Learning Objectives
By the end of this study guide, the student should be able to:
- State Kepler's three laws of planetary motion and the historical and physical context in which they were derived.
- Describe the geometry of an elliptical orbit: semi-major axis, eccentricity, perigee, apogee, and the focus-centred polar equation.
- Explain Kepler's second law as a statement of conservation of angular momentum and predict how orbital speed varies around an elliptical orbit.
- Apply Kepler's third law (T² = 4π²a³/μ) to compute orbital periods and to classify orbits (LEO/MEO/GEO/HEO).
- Connect Kepler's empirical laws to Newton's law of gravitation and the two-body equation of motion.
- Identify the six classical orbital elements and interpret their physical meaning.
Historical Background
Between 1576 and 1601, the Danish astronomer Tycho Brahe accumulated the most precise pre-telescopic positional measurements of the planets — particularly Mars, with errors below 2 arc-minutes. His assistant Johannes Kepler (1571–1630) inherited this dataset in 1600 and spent years trying to fit Mars' orbit to a circle. The data stubbornly refused: after abandoning the circle, Kepler found that an ellipse fit within the measurement noise.
- 1609 — Astronomia Nova: Kepler published the first two laws (ellipse law; equal-areas law).
- 1619 — Harmonices Mundi: the third law relating period to orbit size.
- 1687 — Newton's Principia: Isaac Newton showed that all three laws follow from a single inverse-square law of gravitation plus his laws of motion — elevating Kepler's empirical rules to consequences of universal physics, valid for any two-body system (planets, moons, and artificial satellites alike).
Why this matters to a satellite engineer: every orbit in the EEEN 567 course — LEO, MEO, GEO, Molniya — is an application of these three laws. Kepler's third law is the working formula behind every period/altitude calculation; his second law explains why HEO satellites dwell at apogee; his first law supplies the geometry of every transfer orbit.
KEPLER'S FIRST LAW — THE LAW OF ELLIPSES (1609)
First Law: The Law of Ellipses
Statement: “The orbit of every planet is an ellipse with the Sun at one of the two foci.” For Earth satellites, replace “Sun” with “centre of the Earth.”
1. Ellipse Geometry
- An ellipse is the locus of points whose distances to two fixed points (the foci) sum to a constant, 2a.
- Semi-major axis a: half the longest diameter — the orbit's “size” parameter.
- Semi-minor axis b: b = a√(1 − e²).
- Eccentricity e: e = c/a, where c is the centre-to-focus distance (0 ≤ e < 1). e = 0 gives a circle; e → 1 gives a highly elongated ellipse.
Fig. 1 — Elliptical orbit: Earth occupies one focus; perigee and apogee lie on the major axis.
2. Key Relationships
rp = a(1 − e) ra = a(1 + e) rp + ra = 2a e = (ra − rp)/(ra + rp)
rp, ra measured from Earth's centre. Altitudes: hp = rp − R⊙, ha = ra − R⊙.
Orbit equation (focus-centred): r(ν) = a(1 − e²) / (1 + e cosν)
ν = true anomaly (angle from perigee direction); r = a(1+e) at ν = 0 (perigee), r = a(1−e) at ν = 180° (apogee).
3. Classification by Eccentricity
| e | Orbit | Example |
|---|---|---|
| e = 0 | Circle | Idealised GEO |
| 0 < e < 0.01 | Nearly circular | ISS (e ≈ 0.0005), GPS |
| 0.01 – 0.5 | Moderately elliptical | GTO (e ≈ 0.7 transfer), some science orbits |
| e > 0.5 | Highly elliptical (HEO) | Molniya (e ≈ 0.72), Tundra (e ≈ 0.27) |
Worked Example 1 — Molniya geometry: Given a = 26,578 km and e = 0.72: rp = 26,578 × 0.28 = 7,442 km (hp = 1,064 km); ra = 26,578 × 1.72 = 45,714 km (ha = 39,336 km). The apogee is ~6× farther from Earth's centre than the perigee.
KEPLER'S SECOND LAW — THE LAW OF EQUAL AREAS (1609)
Second Law: The Law of Equal Areas
Statement: “The line joining a planet to the Sun sweeps out equal areas during equal intervals of time.”
In satellite terms: the radius vector from Earth's centre to the satellite sweeps equal areas in equal times.
