Objectives of the Laboratory
After completing this virtual laboratory, the student should be able to:
- Demonstrate, by simulation, Kepler's first law: satellite orbits are ellipses with the Earth at one focus, and quantify the orbit using the semi-major axis a and eccentricity e.
- Verify Kepler's second law: the radius vector sweeps equal areas in equal times, and relate the areal-velocity constancy to the variation of orbital speed between perigee and apogee.
- Verify Kepler's third law numerically for a range of altitudes and hence confirm that T²/a³ = 4π²/μ = constant.
- Compute orbital periods, perigee/apogee radii and velocities using Kepler's laws and the vis-viva equation, and classify orbits as LEO, MEO, GEO or HEO.
- Derive the geostationary radius and altitude from Kepler's third law.
Lab safety note: this is a computational (virtual) laboratory. The only “hazards” are numerical: always check that altitudes are positive, that a = rp + ra is measured from Earth's centre, and that units (km, s, km/s) are consistent. Record your assumptions in the report.
Theory
1. Kepler's First Law — The Law of Ellipses (1609)
The orbit of a satellite is an ellipse with the centre of the Earth at one of the two foci. The ellipse is described by its semi-major axis a (size) and eccentricity e (shape):
rp = a(1 − e) ra = a(1 + e) a = (rp + ra)/2 e = (ra − rp)/(ra + rp)
rp, ra = perigee/apogee radii from Earth's centre; hp = rp − R⊙, ha = ra − R⊙ with R⊙ = 6,378 km. e = 0 gives a circle.
Orbit equation: r(ν) = a(1 − e²)/(1 + e cosν)
ν = true anomaly (angular position measured from perigee).
2. Kepler's Second Law — The Law of Equal Areas (1609)
The radius vector from Earth's centre to the satellite sweeps out equal areas in equal times. Physically this is conservation of angular momentum, because gravity is a central (torque-free) force:
dA/dt = h/2 = constant, h = r² dν/dt = r vθ = √[μ a (1 − e²)]
h = specific angular momentum (km²/s). Consequence: vp/va = ra/rp = (1+e)/(1−e).
3. Kepler's Third Law — The Harmonic Law (1619)
The square of the orbital period is proportional to the cube of the semi-major axis. Newton later supplied the constant through the gravitational parameter μ = GM⊙:
T² = (4π²/μ) a³ ⇔ T = 2π √(a³/μ), μ = 3.986×105 km3/s2
Valid for any ellipse about Earth. T depends only on a — not on satellite mass or eccentricity.
4. Vis-Viva Equation (speed on the orbit)
v² = μ (2/r − 1/a)
At perigee: vp = √[μ(2/rp − 1/a)]; at apogee: va = √[μ(2/ra − 1/a)]. For a circular orbit (r = a) this reduces to v = √(μ/r).
5. Kepler's Equation (position vs. time)
The simulation advances the satellite by solving Kepler's equation for the eccentric anomaly E at each time step (Newton–Raphson iteration):
M = E − e sin E, then x = a(cos E − e), y = a√(1 − e²) sin E
M = mean anomaly (uniform clock: M = 2πt/T). Position measured from the occupied focus, where Earth sits.
6. Worked Example (expected output of Experiment 1, row 2)
Given: circular orbit at h = 1,000 km.
a = 6,378 + 1,000 = 7,378 km; e = 0.
T = 2π√(7,378³/3.986×105) = 2π√(9.283×105) = 2π(963.5) = 6,307 s = 105.1 min.
v = √(μ/a) = √(3.986×105/7,378) = √(54.03) = 7.35 km/s.
a = 6,378 + 1,000 = 7,378 km; e = 0.
T = 2π√(7,378³/3.986×105) = 2π√(9.283×105) = 2π(963.5) = 6,307 s = 105.1 min.
v = √(μ/a) = √(3.986×105/7,378) = √(54.03) = 7.35 km/s.
