Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya: Worked examples

Throughout: $μ = G M_{\oplus} = 3.986 \times 1 0^{14} m^{3} / s^{2}$ and $R_{\oplus} = 6371$ km. Every symbol below was built in the parent topic and in Kepler's Laws of Planetary Motion, Two-Body Problem and the Vis-Viva Equation, J2 Perturbation and Nodal Precession.

The scenario matrix

Before computing, look at the whole space of things a problem can be. Each row is a case class; the last column names the worked example that lands on it.

#	Case class	What is tricky here	Example
A	Low & circular (LEO) — small $r$	large speed, short period	Ex 1
B	High & circular (GEO) — large $r$	period fixed first, solve backwards	Ex 2
C	Sign / direction of inclination ( $cos i < 0$ )	retrograde $i > 9 0^{\circ}$ , SSO	Ex 3
D	Eccentric orbit, both ends (Molniya)	apogee vs perigee, two radii, one $a$	Ex 4
E	Energy comparison across altitudes	which costs more $Δ v$ ? sign of $E$	Ex 5
F	Limiting / degenerate input ( $e \to 0$ , $e \to 1$ , $r \to R_{\oplus}$ )	what the formulas do at the edges	Ex 6
G	Real-world word problem	translate English → the right tool	Ex 7
H	Exam twist (sidereal vs solar, wrong-frame trap)	the deliberate trap	Ex 8

Read the matrix as a checklist: by Ex 8 every cell is filled.

Case A — Low & circular

Worked example Ex 1 (Cell A): Starlink shell at

$h = 550$ km Find the orbital speed and period.

Forecast: Guess before reading — is the speed above or below $7 km/s$ ? Is the period nearer 90 min or 100 min?

Step 1 — Build the orbital radius. $r = R_{\oplus} + h = 6371 + 550 = 6921 km = 6.921 \times 1 0^{6} m$ . Why this step? Every formula uses distance from Earth's centre, not the altitude above the surface. Forgetting $R_{\oplus}$ is the #1 slip.

Step 2 — Speed with $v = μ / r$ . $v = \frac{3.986 \times 1 0 ^{14}}{6.921 \times 1 0 ^{6}} = 5.759 \times 1 0^{7} \approx 7589 m/s \approx 7.59 km/s .$ Why this step? Gravity supplies the exact centripetal pull for a circle, which forced $v = μ / r$ . No mass of the satellite appears — a feather and a bus share the orbit.

Step 3 — Period with $T = 2 π r^{3} / μ$ . $T = 2 π \frac{( 6.921 \times 1 0 ^{6} ) ^{3}}{3.986 \times 1 0 ^{14}} \approx 5731 s \approx 95.5 min .$ Why this step? One lap = circumference $2 π r$ travelled at speed $v$ ; algebra collapses it to Kepler's 3rd.

Verify: Cross-check $T = 2 π r / v = 2 π (6.921 \times 1 0^{6}) /7589 \approx 5731$ s. ✓ Two independent routes agree. Units: $(m^{3} / s^{2}) / m = m/s$ . ✓ Speed sits just under $7.66 km/s$ (ISS at 420 km) — higher orbit, slightly slower, as it must be.

Case B — High & circular (solve backwards)

Worked example Ex 2 (Cell B): a Galileo-style MEO satellite, period 14 h

A navigation satellite has period $T = 14 h = 50400 s$ . Find its altitude and speed.

Forecast: Longer period than GPS (~12 h) — will its altitude be above or below GPS's 20 200 km?

Step 1 — Invert Kepler to get $r$ . $r = (\frac{μ T ^{2}}{4 π ^{2}})^{1/3} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 50400 ) ^{2}}{39.478})^{1/3} .$ Why this step? Here the period is the given, not the altitude. In cell A altitude came first; here we run the same law in reverse. Cubing-then-cube-rooting is the exact undo of $T^{2} \propto r^{3}$ .

Step 2 — Crunch it. Numerator $= 3.986 \times 1 0^{14} \times 2.540 \times 1 0^{9} = 1.0125 \times 1 0^{24}$ ; divide by $39.478 \Rightarrow 2.565 \times 1 0^{22}$ ; cube root $\Rightarrow r \approx 2.948 \times 1 0^{7} m = 29480 km$ . Why this step? We carry the arithmetic out in the same order the formula groups it — square the period, scale by $μ /4 π^{2}$ , then take the cube root — because doing the cube root last keeps every intermediate number at a sane size and lets us sanity-check units ( $m^{3} \to m$ ) at the final step.

