1}{2}\Rightarrow \Omega =120^{\circ }\text{ or }240^{\circ }.$ Why the ambiguity? $\cos$ gives the same value for an angle and its mirror image across the $x$-axis (the plum vs teal arrows). Two angles share $\cos =-1/2$.
3. Quadrant fix: since $n_y=-\sqrt{3}<0$, take $\Omega =360^{\circ }-120^{\circ }=240^{\circ }$. Why this rule? $\Omega$ is measured eastward (counter-clockwise) from the vernal equinox $\widehat{{\rm Y}}$. A node pointing into the lower half of the equatorial plane ($n_y<0$) has already swept past $180^{\circ }$. The sign of $n_y$ is the "which half" flag that $\cos$ threw away.

Verify: $\cos (240^{\circ })=-1/2$ ✓ and $\sin (240^{\circ })=-\sqrt{3}/2<0$, matching $n_y<0$ ✓. (Had we wrongly kept $120^{\circ }$, its sine would be $+\sqrt{3}/2$, contradicting the given $\vec{n}$.)

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Ex 6 — True anomaly quadrant fix, inbound vs outbound (Cell C6)

Forecast: the angle between $\vec{r}$ and $\hat{e}$ (periapsis direction) is $45^{\circ }$ geometrically — but a falling-inward satellite ($\vec{r}\cdot \vec{v}<0$) is on the return leg, so expect $\nu$ near $315^{\circ }$, not $45^{\circ }$.

In the figure, the plum arrow $\hat{e}$ points to periapsis (the reference for measuring $\nu$). The black dot is the satellite with its position vector $\vec{r}$; it sits below the apse line (its $y$ is negative). The orange arrow is the velocity — it has an inward component, so $\vec{r}\cdot \vec{v}<0$. The black arc measures $\nu$ from $\hat{e}$ round to $\vec{r}$ in the direction of motion, landing at $315^{\circ }$, not the naive $45^{\circ }$.

1. $r=|\vec{r}|=\sqrt{5000^2+5000^2}=5000\sqrt{2}\approx 7071$ km. Why? We need the magnitude to normalise the cosine.
2. $\cos \nu =\frac{\vec{e}\cdot \vec{r}}{|\vec{e}|r}=\frac{(1)(5000)+(0)(-5000)+0}{(1)(5000\sqrt{2})}=\frac{5000}{5000\sqrt{2}}=\frac{1}{\sqrt{2}}\Rightarrow \nu =45^{\circ }\text{ or }315^{\circ }.$ Why? $\vec{e}$ points to periapsis; the angle between it and $\vec{r}$ is $\nu$ by definition. But again $\cos$ can't tell above from below the apse line.
3. Velocity-sign fix: $\vec{r}\cdot \vec{v}<0$ means the satellite is moving inward (distance shrinking), i.e. it has already passed apoapsis and is heading toward periapsis, so $180^{\circ }<\nu <360^{\circ }$. Choose $\nu =315^{\circ }$. Why this tool? $\frac{d}{dt}\left(\frac{1}{2}{r}^2\right)=\vec{r}\cdot \vec{v}$. Positive means $r$ growing (outbound, $0<\nu <180^{\circ }$); negative means $r$ shrinking (inbound, $180^{\circ }<\nu <360^{\circ }$). It is the "which half of the trip" flag.

Verify: $\cos (315^{\circ })=1/\sqrt{2}\approx 0.7071$ ✓ and $\sin (315^{\circ })=-1/\sqrt{2}<0$ — consistent with being below the apse line (the $-5000$ in $r_y$) ✓.

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Ex 7 — Word problem: build a geostationary orbit (Cell C7)

Forecast: "stay over one spot" ⇒ its lap time must match Earth's spin. This is a period → size problem, so the tool is Kepler III solved for $a$. Guess a big number — geostationary is famously ~36000 km up.

1. Pick the tool. We know $T$, we want $a$. Invert Kepler III: $T=2\pi \sqrt{\frac{a^3}{\mu }}\Rightarrow a={\left(\frac{\mu T^2}{4{\pi }^2}\right)}^{1/3}.$ Why this and not vis-viva? Vis-viva mixes speed and position; here we only have a time, and Kepler III is the unique bridge between period and size.
2. Plug in. $a={\left(\frac{398600\times 86164^2}{4{\pi }^2}\right)}^{1/3}\approx 42164$ km. Why? Direct substitution of $\mu$ and the sidereal-day period.
3. Altitude. $h=a-R_{\oplus }=42164-6378\approx 35786$ km. Why subtract $R_{\oplus }$? $a$ is measured from Earth's centre; altitude is measured from the surface.