Apparatus and Software
- This HTML virtual laboratory (any modern web browser — Chrome, Edge, Firefox).
- Orbit Simulator (Section 4): canvas-based animation with adjustable perigee/apogee altitudes, live orbital read-outs and equal-area sector shading.
- Period Calculator (Section 6): Kepler's-third-law computation with classification.
- Scientific calculator (or the in-page calculators) and a spreadsheet for the results tables.
- Constants: μ = 3.986×105 km3/s2, R⊙ = 6,378 km.
SECTION 4 — ORBIT SIMULATOR (EXPERIMENTS 1–3)
Orbit Simulator
Set the perigee and apogee altitudes and run the simulation. The satellite's position is advanced in equal time steps by solving Kepler's equation. Observe: (i) the elliptical path (Experiment 1); (ii) the two shaded sectors — equal areas swept in equal times, and the speed change (Experiment 2); (iii) the read-out values (Experiment 3).
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a (km)
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e
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Period (min)
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v perigee (km/s)
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v apogee (km/s)
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Class
Procedure
Experiment 1 — Kepler's First Law: Orbit Geometry (Ellipse Parameters)
1
Open the Orbit Simulator. Set hp = 1,000 km and ha = 39,700 km (Molniya preset).
2
Record the simulated a and e from the read-out. Independently compute a = (rp+ra)/2 and e = (ra−rp)/(ra+rp) with r = R⊙ + h, and compare (they must agree).
3
Repeat for the ISS preset (circular) and the GPS preset (circular). Confirm that e = 0 for circular orbits and that the ellipse collapses to a circle.
4
Sketch (or capture) the ellipse and label the two foci, the occupied focus (Earth), perigee, apogee, and the major axis 2a.
Experiment 2 — Kepler's Second Law: Equal Areas, Varying Speed
1
Set the Molniya preset (e = 0.72). Run the animation and observe the satellite's angular speed: fast at perigee (green sector), slow at apogee (red sector).
2
Verify the two shaded sectors correspond to the same number of equal time steps (6 each) and hence equal areas — by Kepler's second law.
3
Record vp and va from the read-out; compute vp/va and check it equals (1+e)/(1−e) = 6.14.
4
Reduce e by setting hp = 30,000 km, ha = 39,700 km (e ≈ 0.14). Observe how the speed variation shrinks — confirming that a circle (e = 0) has uniform speed.
Experiment 3 — Kepler's Third Law: T² ∝ a³ Verification
1
Using the Period Calculator below, compute T for each altitude in Table 1 (circular orbits, 400 / 1,000 / 2,000 / 20,200 / 35,786 km).
2
Complete the table: a (km), T (s), a³ (km³), T² (s²), and the ratio T²/a³.
3
Show that T²/a³ is constant for all rows and equal to 4π²/μ = 9.899×10-5 s²/km³. This is Kepler's third law verified numerically.
4
Plot T (y-axis) against a (x-axis) and/or T² against a³ in your spreadsheet; fit a power law and confirm the exponent is 1.5.
Experiment 4 — Derivation of the Geostationary Orbit (Design Exercise)
1
Impose T = sidereal day = 86,164 s in Kepler's third law and solve for a = [μ(T/2π)²]1/3.
2
Hence obtain h = a − R⊙ = 35,786 km, and state the additional conditions (i = 0°, prograde) that make the orbit geostationary rather than merely geosynchronous.
3
Using the simulator, set hp = ha = 35,786 km and confirm the computed period is 23 h 56 min.