Step 3 — Altitude and speed. $h = 29480 - 6371 = 23109 km$ → firmly MEO. $v = μ / r = 3.986 \times 1 0^{14} /2.948 \times 1 0^{7} \approx 3.68 km/s$ .

Verify: Feed $r$ back forward: $T = 2 π r^{3} / μ \approx 50400$ s. ✓ Above GPS altitude as forecast (longer period ⇒ bigger orbit). Speed 3.68 km/s < LEO's 7.6 — higher = slower, confirmed.

Case C — Sign of inclination (retrograde)

Definition Two orbit-plane words you need first

$Ω$ (right ascension of the ascending node, RAAN) is the compass-direction, measured in the equator plane, where the orbit crosses the equator going northward. Picture the orbit's tilted hoop stabbing through the equatorial disc: $Ω$ is where that stab-point sits. It fixes the orbit plane's swivel around Earth's axis.
$\dot{Ω}$ (read " $Ω$ -dot") is its rate of change per day — how fast that crossing point drifts around the equator. Sign convention: $\dot{Ω} > 0$ means the node drifts eastward (same sense as Earth's spin); $\dot{Ω} < 0$ means westward. For Sun-synchronism we want the plane to swing eastward at Earth's yearly rate, so we demand $\dot{Ω} > 0$ .
$p$ (semi-latus rectum) is the orbit's "width scale": $p = a (1 - e^{2})$ . For a near-circle $e \approx 0$ , so $p \approx a$ . It appears in the $J_{2}$ formula because the bulge's tug depends on the orbit's actual shape, not just its size.

Worked example Ex 3 (Cell C): Which inclination makes a 700-km orbit Sun-synchronous?

Use $\dot{Ω} = - \frac{3}{2} J_{2} (\frac{R _{\oplus}}{p})^{2} n cos i$ . For a near-circular orbit $e \approx 0$ so $p \approx a = R_{\oplus} + h$ . Require $\dot{Ω} = + 0.985 6^{\circ} / day$ (eastward, to track the Sun).

Forecast: We need $\dot{Ω}$ positive. Look at the formula's leading minus sign — will $cos i$ come out positive or negative? What does a negative $cos i$ say about the angle?

Step 1 — Set up numbers. $a = 6371 + 700 = 7071 km = 7.071 \times 1 0^{6}$ m, and since $e \approx 0$ we take $p \approx a$ . Mean motion $n = μ / a^{3}$ : $n = \frac{3.986 \times 1 0 ^{14}}{( 7.071 \times 1 0 ^{6} ) ^{3}} = 1.128 \times 1 0^{- 6} = 1.062 \times 1 0^{- 3} rad/s .$ Why this step? $n$ is the orbit's angular rate; the precession scales with how fast the satellite laps, so we need it first.

Step 2 — Convert the target rate to rad/s. $0.985 6^{\circ} / day = 0.9856 \cdot \frac{π}{180} \div 86400 = 1.991 \times 1 0^{- 7} rad/s$ . Why this step? Every term in the formula is in SI radians and seconds; mixing degrees/day in silently would give nonsense.

Step 3 — Solve for $cos i$ . $cos i = \frac{Ω ˙}{- \frac{3}{2} J _{2} ( R _{\oplus} / p ) ^{2} n} .$ Denominator (without sign): $\frac{3}{2} (1.0826 \times 1 0^{- 3}) (6371/7071)^{2} (1.062 \times 1 0^{- 3}) = 1.400 \times 1 0^{- 6}$ . With the minus: $cos i = \frac{1.991 \times 1 0 ^{- 7}}{- 1.400 \times 1 0 ^{- 6}} = - 0.1422$ . $i = arccos (- 0.1422) \approx 98. 2^{\circ} .$ Why this step? The required $\dot{Ω}$ is positive but the physics carries a built-in minus, so $cos i$ must be negative. A negative cosine means $i > 9 0^{\circ}$ — the orbit tilts past straight-up-over-the-pole into a retrograde (slightly backwards) path. That negative sign is the whole SSO signature.

