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. Kepler’s Third Law — Jupiter’s moons
The table below shows the four Galilean moons of Jupiter with their orbital data. Use the data to verify Kepler’s Third Law 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. Interpret a graph — T² vs r³ for Solar System planets
The graph below plots T² (in years2) against r³ (in AU3) for the eight planets of the Solar System. Each data point is labelled. 6 marks
Figure 4. T² vs r³ for the Solar System (AU and years units). Inner planets (Mercury, Venus, Earth, Mars) cluster near the origin and are shown as a single cluster. Best-fit dashed line shown.
4.1 Describe the shape of the best-fit line and explain what this tells you about the relationship between T² and r³. 2 marks
4.2 The gradient of this line equals 1 in AU/years units. Use T²/r³ = 4π²/(GM) to explain why the gradient would not equal 1 if you plotted the graph in SI units (seconds and metres). 2 marks
4.3 A newly discovered exoplanet orbits a star with the same mass as our Sun, at a distance of 2 AU. Use the graph trend to predict its orbital period in years. 2 marks
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−3 — consistent 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 — Graph shape
The best-fit line is a straight line through the origin [1]. This shows that T² is directly proportional to r³ — confirming Kepler’s Third Law: T² ∝ r³ with a constant ratio T²/r³ that is the same for all planets orbiting the Sun [1].
Q4.2 — Gradient in SI units
In AU/year units the gradient equals 1 because AU and year are defined so that Earth’s orbital data gives T²/r³ = 1 year²/AU³ by definition [1]. In SI units the gradient is 4π²/(GMSun) = 4π² / (6.67 × 10−11 × 1.99 × 1030) ≈ 2.97 × 10−19 s2 m−3, which is a very different numerical value [1].
Q4.3 — Exoplanet period prediction
In the same stellar system (same Sun-like mass), T² = r³ (AU, years). For r = 2 AU: T² = (2)3 = 8; T = √8 = 2√2 ≈ 2.83 years [1]. Award 1 mark for correct setup, 1 mark for final answer with units.