🪐 Identify and describe the objects in our Solar System: planets, dwarf planets, moons, asteroids and comets
☀️ Explain how the Sun's gravity keeps all objects in orbit and how orbital speed varies with distance
📐 Compare the relative sizes of planets, the Sun, and distances within the Solar System using orders of magnitude
🌌 Describe the highly elliptical orbits of comets and how speed changes along the orbit
🔭 Use the relationship between orbital period, orbital radius and gravitational force qualitatively and quantitatively
📏 Understand the use of astronomical units (AU) and light-years to measure distances in space
🌍 Objects in the Solar System
The Solar System consists of the Sun at its centre, and everything bound to it by gravity. Our Sun is a medium-sized star roughly 1.4 million km in diameter — about 109 times the diameter of Earth.
Planet: A large body that orbits the Sun, has sufficient gravity to be roughly spherical, and has cleared its orbital neighbourhood of other debris.
There are eight planets in order from the Sun: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune. The first four are rocky (terrestrial) planets; the outer four are gas or ice giants.
Moon (natural satellite): A body that orbits a planet rather than the Sun directly. Earth has one Moon; Jupiter has over 90 known moons.
Asteroid: A rocky body, much smaller than a planet, that orbits the Sun. Most asteroids lie in the asteroid belt between Mars and Jupiter. They range from a few metres to ~900 km across (Ceres).
Comet: A small icy body that orbits the Sun in a highly elliptical orbit. When a comet approaches the Sun, ices vaporise to form a bright coma and tail pointing away from the Sun due to solar wind and radiation pressure.
Dwarf Planet: A body that orbits the Sun and is roughly spherical but has not cleared its orbital neighbourhood. Examples: Pluto, Eris, Ceres.
The Solar System formed about 4.6 billion years ago from a collapsing cloud of gas and dust (nebula). Gravity pulled material together to form the Sun; remaining material formed a rotating disc from which planets condensed.
🔵 Orbits and Gravitational Force
All objects in the Solar System are held in orbit by gravity. Gravity is a long-range attractive force between any two masses. Isaac Newton showed that the gravitational force depends on both masses and the distance between them:
F = G × (m₁ × m₂) ÷ r²
where G = 6.67 × 10⁻¹¹ N m² kg⁻²
Symbol
Quantity
SI Unit
F
Gravitational force
Newton (N)
G
Gravitational constant
N m² kg⁻²
m₁, m₂
Masses of the two objects
kilogram (kg)
r
Distance between centres
metre (m)
For circular orbits, gravity provides the centripetal force needed to keep the planet curving around the Sun. A planet in a circular orbit moves at constant speed but is constantly changing direction — so it is always accelerating toward the Sun.
As distance from the Sun increases, gravitational force decreases (inverse square law). This means outer planets orbit more slowly and have longer orbital periods than inner planets.
Orbital speed and period: For a circular orbit, centripetal force equals gravitational force:
F = m × v² ÷ r and F = G × M × m ÷ r²
Combining: v² = G × M ÷ r, so v = √(G × M ÷ r)
This means orbital speed v decreases as orbital radius r increases — Mercury moves at ~48 km/s; Neptune at only ~5 km/s.
The orbital period T (time for one full orbit) is related to radius by Kepler's Third Law:
T² ∝ r³ (for orbits around the same central body)
☄️ Comets and Elliptical Orbits
Unlike planets (whose orbits are nearly circular), comets follow highly elliptical orbits. The Sun sits at one focus of the ellipse.
As a comet approaches the Sun:
Gravitational force on the comet increases (F ∝ 1/r²)
The comet speeds up (gaining kinetic energy from gravitational potential energy)
At perihelion (closest point), speed is maximum
After perihelion, it slows as it moves away (gaining GPE, losing KE)
At aphelion (farthest point), speed is minimum
Perihelion: The point in a comet's (or planet's) orbit closest to the Sun. Speed is at its maximum here.
Aphelion: The point farthest from the Sun. Speed is at its minimum here.
Energy is conserved throughout the orbit: total mechanical energy (KE + GPE) remains constant (ignoring air resistance, which doesn't exist in space). This is why comets speed up near the Sun and slow down far away.
Halley's Comet has an orbital period of about 75 years. It reaches as close as 0.59 AU and as far as 35 AU from the Sun — a dramatically stretched ellipse compared to Earth's nearly circular orbit.
📏 Sizes and Distances: Orders of Magnitude
The scale of the Solar System is enormous. We use scientific notation and special units to express these distances meaningfully.