Verify: Plug $i = 98.1 8^{\circ}$ , $cos i = - 0.1422$ back: $\dot{Ω} = - 1.400 \times 1 0^{- 6} \times (- 0.1422) = 1.991 \times 1 0^{- 7}$ rad/s $= 0.985 6^{\circ} / day$ . ✓ And $9 8^{\circ}$ matches the real SSO band quoted in the parent note. See J2 Perturbation and Nodal Precession.

Case D — Eccentric orbit, both ends

Figure — Orbit types — LEO, MEO, GEO, HEO, SSO, Molniya

The figure above is the derivation: Earth sits at the green focus (origin), not at the centre of the ellipse. The yellow perigee point is the nearest end, the red apogee point the farthest, and the white dashed line is the full long axis whose half-length is $a$ . Notice how far Earth's focus is offset from the ellipse's centre — that offset is exactly why the two radii are so unequal. We read the numbers straight off this picture in Step 1.

Worked example Ex 4 (Cell D): Molniya apogee & perigee speeds

Molniya has period = half a sidereal day, so $a = 26560$ km, and perigee altitude $h_{p} = 600$ km. Find both radii and the speeds at apogee and perigee.

Forecast: Kepler's 2nd law (equal areas) says the satellite is slowest where? Which end will "loiter"? Guess the speed ratio before computing.

Step 1 — Perigee radius, then apogee from the ellipse identity. $r_{p} = 6371 + 600 = 6971$ km. In the figure the perigee (yellow) and apogee (red) sit at the two ends of the white dashed long axis; the whole axis has length $2 a$ , so the two extreme radii, measured from the green focus, must add to it: $2 a = r_{p} + r_{a}$ . $r_{a} = 2 a - r_{p} = 2 (26560) - 6971 = 46149 km .$ Why this step? An orbit's size is fixed by $a$ alone; $r_{p}$ and $r_{a}$ are just the two ends sharing that long axis. This is a geometry fact you can literally see in the figure, not extra physics.

Step 2 — Speeds via vis-viva. For any point at radius $r$ , energy conservation gives (see Two-Body Problem and the Vis-Viva Equation): $v = μ (\frac{2}{r} - \frac{1}{a}) .$ At perigee ( $r_{p} = 6.971 \times 1 0^{6}$ m, $a = 2.656 \times 1 0^{7}$ m): $v_{p} = 3.986 \times 1 0^{14} (\frac{2}{6.971 \times 1 0 ^{6}} - \frac{1}{2.656 \times 1 0 ^{7}}) \approx 9.85 km/s .$ At apogee ( $r_{a} = 4.6149 \times 1 0^{7}$ m): $v_{a} = 3.986 \times 1 0^{14} (\frac{2}{4.6149 \times 1 0 ^{7}} - \frac{1}{2.656 \times 1 0 ^{7}}) \approx 1.49 km/s .$ Why this step? On an ellipse the speed is not constant, so $v = μ / r$ (circles only) cannot be used. Vis-viva subtracts the $1/ a$ term to account for the shape — the correct tool for the eccentric case.

Verify (Kepler's 2nd law): areal speed is constant, so $r_{p} v_{p}$ should equal $r_{a} v_{a}$ : $6.971 \times 1 0^{6} \times 9850 \approx 6.87 \times 1 0^{10}$ ; $4.6149 \times 1 0^{7} \times 1489 \approx 6.87 \times 1 0^{10}$ . ✓ Match. The apogee speed is ~ $1/6.6$ of perigee — the satellite crawls over the north (the red end of the figure), exactly the loitering the parent note promised.

Case E — Energy comparison

Worked example Ex 5 (Cell E): Does raising LEO→GEO cost energy? By how much per kg?

Specific orbital energy (energy per kg) of a circular orbit is $ε = - \frac{μ}{2 r}$ . Compare $r_{1} = 6791$ km (ISS, 420 km) with $r_{2} = 42164$ km (GEO).

Forecast: Both energies are negative (bound orbits). Which is less negative — i.e. closer to zero, i.e. higher energy? Predict which orbit is "richer".

Step 1 — Compute each specific energy. $ε_{1} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.791 \times 1 0 ^{6} )} = - 2.934 \times 1 0^{7} J/kg,$ $ε_{2} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 4.2164 \times 1 0 ^{7} )} = - 4.727 \times 1 0^{6} J/kg .$ Why this step? Energy fixes "how hard to reach". Using the signed formula keeps the sign truth: bound orbits are negative, and higher orbit = less negative = more energy.

