Physics • Year 12 • Module 5 • Lesson 11
Gravitational Orbits
Apply orbital velocity, Kepler’s Third Law, and geostationary conditions to real satellite data, calculations, and multi-step reasoning tasks.
1. Satellite orbital speed calculations
The table below lists four Earth-orbiting satellites with their orbital altitudes. Use the constants provided to calculate the orbital speed of each. Show all working including r = RE + h. 8 marks (2 each)
Constants: G = 6.67 × 10−11 N m2 kg−2, ME = 5.97 × 1024 kg, RE = 6.37 × 106 m
| Satellite | Altitude h (km) | Orbital radius r (m) | Orbital speed v (m s−1) |
|---|---|---|---|
| ISS | 400 | ||
| Hubble Space Telescope | 547 | ||
| GPS satellite | 20 200 | ||
| Geostationary satellite | 35 900 |
1.1 Describe the pattern between orbital altitude and orbital speed. Explain this relationship using the formula v = √(GM/r). 2 marks
2. Circular-period relation, Jupiter’s moons
The table below shows the four Galilean moons of Jupiter. Treat their orbits as circular approximations and test whether T²/r³ is constant for the Jovian system. 7 marks
| Moon | Period T (days) | T (s) | Orbital radius r (m) | T² / r³ (s² m−3) |
|---|---|---|---|---|
| Io | 1.769 | 4.22 × 108 | ||
| Europa | 3.551 | 6.71 × 108 | ||
| Ganymede | 7.155 | 1.07 × 109 | ||
| Callisto | 16.69 | 1.88 × 109 |
2.1 Convert each period to seconds and calculate T²/r³ for each moon. Record your values in the table. 4 marks
2.2 Are the T²/r³ values consistent with Kepler’s Third Law? Explain your reasoning, including what the constant represents physically. 2 marks
2.3 Use the constant T²/r³ = 4π²/(GMJ) to calculate the mass of Jupiter MJ. Show your working. 1 mark
3. Predict and justify, changing orbits
Answer each scenario question using physics reasoning and relevant formulas. 9 marks (3 each)
3.1 A satellite at radius r1 is boosted to a new orbit at radius r2 = 4r1. By what factor does its orbital speed change? By what factor does its period change? Show your working clearly.
3.2 A student claims: “Satellites in orbit need to fire engines continuously to stop falling into Earth.” Identify the physics error in this claim and write a corrected explanation of why a satellite does not fall to Earth.
3.3 Two satellites A and B orbit Earth at 400 km and 800 km altitude respectively. Satellite A has mass 500 kg and satellite B has mass 5000 kg. Compare their orbital speeds. Explain whether the mass difference affects the result.
4. Select an Earth-orbit region
For each mission, select LEO, MEO or GEO and justify the choice using orbital properties and limitations. 6 marks (2 each)
4.1 Continuous weather images of the same Australian region using a fixed ground antenna.
4.2 A global navigation constellation whose satellites have periods of approximately 12 hours.
4.3 High-resolution Earth observation with a short signal path and frequent passes over different regions.
Q1, Orbital speed calculations
Using r = RE + h and v = √(GM/r):
ISS: r = 6.37 × 106 + 4.00 × 105 = 6.77 × 106 m. v = √(6.67 × 10−11 × 5.97 × 1024 / 6.77 × 106) = 7.66 × 103 m s−1.
Hubble: r = 6.37 × 106 + 5.47 × 105 = 6.917 × 106 m. v = √(3.981 × 1014 / 6.917 × 106) = 7.59 × 103 m s−1.
GPS: r = 6.37 × 106 + 2.02 × 107 = 2.657 × 107 m. v = √(3.981 × 1014 / 2.657 × 107) = 3.87 × 103 m s−1.
Geostationary: r = 6.37 × 106 + 3.59 × 107 = 4.227 × 107 m. v = √(3.981 × 1014 / 4.227 × 107) = 3.07 × 103 m s−1.
