Cambridge IGCSE Physics 0625 – Unit 6: Space Physics
6.1.2 The Solar System – Core Content
Learning Objective
Explain why the Sun’s gravitational field becomes weaker with distance and why the orbital speeds of the planets decrease as their distance from the Sun increases.
Key Concepts
- Inverse‑square law for the Sun’s gravitational field
Field strength g = GM☉/r²
(where G = 6.67 × 10⁻¹¹ N m² kg⁻², M☉ = 1.99 × 10³⁰ kg, r = distance from the Sun’s centre).
Because the same total “pull” is spread over the surface of a sphere (area = 4πr²), the field strength falls as 1/r².
- Orbital speed for a circular orbit
Required centripetal force = gravitational force → mv²/r = GM☉m/r²
Hence v = √(GM☉/r) → v ∝ r⁻¹ᐟ².
A weaker pull at larger r means a slower speed is sufficient to keep the planet in orbit.
- Orbital period
Period T = circumference / speed = 2πr / v → T = 2π√(r³/GM☉) → T ∝ r³ᐟ² (Kepler’s 3rd law).
Why the Gravitational Field Decreases with Distance
The Sun’s mass creates a field that radiates outward uniformly. At distance r the field lines intersect a spherical surface of area 4πr². Since the total flux is constant, the field strength per unit area must fall as the area grows – the classic inverse‑square behaviour.
Why Orbital Speed Decreases with Distance
In a stable circular orbit the inward gravitational force exactly balances the outward “centrifugal” tendency of the moving planet. If the pull is weaker (larger r), a smaller speed provides the required centripetal force, so the planet moves more slowly.
Illustrative Data – Distance vs. Orbital Speed
| Planet | Mean distance from Sun (×10⁸ km) | Average orbital speed (km s⁻¹) |
|---|
| Mercury | 0.579 | 47.4 |
| Venus | 1.082 | 35.0 |
| Earth | 1.496 | 29.8 |
| Mars | 2.279 | 24.1 |
| Jupiter | 7.785 | 13.1 |
| Saturn | 14.34 | 9.7 |
| Uranus | 28.73 | 6.8 |
| Neptune | 44.95 | 5.4 |
The table shows a clear trend: the farther a planet is from the Sun, the slower it moves in its orbit.
Related Core Topics in Unit 6
- Earth’s rotation – 24 h period produces day and night.
- Earth’s axial tilt (≈23.5°) – causes the seasons.
- Orbital period – time for one complete revolution (e.g., Earth = 365 days).
- Escape velocity – minimum speed to leave a planet’s gravitational field:
ve = √(2GM/r)
- Artificial satellites – low‑Earth orbit (LEO) vs. geostationary orbit (GEO); uses such as communication, weather monitoring, navigation.
- Space‑probe navigation – matching the target planet’s orbital speed and using gravity‑assist manoeuvres.
Practical Implications for Space Missions
- To travel from Earth to another planet a spacecraft must first reach Earth’s escape velocity (≈11.2 km s⁻¹). After leaving Earth’s sphere of influence it must adjust its speed to match the slower orbital speed of the destination planet.
- Because orbital periods increase with distance, missions to the outer planets take many years (e.g., Voyager 2: 12 years to reach Neptune).
- Gravity‑assist fly‑bys exploit the relative motion of a planet to increase a spacecraft’s heliocentric speed without using extra fuel.
Sample Calculations (Extended Material)
- Orbital speed of a satellite 300 km above Earth’s surface
r = RE + 300 km = 6.371 × 10⁶ m + 3.0 × 10⁵ m = 6.671 × 10⁶ m
v = √(GME/r) = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴)/6.671 × 10⁶] ≈ 7.73 km s⁻¹
- Escape velocity from the Moon (radius = 1.74 × 10⁶ m, mass = 7.35 × 10²² kg)
ve = √(2GMMoon/r) ≈ 2.38 km s⁻¹
Kepler’s Laws (Supplementary)
- Planets move in ellipses with the Sun at one focus.
- A line joining a planet to the Sun sweeps out equal areas in equal times – explains why a planet moves faster at perihelion and slower at aphelion.
- Square of the orbital period is proportional to the cube of the semi‑major axis: T² ∝ r³ (derived from the v ∝ r⁻¹ᐟ² relationship).
Quick Reference Table – Earth‑Moon‑Sun System
| Object | Key Property | Relevance to Topic |
|---|
| Sun | Mass = 1.99 × 10³⁰ kg | Source of the gravitational field (g = GM/r²) |
| Earth | Rotation = 24 h; axial tilt ≈ 23.5° | Day/night, seasons; reference for escape velocity (11.2 km s⁻¹) |
| Moon | Orbital period ≈ 27.3 days; distance ≈ 3.84 × 10⁵ km | Example of orbital period and gravitational balance |
Suggested Diagrams (for classroom or revision notes)
- A schematic of the Solar System showing concentric circular orbits, arrows indicating relative orbital speeds (longer arrows for inner planets), and a colour gradient from bright (strong field) near the Sun to faint (weak field) outward.
- Cross‑section of Earth illustrating rotation, axial tilt, and the resulting variation in solar illumination (seasons).
- Force diagram for a planet in circular orbit: gravitational force directed toward the Sun balanced by centripetal “requirement”.
- Plot of orbital speed (v) versus distance (r) on a log‑log scale to highlight the v ∝ r⁻¹ᐟ² relationship.
6.1 The Solar System – Additional Core Topics (required for the IGCSE)
- Earth’s rotation – 24 h day/night cycle; angular speed ω = 2π/86400 s⁻¹.
- Earth’s tilt (≈23.5°) – produces seasons; variation in solar altitude.
- Moon’s orbit – period 27.3 days; causes lunar phases; tidal effects.
- Definition of orbital period – time for one complete revolution around a central body.
- Escape velocity – ve = √(2GM/r); examples for Earth, Moon, and the Sun.
- Artificial satellites – LEO (≈200–2 000 km), GEO (≈35 786 km); applications.
- Space‑probe navigation – matching target orbital speed, Hohmann transfer orbits, gravity assists.
Key Take‑away
The Sun’s gravitational pull weakens with the square of the distance, so planets farther out experience a smaller pull and consequently travel more slowly in their orbits. This inverse‑square relationship underlies the observed pattern of orbital speeds, orbital periods, and the design of space‑flight trajectories.