Verify: feed $a=42164$ km back into $T=2\pi \sqrt{a^3/\mu }$ and you recover $\approx 86164$ s ✓.

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Ex 8 — Exam twist: same energy, different shapes (Cell C8)

Forecast: the parent-note mistake box warns: energy depends only on $a$. Both have the same $a$, so guess equal energy — the student is wrong.

1. Use the energy–size identity. $\epsilon =-\frac{\mu }{2a}$. Why this tool? It is the one relation that isolates energy, and $e$ never appears in it.
2. Orbit A: ${\epsilon }_A=-\frac{398600}{2(10000)}=-19.93{\text{km}}^2/{\text{s}}^2$. Why this step? Substitute A's $a$ into the identity; its $e=0.1$ is deliberately ignored because the identity has no slot for $e$.
3. Orbit B: ${\epsilon }_B=-\frac{398600}{2(10000)}=-19.93{\text{km}}^2/{\text{s}}^2$. Identical. Why this step? Substitute B's $a$ — which is the same $10000$ km — so we land on the identical number; B's larger $e=0.6$ again has nowhere to enter. Why the student's intuition misleads: B is faster at periapsis — but it is also slower at apoapsis and swings out farther, deeper out of the potential well. The kinetic gain and potential loss trade off exactly. Speed is not conserved; total energy is.

Recall What actually differs between A and B?

Same $a$ ⇒ same energy and same period. What differs ::: the shape (B is far more squashed), the periapsis/apoapsis distances, and the periapsis/apoapsis speeds — but never the total energy or the period.

Verify: ${\epsilon }_A={\epsilon }_B=-19.93{\text{km}}^2/{\text{s}}^2$, difference $=0$ ✓. Cross-check with vis-viva at periapsis: $v_p^2=\mu (2/r_p-1/a)$, then $v_p^2/2-\mu /r_p$ must also equal $-\mu /2a$ for each orbit ✓ (checked numerically below).

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Ex 9 — Unbound case: a hyperbolic flyby, $e>1$ (Cell C9)

Forecast: positive energy means it escapes and never returns — a hyperbola. From the spectrum in Ex 3, guess $a<0$ and $e>1$.

1. Semi-major axis from energy. Rearrange $\epsilon =-\frac{\mu }{2a}$ to $a=-\frac{\mu }{2\epsilon }=-\frac{398600}{2(2.0)}=-99650$ km. Why this step? The energy–size identity holds for all conics, not just ellipses. A positive $\epsilon$ forces a negative $a$ — the sign convention that keeps the identity true past escape.
2. Eccentricity from $h$ and $a$. The general relation $h^2=\mu a(1-e^2)$ gives $e=\sqrt{1-\frac{h^2}{\mu a}}=\sqrt{1-\frac{80000^2}{398600\times (-99650)}}=\sqrt{1+0.16116\dots }\approx \mathrm{1.0776.}$ Why this step? With $a<0$ the fraction $h^2/(\mu a)$ is negative, so we are taking the square root of something bigger than 1 — that is exactly what makes $e>1$. The formula automatically reports a hyperbola.
3. Shape. $e=1.0776>1\Rightarrow$ hyperbola (an open flyby trajectory, not a closed loop). Why this step? $e>1$ is the definition of a hyperbolic orbit; the spacecraft passes once and leaves forever.

Verify: plug back — $\epsilon =-\mu /(2a)=-398600/(2\times (-99650))=+2.0{\text{km}}^2/{\text{s}}^2$ ✓ (positive, unbound), and $h^2=\mu a(1-e^2)$ reproduces $h=80000$ ✓ (checked numerically below).

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Recall Degeneracy cheat sheet (which element dies when)

$e=0$ (circle) kills ::: $\omega$ and $\nu$ — replace with argument of latitude $u=\omega +\nu$ $i=0$ or $180^{\circ }$ (equatorial) kills ::: $\Omega$ (no node) — replace with longitude of periapsis $\varpi =\Omega +\omega$

Related tools used on this page: Vis-viva equation, Kepler's Laws, Angular momentum in orbits, Eccentricity vector, State vector to orbital elements conversion, Orbital perturbations.