Physics • Year 12 • Module 5 • Lesson 11
Gravitational Orbits
Lock in the orbital velocity formula, Kepler’s Third Law, and geostationary orbit conditions before tackling harder problems.
1. Formula recall card
Complete the table below for the three core formulas from this lesson. For each, write the SI units of every variable and a one-line note on when to use it. 9 marks (3 each)
| Name | Formula | Variables & SI units | When to use it |
|---|---|---|---|
| Orbital velocity | v = √(GM / r) | v = _____ (m s−1), G = _____ (_____), M = _____ (_____), r = _____ (_____) | |
| Kepler’s Third Law | T2 = 4π2r3 / (GM) | T = _____ (_____), r = _____ (_____), G = _____ (_____), M = _____ (_____) | |
| Orbital radius from surface | r = R + h | r = _____ (_____), R = _____ (_____), h = _____ (_____) |
2. Term–definition match
The definitions below are shuffled. Write the matching term from this list in the right-hand column: geostationary orbit, orbital velocity, Kepler’s Third Law, centripetal force, LEO, MEO, orbital radius, period. 8 marks (1 each)
| # | Definition | Matching term |
|---|---|---|
| 2.1 | The speed required for a satellite to maintain a stable circular orbit at a given radius. | |
| 2.2 | The time taken for one complete revolution around the central body. | |
| 2.3 | The law stating that T² ∝ r³ for all satellites orbiting the same central body. | |
| 2.4 | The inward force that keeps a satellite moving in a circular path; provided by gravity in orbital mechanics. | |
| 2.5 | An orbit with a period of 24 hours, directly above the equator, so the satellite appears stationary from Earth’s surface. | |
| 2.6 | The centre-to-centre distance between the satellite and the central body; equals R + h. | |
| 2.7 | Low Earth Orbit — altitude below ~2000 km; used by the ISS (~400 km) and weather satellites. | |
| 2.8 | Medium Earth Orbit — altitude ~2000–35 000 km; used by GPS satellites (~20 200 km). |
3. True or false — with correction
Circle T or F for each statement. If the statement is false, write the corrected version on the line below it. 12 marks (1 T/F + 1 correction each)
3.1 Orbital speed depends on the mass of the satellite — a heavier satellite moves slower in the same orbit. T / F
3.2 The orbital radius r in the formula v = √(GM/r) is measured from the surface of the planet to the satellite. T / F
3.3 A geostationary satellite must orbit directly above the equator to remain stationary over one point on Earth. T / F
3.4 A satellite in a higher orbit moves faster than one in a lower orbit around the same planet. T / F
3.5 Geostationary satellites can provide continuous communication coverage over polar regions. T / F
3.6 The ratio T²/r³ is the same for all satellites orbiting the same central body. T / F
4. Fill-in-the-blank derivation
Use the word/symbol bank to complete the step-by-step derivation of orbital velocity. Each term is used once. 6 marks (1 per blank)
Word/symbol bank:
cancels · centripetal · GM/r · m · mv²/r · √(GM/r)
For a satellite in circular orbit, the gravitational force provides the ___________ force required to keep it in orbit. We equate: GMm/r² = ___________. The satellite mass ___________ from both sides of the equation (the m in GMm and the ___________ in mv²/r both cancel). After simplification: v² = ___________. Taking the square root: v = ___________.
5. Geostationary orbit conditions
Answer each question in 1–2 sentences using precise physics terms. 8 marks (2 each)
5.1 State the orbital period of a geostationary satellite and explain why this specific period is required.
5.2 What altitude above Earth’s surface must all geostationary satellites orbit at, and why is this altitude unique?
5.3 Why must a geostationary orbit be equatorial (above the equator) rather than above some other latitude?
5.4 Name one real-world application of geostationary satellites and explain why geostationary orbit is suited to it.
Q1 — Formula recall card
Orbital velocity v = √(GM/r): v in m s−1; G = 6.67 × 10−11 N m2 kg−2; M in kg (mass of central body); r in m (centre-to-centre orbital radius). Use when the orbital speed of a satellite at a known radius is required.
Kepler’s Third Law T² = 4π²r³ / (GM): T in s; r in m; G in N m2 kg−2; M in kg. Use when the period is known and orbital radius is unknown (or vice versa), or when comparing orbits around the same central body.
Orbital radius r = R + h: r in m; R in m (radius of planet); h in m (altitude above surface). Always use before substituting into orbital formulas to convert altitude to centre-to-centre distance.
Q2 — Term–definition match
2.1 orbital velocity • 2.2 period • 2.3 Kepler’s Third Law • 2.4 centripetal force • 2.5 geostationary orbit • 2.6 orbital radius • 2.7 LEO • 2.8 MEO.
Q3 — True / false with correction
3.1 False. The satellite mass m cancels from the derivation; orbital speed depends only on the central body’s mass M and the orbital radius r: v = √(GM/r). A heavier satellite orbits at exactly the same speed as a lighter one at the same altitude.
3.2 False. The orbital radius r is the centre-to-centre distance from the centre of the planet to the satellite: r = R + h, where R is the planet’s radius and h is the altitude above the surface.
3.3 True.
3.4 False. A satellite in a higher orbit moves slower. From v = √(GM/r), increasing r decreases v; v ∝ 1/√r.
3.5 False. Geostationary satellites orbit in the equatorial plane and cannot provide good coverage at high latitudes or polar regions.
3.6 True.
Q4 — Fill-in-the-blank derivation
In order: centripetal / mv²/r / cancels / m / GM/r / √(GM/r).
Q5.1 — Geostationary period
A geostationary satellite has an orbital period of exactly 24 hours (86 400 s). This specific period is required so that the satellite completes exactly one orbit for every rotation of Earth, keeping it above the same point on the surface.
Q5.2 — Geostationary altitude
All geostationary satellites orbit at approximately 35 900 km above Earth’s surface (orbital radius ≈ 42 200 km from Earth’s centre). This altitude is unique because it is the only radius at which Kepler’s Third Law gives T = 24 h; any other altitude produces a different period, causing the satellite to drift.
Q5.3 — Equatorial requirement
A circular orbit must be centred on the centre of the Earth. The only circle centred on Earth’s centre that stays above a fixed point on the equator is the equatorial plane. A satellite in an inclined orbit will trace a figure-eight path (analemma) across the sky as seen from the ground, so it cannot remain stationary over one point.
Q5.4 — Real-world application
Accept any valid application. Example: TV broadcast satellites (e.g. Foxtel in Australia). The satellite appears stationary above the equator, allowing home dish antennas to be fixed in one direction and continuously receive the signal without tracking motors — making the system simple and inexpensive to operate.