Period Calculator & Results Tables
Period Calculator (circular orbit):
Table 1 — Kepler's Third-Law Verification (fill in from the calculator)
| h (km) | a (km) | T (s) | T (min) | a³ (km³) | T² (s²) | T²/a³ (s²/km³) | Orbit class |
|---|---|---|---|---|---|---|---|
| 400 | |||||||
| 1,000 | |||||||
| 2,000 | |||||||
| 20,200 | |||||||
| 35,786 |
Table 2 — Kepler's Second Law / Vis-Viva (fill in from the simulator)
| Case | hp (km) | ha (km) | e | vp (km/s) | va (km/s) | vp/va | (1+e)/(1−e) |
|---|---|---|---|---|---|---|---|
| Molniya | 1,000 | 39,700 | |||||
| Low-e case | 30,000 | 39,700 | |||||
| Circular | 1,000 | 1,000 |
Reference values for checking your tables (instructors may withhold): T at 400 km = 5,554 s (92.6 min); at 2,000 km = 7,632 s (127.2 min); at 20,200 km = 43,122 s (11 h 59 min); Molniya vp = 9.60 km/s, va = 1.56 km/s.
Guidelines for Writing the Laboratory Report
The report must be an individual submission, typed, and organised exactly in the following structure. Use SI units with km/km/s for orbital quantities, show all formula substitutions, and quote results to appropriate precision (T to 0.1 min or better).
| Section | Contents | Marks |
|---|---|---|
| 1. Title page | Lab title, course code (EEEN 567), student name & registration number, department (Electrical & Electronic Engineering, Egerton University), date, group number | 5 |
| 2. Objectives | The lab objectives, restated in your own words | 5 |
| 3. Theory | Concise summary of the four theory blocks: first-law geometry, second law/angular momentum, third law, vis-viva. Include the governing equations with defined symbols. | 15 |
| 4. Apparatus | Simulator description, constants used (μ, R⊙) | 5 |
| 5. Procedure | Brief description of what you actually did per experiment (not a copy of the handout), noting any deviations | 10 |
| 6. Results | Completed Tables 1 and 2 with units; the T vs. a (and T² vs. a³) plot with fitted power law and exponent; screenshots of the simulator for each preset | 25 |
| 7. Analysis & Discussion | Verification statements: (a) T²/a³ constant ≈ 9.9×10-5 s²/km³ with % deviation; (b) vp/va = (1+e)/(1−e); (c) GEO derivation result; discussion of discrepancies and their sources (rounding, solver tolerance, unit slips); physical interpretation of each law for satellite missions | 20 |
| 8. Conclusion | Short paragraph: which laws were verified, the key numerical findings, and one engineering implication (e.g., HEO apogee dwell) | 10 |
| 9. References | Course notes, Curtis (Orbital Mechanics), ITU/standard constants — any consistent citation style | 5 |
| Total | 100 | |
Quality Requirements
- Show every numerical substitution explicitly (e.g., T = 2π√(7,378³/3.986×105) = 6,307 s) — answers without working earn no method marks.
- Report percentages to 1 decimal; state agreement criteria (e.g., simulated vs. computed within 0.5%).
- Figures must be numbered, titled, and referenced in the text; tables likewise.
- Plagiarism and copied tables between students attract zero marks for the results section.
Review Questions
Q1. From your completed Table 1, state the average value of T²/a³ and its percentage deviation from 4π²/μ.
Show answer
Expect T²/a³ ≈ 9.899×10-5 s²/km³ for every row; deviations should be < 1% if full precision is carried (round-off in a³ and T² is the main error source).Q2. Why does the simulated satellite move fastest at perigee even though no force pushes it there?
Show answer
Gravity is a central force (zero torque), so angular momentum h = r vθ is conserved; at minimum r the transverse velocity must be maximum (Kepler's second law / energy exchange: kinetic ↔ potential along the ellipse).Q3. What would happen to the period if the satellite's mass were doubled?
Show answer
Nothing — T = 2π√(a³/μ) contains no satellite mass: both inertia and gravitational force scale with m and cancel. (This fails only if the satellite were comparable to Earth's mass.)Q4. Using your GEO derivation, explain why a geosynchronous satellite with i = 10° cannot serve fixed DTH dishes.