Step 2 — Energy that must be added. $Δ ε = ε_{2} - ε_{1} = (- 4.727 \times 1 0^{6}) - (- 2.934 \times 1 0^{7}) = + 2.461 \times 1 0^{7} J/kg .$ Why this step? The change is positive, so you must spend energy to climb — confirming the parent's "high = expensive to reach".

Verify: GEO energy is less negative than LEO ( $- 4.7 \times 1 0^{6} > - 2.9 \times 1 0^{7}$ ), so GEO is the higher-energy orbit — matches forecast. Order of magnitude: ~ $25$ MJ/kg extra, i.e. the reason a Hohmann transfer to GEO carries a hefty $Δ v$ budget. See Hohmann Transfer and Delta-v Budgets.

Case F — Limiting / degenerate inputs

Read this figure before the numbers: the horizontal axis is orbit radius, the vertical axis is speed. The blue curve is circular speed $μ / r$ and the red curve is escape speed $2 μ / r$ ; both slope downward — farther out is always slower, no exceptions. The three marked dots are the three edge cases we test below: the yellow dot is the lowest possible circular orbit (grazing the surface), the green dot the escape speed at the same radius, and the white dot is GEO far out to the right. Notice the red curve always sits exactly $2$ above the blue one — that constant gap is Case F3.

Worked example Ex 6 (Cell F): What the formulas do at the edges

Test three extreme inputs. No new numbers to memorise — just check the tools behave.

Forecast: As eccentricity $e \to 0$ should vis-viva collapse into $v = μ / r$ ? As $r \to R_{\oplus}$ (skim the surface) should speed blow up or settle to a finite value?

Case F1 — $e \to 0$ (ellipse becomes a circle). For a circle $r = a$ , so vis-viva gives $v = μ (\frac{2}{a} - \frac{1}{a}) = \frac{μ}{a} .$ Why this step? The eccentric formula must reduce to the circular one, or one of them is wrong. It does — the $2/ r$ and $1/ a$ terms leave exactly $μ / r$ . The two tools are consistent. (This is why blue and red are the only two curves in the figure — every ellipse lives between them.)

Case F2 — $r \to R_{\oplus}$ (grazing orbit, ignoring air). Set $r = 6.371 \times 1 0^{6}$ m — the yellow dot at the far left of the figure: $v = 3.986 \times 1 0^{14} /6.371 \times 1 0^{6} \approx 7.91 km/s .$ Why this step? This is the maximum possible circular speed — the blue curve peaks here and no orbit can be lower. It is finite, not infinite: the formula caps out at ~7.9 km/s (first cosmic velocity). Real satellites can't sit here because Atmospheric Drag and Orbital Decay would erase them in minutes.

Case F3 — $e \to 1$ (parabolic escape limit). At $e = 1$ , $a \to \infty$ , so $ε = - μ /2 a \to 0$ and vis-viva → $v = 2 μ / r$ (escape speed) — the red curve. At $r = R_{\oplus}$ (green dot): $v_{esc} = \frac{2 μ}{R _{\oplus}} = \frac{2 ( 3.986 \times 1 0 ^{14} )}{6.371 \times 1 0 ^{6}} = 2 \times 7.91 \approx 11.19 km/s .$ Why this step? The eccentricity knob, pushed to its limit ( $a \to \infty$ makes the $1/ a$ term vanish), hands you the escape-velocity formula for free — showing the ellipse family smoothly becomes an open (unbound) trajectory. The $2$ vertical gap in the figure is precisely this limit: escape always sits a factor $2$ above circular at the same radius.

Verify: F1 reduces algebraically to $μ / a$ (checked). F2: $7.91$ km/s is the textbook first cosmic velocity. ✓ F3: $v_{esc} / v_{c i r c} = 2 \approx 1.414$ , and $7.91 \times 1.414 = 11.19$ km/s = known Earth escape speed. ✓ All three edges behave — no blow-ups, no contradictions.

Case G — Real-world word problem

Worked example Ex 7 (Cell G): "A weather agency wants one satellite to hover over the same equatorial city forever." What orbit, altitude, and speed?