1. Physical Content — Conservation of Angular Momentum
The areal velocity is constant because gravity is a central force — it acts along the radius vector and therefore exerts zero torque about Earth's centre:
dA/dt = h/2 = constant, where h = r²ω = r × vθ = r v cosφ = √(μa(1−e²))
h = specific angular momentum (km²/s), φ = flight-path angle. No torque ⇒ h conserved ⇒ equal areas in equal times.
2. Consequence — Speed Varies Around the Orbit
rp vp = ra va ⇒ vp/va = ra/rp = (1+e)/(1−e)
The satellite moves fastest at perigee and slowest at apogee. Combined with the vis-viva equation v² = μ(2/r − 1/a).
vp = 10.0 km/s
Molniya at perigee (1,000 km alt.)
va = 1.6 km/s
Molniya at apogee (39,700 km alt.)
~6.3× ratio
Speed swing per orbit (e = 0.72)
~8 of 12 h
Time spent near apogee — the HEO dwell
Worked Example 2 — Speed ratio: For the Molniya orbit, vp/va = (1+0.72)/(1−0.72) = 1.72/0.28 = 6.14. Indeed vis-viva gives vp = √(μ(2/7,378 − 1/26,578)) = 9.97 km/s and va = √(μ(2/45,714 − 1/26,578)) = 1.62 km/s — ratio 6.15. ✓
Engineering significance: the apogee dwell is the entire design concept of Molniya/HEO communications (8 hours of high-elevation visibility over high latitudes per 12-hour orbit). Conversely, launch vehicles exploit the perigee speed maximum: adding velocity at perigee is the most efficient way to raise apogee (the Hohmann transfer logic).
KEPLER'S THIRD LAW — THE HARMONIC LAW (1619)
Third Law: The Harmonic Law
Statement: “The squares of the orbital periods are proportional to the cubes of the semi-major axes.”
T² = (4π²/μ) a³ or T = 2π√(a³/μ)
μ = GM⊙ = 3.986×105 km3/s2 (Earth). Valid for any orbit about the primary body — circular or elliptical (a = semi-major axis).
Newton later explained the proportionality constant: it contains the mass of the central body. For two satellites of the same planet, T²/a³ is identical — period depends only on orbit size, not on satellite mass or eccentricity.
Applications Across the Orbit Classes
| Orbit | a (km) | T (computed) | T (known) |
|---|---|---|---|
| LEO, 400 km (ISS) | 6,778 | 5,557 s = 92.6 min | 92.6 min ✓ |
| LEO, 1,000 km | 7,378 | 6,307 s = 105.1 min | 105.1 min ✓ |
| MEO, 20,200 km (GPS) | 26,578 | 43,080 s = 11 h 58 min | 11 h 58 min ✓ |
| GEO | 42,164 | 86,164 s = 23 h 56 min 4 s | sidereal day ✓ |
| Molniya HEO | 26,578 | 43,080 s = 11 h 58 min | 11 h 58 min ✓ |
Worked Example 3 — Deriving GEO altitude: Impose T = 86,164 s: a = [μ(T/2π)²]1/3 = [3.986×105 × (86,164/2π)²]1/3 = 42,164 km ⇒ h = 35,786 km. Kepler's third law derives the geostationary radius — it is not an arbitrary definition.
Worked Example 4 — Orbit identification from period: A satellite's period is measured as 717.7 min. a = [μ(T/2π)²]1/3 = [3.986×105 × (43,060/2π)²]1/3 = 26,560 km ⇒ h = 20,182 km — the orbit is MEO, and the half-sidereal-day period identifies it as a navigation (GPS-type) orbit.
NEWTON'S SYNTHESIS — FROM EMPIRICISM TO MECHANICS (1687)
Newton's Synthesis: The Two-Body Problem
Newton combined his second law (F = ma) with the inverse-square gravitational force to produce the general equation of orbital motion:
r̈ + (μ/r³)r = 0
Two-body equation: relative acceleration of satellite (mass m) about primary (mass M), valid when m << M and no other forces act.
Four constants of integration emerge (energy E and angular momentum h), and solving yields exactly Kepler's ellipse with:
p = h²/μ = a(1 − e²) (semi-latus rectum); specific orbital energy ε = −μ/2a
ε depends only on a: all orbits with the same a have the same period and the same energy per kg — circular LEO and GEO differ only in a.