Astronomical Unit (AU): The mean distance from Earth to the Sun. 1 AU ≈ 1.5 × 10¹¹ m (150 million km).
Light-year (ly): The distance light travels in one year. 1 ly ≈ 9.46 × 10¹⁵ m. Used for distances to other stars and galaxies.
Object
Diameter
Distance from Sun
Sun
1.4 × 10⁹ m
—
Earth
1.3 × 10⁷ m
1.5 × 10¹¹ m (1 AU)
Jupiter
1.4 × 10⁸ m
7.8 × 10¹¹ m (5.2 AU)
Neptune
4.9 × 10⁷ m
4.5 × 10¹² m (30 AU)
Nearest star (Proxima Centauri)
—
4.0 × 10¹⁶ m (4.2 ly)
Comparing Earth and Sun: The Sun's diameter is about 100× Earth's diameter. In volume, the Sun could contain about 1.3 million Earths. Neptune is about 30 times farther from the Sun than Earth is — so light takes roughly 4 hours to reach Neptune, compared to 8 minutes to reach Earth.
When comparing sizes and distances, always calculate the ratio and express it as a power of 10 (order of magnitude). The Solar System spans roughly 10¹³ m; the nearest star is 10¹⁶ m away — three orders of magnitude farther.
🌑 Moons and Their Orbits
Moons are natural satellites that orbit planets. The same gravitational principles apply: a moon orbits its planet because gravity provides the centripetal force.
Key facts about moons:
Earth's Moon: diameter ≈ 3,474 km, orbital radius ≈ 3.84 × 10⁸ m, period ≈ 27.3 days
Jupiter's moon Io is volcanically active due to tidal heating from Jupiter's gravity
Europa (Jupiter's moon) may have a liquid water ocean beneath its icy surface
Saturn's moon Titan has a thick nitrogen atmosphere and lakes of liquid methane
Kepler's Third Law applies to moon systems too — moons closer to their planet have shorter orbital periods. For Earth's Moon, using T² ∝ r³ we can compare it with artificial satellites to find their orbital radii.
The International Space Station (ISS) orbits at about 400 km altitude with a period of ~92 minutes. The Moon orbits at 384,000 km with a period of 27.3 days. Their ratio of orbital radii ≈ 960, and T² ∝ r³ predicts (960)³/² ≈ 29,700 minutes for the Moon — consistent with ~27.3 days.
For any two moons of the same planet: T₁² ÷ T₂² = r₁³ ÷ r₂³
Example 1: Calculate the gravitational force between the Sun (mass = 2.0 × 10³⁰ kg) and Earth (mass = 6.0 × 10²⁴ kg), given that the mean orbital radius of Earth is 1.5 × 10¹¹ m. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)
1 Write out the formula for gravitational force: F = G × m₁ × m₂ ÷ r²
2 Identify the values: G = 6.67 × 10⁻¹¹, m₁ = 2.0 × 10³⁰ kg, m₂ = 6.0 × 10²⁴ kg, r = 1.5 × 10¹¹ m
5 Divide: F = 8.004 × 10⁴⁴ ÷ 2.25 × 10²² = 3.56 × 10²² N
F ≈ 3.6 × 10²² N (to 2 s.f.) — this enormous force keeps Earth in its orbit around the Sun.
Example 2: Earth's orbital radius is 1.5 × 10¹¹ m and its orbital period is 1 year. Mars has an orbital radius of 2.28 × 10¹¹ m. Use Kepler's Third Law (T² ∝ r³) to find the orbital period of Mars in Earth years.
1 Write Kepler's Third Law for two planets: T_M² ÷ T_E² = r_M³ ÷ r_E³
2 Rearrange: T_M² = T_E² × (r_M ÷ r_E)³
3 Calculate the ratio of radii: r_M ÷ r_E = 2.28 × 10¹¹ ÷ 1.5 × 10¹¹ = 1.52
6 Take the square root: T_M = √3.512 = 1.874 years
T_Mars ≈ 1.87 Earth years (actual value ≈ 1.88 years — excellent agreement!)
Example 3: The Sun has a diameter of 1.4 × 10⁹ m. Earth has a diameter of 1.27 × 10⁷ m. (a) How many times larger is the Sun's diameter than Earth's? (b) Approximately how many Earths could fit inside the Sun by volume?