Forecast: Which named family? What must the period equal? Sidereal or solar day?

Step 1 — Translate English to constraints. "Hover over the same spot" ⇒ appears motionless in the sky ⇒ geostationary: circular ( $e = 0$ ), equatorial ( $i = 0$ ), and period = one sidereal day $= 86164$ s. Why this step? Word problems are won by extracting the numeric requirement hidden in the phrasing. "Same spot always" is code for period-match-Earth-rotation.

Step 2 — Invert Kepler for the altitude. $r = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86164 ) ^{2}}{39.478})^{1/3} \approx 4.216 \times 1 0^{7} m = 42164 km .$ Why this step? Same inversion tool as Ex 2, now driven by the sidereal-day period we just extracted from the words. $h = 42164 - 6371 = 35793 km$ (≈35 786 km to standard precision).

Step 3 — Speed. $v = μ / r = 3.986 \times 1 0^{14} /4.216 \times 1 0^{7} \approx 3.07 km/s$ . Why this step? Once the orbit is circular and $r$ is known, the circular-speed tool gives the cruise speed directly.

Verify: $T = 2 π r^{3} / μ$ with this $r$ returns $86164$ s. ✓ 3.07 km/s matches the parent's quoted GEO speed. Answer: GEO, ~35 800 km, ~3.07 km/s.

Case H — Exam twist (the deliberate trap)

Worked example Ex 8 (Cell H): "Use a 24-hour day for GEO." How wrong is that?

A student computes GEO using the solar day $86400$ s instead of the sidereal $86164$ s. Find the altitude they get and how far off it is from the correct value.

Forecast: Longer assumed period ⇒ bigger or smaller radius? By roughly how many km will the "GEO" altitude be off?

Step 1 — Radius with the wrong (solar) period. $r_{wrong} = (\frac{μ T _{solar}^{2}}{4 π ^{2}})^{1/3} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86400 ) ^{2}}{39.478})^{1/3} \approx 4.2241 \times 1 0^{7} m .$ Why this step? This is the same Kepler inversion used in Ex 2 and Ex 7, but fed the trap value $86400$ s, so we can measure the damage the wrong frame does.

Step 2 — Radius with the correct (sidereal) period. $r_{correct} = (\frac{3.986 \times 1 0 ^{14} \cdot ( 86164 ) ^{2}}{39.478})^{1/3} \approx 4.2164 \times 1 0^{7} m .$ Why this step? We already know this is the true GEO radius (Ex 7); computing both lets us subtract and see the gap directly.

Step 3 — The error. $Δ r = r_{wrong} - r_{correct} \approx (4.2241 - 4.2164) \times 1 0^{7} = 0.0077 \times 1 0^{7} m \approx 77 km too high .$ Why this step? $T^{2} \propto r^{3}$ means a $0.27%$ period error grows into only a $\frac{2}{3} \times 0.27% \approx 0.18%$ radius error — but $0.18%$ of 42 000 km is still ~77 km, enough to matter.

Verify: $Δ r \approx 77$ km — small in percentage but mission-fatal: a satellite parked 77 km too high has a period longer than one sidereal day, so its "fixed" spot drifts steadily eastward across the sky and it is no longer geostationary. The fix (from the parent note): always use the sidereal day $86164$ s, because GEO must sync with Earth's rotation relative to the stars, not the Sun.

Recall Self-test — cover the answers

Which cell needs vis-viva rather than $μ / r$ ? ::: The eccentric case D (and F3) — speed varies with $r$ on an ellipse. Why does SSO force $i > 9 0^{\circ}$ ? ::: The precession formula has a leading minus, so a positive required $\dot{Ω}$ demands $cos i < 0$ . What does $Ω$ mean and what does $\dot{Ω} > 0$ signify? ::: $Ω$ is the equator-crossing (ascending-node) direction of the orbit plane; $\dot{Ω} > 0$ means that node drifts eastward. What does $J_{2}$ represent? ::: The strength of Earth's equatorial-bulge term in its gravity field, ~ $1.08 \times 1 0^{- 3}$ . Grazing circular speed and escape speed at the surface? ::: 7.91 km/s and 11.19 km/s (ratio $2$ ). Which day length for GEO, and the penalty for the wrong one? ::: Sidereal 86 164 s; using 86 400 s over-shoots altitude by ~77 km.