- Ellipse (bound): ε < 0, e < 1 — all operational satellites.
- Parabola: ε = 0, e = 1 — escape boundary (v = vesc = √(2μ/r)).
- Hyperbola: ε > 0, e > 1 — interplanetary flybys.
Validity and corrections: Kepler/Newton two-body motion assumes a point-mass (or spherically symmetric) Earth. Real orbits are perturbed by Earth's oblateness (J2 — causes nodal regression and apsidal precession), the Moon/Sun (third body), solar radiation pressure and drag. These are small corrections for analysis, but they dominate long-term station-keeping — the subject of orbital perturbations later in this course.
The Six Classical Orbital Elements
Kepler's first law gives the shape of the orbit in its own plane; three more angles are needed to orient that plane in 3-D space and to locate the satellite on the ellipse. Together they form the six classical (Keplerian) orbital elements:
| # | Element | Symbol | Physical meaning |
|---|---|---|---|
| 1 | Semi-major axis | a | Orbit size (sets period via third law; sets energy) |
| 2 | Eccentricity | e | Orbit shape (0 = circle; sets perigee/apogee split) |
| 3 | Inclination | i | Tilt of orbital plane relative to the equator (0–180°) |
| 4 | Right ascension of ascending node | Ω | Orientation of the orbital plane in inertial space (swivel about polar axis) |
| 5 | Argument of perigee | ω | Orientation of the ellipse within its plane (where perigee sits) |
| 6 | True anomaly | ν | Position of the satellite on the ellipse at epoch time |
- Elements 1–2 (a, e) come straight from Kepler's first law; element 6 positions the satellite in time (linked to the second law through the equal-area sweep); elements 3–5 orient the orbit.
- Alternate forms: true anomaly may be replaced by mean anomaly M (uniform clock tied to period) or by time since perigee passage; for near-circular orbits ω and ν are sometimes merged into “argument of latitude” u = ω + ν.
- Real-world data: TLE (two-line element) sets from NORAD encode a modified version of these elements and are the input to every tracking/antenna-pointing calculation.
Interactive Animation — Kepler's Second Law
The animation shows a satellite on an elliptical orbit (e = 0.6) with Earth at the focus. Watch the speed: the satellite races through perigee and crawls near apogee. The two shaded sectors are swept during two equal time intervals — by Kepler's second law their areas are equal.
Fig. 2 — Equal-area sectors over equal time steps (shaded). Position solved by Newton-iteration on Kepler's equation M = E − e sin E.
Interactive Orbital Calculator (Kepler's Laws)
Enter the perigee and apogee altitudes. The tool applies Kepler's first-law geometry (a, e, rp, ra), the vis-viva/second-law speeds at perigee and apogee, and Kepler's third-law period.
Summary of Key Points
- First law (ellipses, 1609): every satellite orbit is an ellipse with Earth's centre at one focus; rp = a(1−e), ra = a(1+e), r(ν) = a(1−e²)/(1+e cosν).
- Second law (equal areas, 1609): areal velocity is constant — a consequence of angular-momentum conservation under a central force; satellites move fastest at perigee (vp/va = (1+e)/(1−e)) and dwell near apogee — the HEO design principle.
- Third law (harmonic, 1619): T² = 4π²a³/μ; period depends only on semi-major axis. It derives the GEO radius (42,164 km) and classifies all orbits.
- Newton (1687) showed all three laws follow from F = ma plus the inverse-square law; the two-body equation r̈ + (μ/r³)r = 0 is their modern statement.
- Orbital energy ε = −μ/2a and angular momentum h = √(μa(1−e²)) are conserved; ε < 0 bound, = 0 parabolic escape, > 0 hyperbolic.
- Six classical elements (a, e, i, Ω, ω, ν) fully describe an orbit: two define the ellipse, three orient it, one positions the satellite.
- Real-world orbits include perturbations (J2 oblateness, third body, drag, SRP) — Keplerian motion is the essential first approximation of satellite engineering.
Review Questions
Q1. State Kepler's three laws and identify which one is used to compute (i) a satellite's period from its altitude, (ii) the variation of speed around an elliptical orbit.