1 Part (a): Ratio of diameters = 1.4 × 10⁹ ÷ 1.27 × 10⁷
2 = (1.4 ÷ 1.27) × 10⁽⁹⁻⁷⁾ = 1.10 × 10² ≈ 110
3 Part (b): Volume ∝ diameter³, so ratio of volumes = (diameter ratio)³ = 110³
4 110³ = 110 × 110 × 110 = 1,331,000 ≈ 1.3 × 10⁶
(a) The Sun's diameter is about 110 times Earth's diameter. (b) Approximately 1.3 million Earths could fit inside the Sun by volume.
Example 4: A comet is at perihelion (closest approach to the Sun) at a distance of 8.9 × 10¹⁰ m and moving at 54 km/s. At aphelion it is 6.0 × 10¹² m from the Sun. Explain qualitatively why the comet moves much more slowly at aphelion, and estimate its approximate speed there using conservation of angular momentum (v₁r₁ = v₂r₂ for elliptical orbits).
1 Qualitative explanation: At perihelion, gravitational PE is minimum (close to Sun), so KE is maximum → maximum speed. At aphelion, GPE is maximum, KE is minimum → minimum speed. Total energy (KE + GPE) is conserved throughout the orbit.
2 Using conservation of angular momentum: v₁ × r₁ = v₂ × r₂ (valid at perihelion and aphelion where velocity is perpendicular to radius)
3 Rearrange for v₂: v₂ = v₁ × r₁ ÷ r₂
4 v₂ = 54,000 m/s × (8.9 × 10¹⁰ ÷ 6.0 × 10¹²)
5 v₂ = 54,000 × 0.01483 = 800 m/s ≈ 0.80 km/s
At aphelion the comet moves at approximately 0.80 km/s — about 68 times slower than at perihelion — because it has converted almost all its kinetic energy into gravitational potential energy.
Q1. Which of the following correctly lists the planets in order from the Sun outward?
Q2. The gravitational force on a planet doubles. According to the inverse square law, by what factor has the distance between the planet and the Sun changed?
Q3. A comet has an orbital period of 76 years. Using Kepler's Third Law (T² ∝ r³), if Earth's orbital radius is 1.5 × 10¹¹ m, calculate the comet's mean orbital radius in metres. Enter your answer in the form A × 10ⁿ (e.g. 2.7 × 10¹² for 2700000000000).
Q4. Which statement about a comet at perihelion is correct?
Q5. The distance from Earth to the Sun is 1.5 × 10¹¹ m. Light travels at 3.0 × 10⁸ m/s. Calculate the time in minutes for light to travel from the Sun to Earth. Enter a number in minutes (to 1 d.p.).
Challenge Q1 — Extended Calculation: Saturn has an orbital radius of 1.43 × 10¹² m and an orbital period of 29.5 Earth years. Jupiter has an orbital radius of 7.78 × 10¹¹ m. Use Kepler's Third Law to verify this relationship for Saturn and Jupiter, then predict Jupiter's orbital period in Earth years. Show all working.
Step 1: Kepler's Third Law: T² ÷ r³ = constant for all planets around the Sun. Step 2: For Saturn: T_S² ÷ r_S³ = (29.5)² ÷ (1.43×10¹²)³ = 870.25 ÷ 2.924×10³⁶ = 2.976×10⁻³⁴ yr² m⁻³ Step 3: For Jupiter using the same constant: T_J² = 2.976×10⁻³⁴ × (7.78×10¹¹)³ Step 4: (7.78×10¹¹)³ = 471.0×10³³ = 4.71×10³⁵ m³ Step 5: T_J² = 2.976×10⁻³⁴ × 4.71×10³⁵ = 140.2 yr² Step 6: T_J = √140.2 ≈ 11.8 years Answer: Jupiter's orbital period ≈ 11.8 Earth years (actual: 11.86 years ✓). The ratio T²/r³ is confirmed consistent between Saturn and Jupiter, verifying Kepler's Third Law.