Q1.1: As altitude increases, orbital speed decreases. From v = √(GM/r), speed is inversely proportional to the square root of orbital radius: v ∝ 1/√r. A larger radius means weaker gravitational pull and a lower speed needed to maintain orbit. Award 1 mark for identifying the inverse relationship, 1 mark for the formula-based explanation.
Q2, Jupiter’s moons
Io: T = 1.769 × 86 400 = 1.528 × 105 s; T² = 2.335 × 1010 s2; r³ = (4.22 × 108)3 = 7.52 × 1025 m3; T²/r³ = 3.10 × 10−16 s2 m−3.
Europa: T = 3.551 × 86 400 = 3.068 × 105 s; T² = 9.41 × 1010 s2; r³ = (6.71 × 108)3 = 3.02 × 1026 m3; T²/r³ = 3.11 × 10−16 s2 m−3.
Ganymede: T = 7.155 × 86 400 = 6.182 × 105 s; T² = 3.82 × 1011 s2; r³ = (1.07 × 109)3 = 1.225 × 1027 m3; T²/r³ = 3.12 × 10−16 s2 m−3.
Callisto: T = 16.69 × 86 400 = 1.442 × 106 s; T² = 2.079 × 1012 s2; r³ = (1.88 × 109)3 = 6.645 × 1027 m3; T²/r³ = 3.13 × 10−16 s2 m−3.
Q2.2: Yes, all four values are approximately 3.1 × 10−16 s2 m−3consistent with Kepler’s Third Law. The constant equals 4π²/(GMJ) and depends only on the mass of Jupiter, not on which moon is used [1]. This confirms that T²/r³ = constant for all bodies orbiting the same central mass [1].
Q2.3: MJ = 4π² / (G × 3.11 × 10−16) = 39.48 / (6.67 × 10−11 × 3.11 × 10−16) ≈ 1.90 × 1027 kg (actual: 1.898 × 1027 kg). Award 1 mark for correct rearrangement and substitution with a value in the correct order of magnitude.
Q3.1, Changing orbits
From v = √(GM/r): if r → 4r, then v → √(GM/4r) = v/2. Speed halves [1]. From T² = 4π²r³/(GM): if r → 4r, then T² → 4π²(4r)3/(GM) = 64 × 4π²r³/(GM), so T → 8T. Period increases by a factor of 8 [1]. Award 1 mark per correct factor with working shown.
Q3.2, Satellite misconception
Error: the claim assumes a satellite needs continuous thrust to maintain orbit, as if it were “fighting gravity” [1]. Correct explanation: in a stable circular orbit, the gravitational force provides exactly the centripetal force needed to deflect the satellite’s path in a circle, no tangential thrust is required. The satellite is continuously “falling” toward Earth but its horizontal speed is large enough that it keeps missing the curved Earth beneath it. This is Newton’s concept of an orbit as perpetual free-fall [1]. Award 1 mark for identifying the error, 1 mark for the correct explanation involving gravity as centripetal force.
Q3.3, Mass comparison
Satellite A (400 km): r = 6.77 × 106 m, vA ≈ 7.66 km s−1. Satellite B (800 km): r = 6.37 × 106 + 8.00 × 105 = 7.17 × 106 m; vB = √(3.981 × 1014/7.17 × 106) ≈ 7.45 km s−1 [1]. Satellite A is faster than B because it is at a lower orbit [1]. The mass difference (500 kg vs 5000 kg) has no effect on orbital speed; both masses cancel in the derivation GMm/r² = mv²/r, so orbital speed is independent of satellite mass [1].
Q4.1, Continuous regional weather monitoring
GEO. A geostationary satellite remains above one equatorial longitude, so it can continuously observe the same broad region and use a fixed ground antenna. Its limitations include greater latency and weak coverage at high latitudes.
Q4.2, Navigation constellation
MEO. GPS-like satellites operate near 20 200 km altitude with periods close to 12 h. MEO gives wider coverage per satellite than LEO while avoiding the very large radius and 24 h period of GEO.
Q4.3, Earth observation
LEO. Its relatively small radius gives a short period, a shorter signal path and closer imaging. A single satellite does not continuously cover one region, so repeated passes or a constellation are required.