Show answer
(i) Third law, T = 2π√(a³/μ); (ii) second law (equal areas ⇔ constant areal velocity ⇔ h conserved ⇒ v varies inversely with r).Q2. A satellite has perigee altitude 500 km and apogee altitude 1,500 km. Compute a, e, and the period.
Show answer
rp = 6,878, ra = 7,878 km; a = 7,378 km; e = 1,000/14,756 = 0.0677; T = 2π√(7,378³/3.986×105) = 6,307 s ≈ 105 min.Q3. Using Kepler's third law, show that the geostationary semi-major axis is 42,164 km and hence the altitude is 35,786 km.
Show answer
Set T = 86,164 s: a = [μ(T/2π)²]1/3 = [3.986×105 × (13,715)²]1/3 = 42,164 km; h = a − R⊙ = 42,164 − 6,378 = 35,786 km.Q4. Explain physically why a satellite moves fastest at perigee, referencing torque and angular momentum.
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Gravity is a central force (no transverse component) so torque τ = r × F = 0 about Earth's centre; hence h = r²ω = r vθ is conserved. At perigee r is minimum, so v must be maximum. Kepler's second law is exactly this statement in geometric form.Q5. A Molniya orbit has a = 26,578 km, e = 0.72. Find perigee and apogee altitudes, the period, and the perigee-to-apogee speed ratio.
Show answer
rp = 7,442 km → hp ≈ 1,064 km; ra = 45,714 km → ha ≈ 39,336 km; T = 2π√(a³/μ) = 43,080 s = 11 h 58 min; vp/va = (1+e)/(1−e) = 6.14.Q6. Two satellites orbit Earth: one in a circular 800 km orbit, one in an ellipse with a = 7,178 km. Which has the longer period, and which the higher energy per kilogram?
Show answer
Equal a ⇒ equal period (≈ 6,054 s) and equal specific energy (ε = −μ/2a). The circular orbit has e = 0; the elliptical one has constant a but r and v vary around the orbit — Kepler's third law depends on a only.Q7. What is the escape speed from a 400 km circular orbit, and how does it relate to the circular speed there (vis-viva)?
Show answer
vcirc = √(μ/r) = √(398,600/6,778) = 7.67 km/s; vesc = √(2μ/r) = √2 × vcirc = 10.84 km/s. The √2 factor is the energy condition ε = 0 (parabola).Q8. List the six classical orbital elements and group them by function: size/shape, orientation, position-in-orbit.
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Size/shape: a, e. Orientation: i, Ω, ω. Position in orbit: ν (or M / time since perigee).Q9. Why do real satellite orbits deviate from Keplerian ellipses? Name the four principal perturbations for a LEO satellite.
Show answer
Earth is not a point mass nor a perfect sphere; other bodies and forces act. Perturbations: (1) J2 oblateness (nodal regression, apsidal precession), (2) atmospheric drag (decay), (3) third-body gravity (Moon, Sun), (4) solar radiation pressure. These are handled as corrections to the two-body ellipse.Q10. A tracking station measures a satellite's period as 23 h 56 min. What orbit is it in, and what additional conditions make it geostationary rather than merely geosynchronous?
Show answer
a = 42,164 km ⇒ geosynchronous. For geostationary also require e = 0 and i = 0° with prograde motion — only then is the satellite fixed over one equatorial longitude. Non-zero i or e gives a figure-8 drift.References and Further Reading
- Curtis, H. D. (2020). Orbital Mechanics for Engineering Students (4th ed.). Butterworth-Heinemann — Ch. 1–3 on two-body motion and Kepler's equation.
- Bate, R. R., Mueller, D. D., & White, J. E. (1971). Fundamentals of Astrodynamics. Dover — the classic concise treatment.
- Prussing, J. E., & Conway, B. A. (2012). Orbital Mechanics (2nd ed.). Oxford University Press.
- Fortescue, P., Swinerd, G., & Stark, J. (2011). Spacecraft Systems Engineering (4th ed.). Wiley — orbital context for subsystem design.
- Kepler, J. (1609). Astronomia Nova; (1619). Harmonices Mundi — original statements of the three laws.
- EEEN 567 Course Lecture Notes, Department of Electrical and Electronic Engineering, Egerton University.