Challenge Q2 — Gravitational Force Comparison: The mass of Mars is 6.4 × 10²³ kg and its orbital radius is 2.28 × 10¹¹ m. The mass of Jupiter is 1.9 × 10²⁷ kg and its orbital radius is 7.78 × 10¹¹ m. The Sun's mass is 2.0 × 10³⁰ kg. (a) Calculate the gravitational force on Mars due to the Sun. (b) Calculate the gravitational force on Jupiter due to the Sun. (c) Explain why Jupiter experiences a greater force despite being much further away. (G = 6.67 × 10⁻¹¹ N m² kg⁻²)
(a) Force on Mars:
F = G×M_Sun×M_Mars ÷ r_Mars² = 6.67×10⁻¹¹ × 2.0×10³⁰ × 6.4×10²³ ÷ (2.28×10¹¹)²
Numerator: 6.67×10⁻¹¹ × 2.0×10³⁰ × 6.4×10²³ = 8.538×10⁴³
Denominator: (2.28×10¹¹)² = 5.198×10²² m²
F_Mars = 8.538×10⁴³ ÷ 5.198×10²² = 1.64 × 10²¹ N
(b) Force on Jupiter:
F = 6.67×10⁻¹¹ × 2.0×10³⁰ × 1.9×10²⁷ ÷ (7.78×10¹¹)²
Numerator: 6.67×10⁻¹¹ × 2.0×10³⁰ × 1.9×10²⁷ = 2.534×10⁴⁷
Denominator: (7.78×10¹¹)² = 6.053×10²³ m²
F_Jupiter = 2.534×10⁴⁷ ÷ 6.053×10²³ = 4.19 × 10²³ N
(c) Explanation: Although Jupiter is ~3.4× farther away (reducing force by ~11.6×), Jupiter's mass is ~3,000× greater than Mars's mass. This enormous mass difference far outweighs the distance effect, resulting in Jupiter experiencing a gravitational force about 250× larger than Mars.
Challenge Q3 — Orbital Speed: Using the formula v = √(G × M ÷ r), calculate Earth's orbital speed around the Sun. Then explain why astronauts in the ISS (orbiting at r = 6.77 × 10⁶ m from Earth's centre) experience apparent weightlessness, even though Earth's gravitational field is still ~8.7 N/kg there. (Mass of Sun = 2.0 × 10³⁰ kg; Mass of Earth = 6.0 × 10²⁴ kg; G = 6.67 × 10⁻¹¹ N m² kg⁻²)
Earth's orbital speed:
v = √(G × M_Sun ÷ r_Earth) = √(6.67×10⁻¹¹ × 2.0×10³⁰ ÷ 1.5×10¹¹)
= √(1.334×10²⁰ ÷ 1.5×10¹¹) = √(8.893×10⁸) = 2.98 × 10⁴ m/s ≈ 29.8 km/s
ISS weightlessness explanation:
Gravity at ISS altitude (r = 6.77×10⁶ m): g = G×M_Earth ÷ r² = 6.67×10⁻¹¹ × 6.0×10²⁴ ÷ (6.77×10⁶)² = 8.74 N/kg — so gravity is definitely present.
Astronauts experience apparent weightlessness because the ISS and everything inside it are in free fall together around Earth. Gravity provides exactly the centripetal force needed for the orbital speed (~7.7 km/s). There is no normal (contact) force between astronaut and floor — both fall at the same rate. This is sometimes called "microgravity" — not zero gravity, but zero apparent weight.
Challenge Q4 — Scale and Reasoning: A model of the Solar System is built where Earth is represented by a ball 1.27 cm in diameter. (a) What diameter sphere represents the Sun in this model? (b) How far away would Neptune be placed, given its actual distance is 30 AU? (c) How far away would Proxima Centauri (4.2 light-years = 4.0 × 10¹⁶ m) be? Express your answer in km. Comment on the challenge of representing the Solar System and nearby stars on the same scale.
Scale factor: Model Earth diameter = 1.27 cm = 1.27×10⁻² m; Real Earth diameter = 1.27×10⁷ m
Scale = 1.27×10⁻² ÷ 1.27×10⁷ = 1×10⁻⁹ (1 nanometre represents 1 metre, or 1 cm = 10⁷ m)
(a) Sun diameter: Real = 1.4×10⁹ m → Model = 1.4×10⁹ × 10⁻⁹ = 1.4 m diameter sphere
(b) Neptune distance: Real = 30 AU = 30 × 1.5×10¹¹ = 4.5×10¹² m
Model = 4.5×10¹² × 10⁻⁹ = 4,500 m = 4.5 km
(c) Proxima Centauri: Real = 4.0×10¹⁶ m
Model = 4.0×10¹⁶ × 10⁻⁹ = 4.0×10⁷ m = 40,000 km
Comment: On this scale, where Earth is 1.27 cm, you could just fit the Solar System (to Neptune) into a 9 km circle — but Proxima Centauri would be 40,000 km away (over 3× the Earth's diameter in reality!). This illustrates how vast interstellar distances are compared even to our enormous Solar System. The nearest star is about 8,900 times farther than Neptune.