Space & Stars · 28.1

The Earth & the Solar System

The Sun rises because the Earth spins; summer comes because the spin axis leans; the Moon changes shape because we watch its sunlit half from different angles. None of it is mysterious — it is one tilted, spinning planet riding a gravitational clockwork that Newton could wind forward or back for centuries.

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Connect the measurement to the mechanism.

The Earth turns once on its axis every 24 hours, giving day and night, and orbits the Sun once every 365¼ days. Seasons come from the 23.5° tilt of the spin axis — in June the northern hemisphere leans toward the Sun, catching steeper, longer sunlight — not from any change in distance. The Moon orbits the Earth in about a month, and its phases are simply the fraction of its sunlit half we can see.

The Sun is an ordinary star, a ball of mostly hydrogen and helium containing 99.8% of the Solar System’s mass, and its gravity is the glue: g = GM/r² weakens with distance, so each planet moves on a near-circular orbit with speed v = 2πr/T. Because gravity is stronger closer in, the inner planets must move faster — Mercury tears round at 48 km/s while Neptune drifts at 5.4 km/s — and comets on stretched elliptical orbits obey the same rule moment by moment, sprinting at perihelion and crawling at aphelion.

Simple definition: The Solar System is the Sun plus everything its gravity holds — eight planets, moons, asteroids, and comets — with each body’s orbital speed set by v = 2πr/T, faster close to the Sun where gravity is stronger and slower far out.
Example: Mercury (r = 0.39 AU) completes an orbit in 88 days at about 48 km/s; Neptune (r = 30 AU) needs 165 years at 5.4 km/s — thirty times farther, nine times slower.

Orbital speedv = 2πr/T

A near-circular orbit is just a circle traversed once per period. Know any two of r, T, and v and the third follows — this one equation handles planets, moons, and satellites alike.

Circumference of the orbit divided by the period

Gravity as the glueg = GM/r²

The Sun’s gravitational field weakens with the square of distance. That inverse square is why inner planets must orbit fast to keep falling around the Sun rather than into it.

G = 6.67 × 10⁻¹¹ N m² kg⁻²

Kepler’s patternT² ∝ r³

Farther orbits are both longer and slower, so the period grows faster than the radius. Jupiter at 5.2 AU takes not 5.2 years but √(5.2³) ≈ 12 years.

In AU and years: T² = r³ for the Sun

Days, years, and seasons

Day and night come from the Earth’s rotation, the year from its orbit. Seasons come from the 23.5° axial tilt: the hemisphere leaning toward the Sun gets longer days and steeper, more concentrated sunlight. The proof that distance is not the cause: the Earth is actually closest to the Sun in early January, in the middle of the northern winter — and the two hemispheres always have opposite seasons at the same distance.

The Sun’s family

Four small rocky planets orbit close in — Mercury, Venus, Earth, Mars — then the asteroid belt, then four large, cold gas and ice giants: Jupiter, Saturn, Uranus, Neptune. The split is a fossil of formation: near the young Sun it was too hot for ices to condense, so only rock and metal remained, while beyond the frost line ice-rich cores grew massive enough to capture thick hydrogen envelopes. Moons orbit planets; the whole ensemble orbits the Sun.

Comets: the eccentric relatives

A comet rides a highly elliptical orbit with the Sun at one focus. Far out it is a dormant lump of ice and dust moving slowly; falling inward it trades gravitational potential energy for kinetic energy, so it moves fastest at perihelion — exactly where solar heating boils off gas and dust into the tail, which always streams away from the Sun. Then it climbs back out, slowing all the way, and the loop repeats: Halley’s comet has done so every 76 years for millennia.

Change one variable at a time

Make the relationship visible.

Orbital radius r1.05 AU

Gravity g = GM/r² weakens with distance, so closer orbits must be faster: v = 29.8 km/s ÷ √(r/AU). Slide from Mercury’s scorched sprint to Neptune’s century-long drift — every planet sits on this one curve.

Orbital radius1.05 AU = 1.58e+11 m

Period T² = r³1.1 years

Speed v = 2πr/T29 km/s

Locationnear Earth’s orbit

Catch the common trap

Explain before calculating.

Why does Mercury orbit the Sun faster than Neptune?

Choose an answer to test the model.

Worked examples

State the rule, substitute, then check units.

EasyThe Earth orbits the Sun at radius 1.5 × 10¹¹ m with a period of 1 year (3.15 × 10⁷ s). Find its orbital speed.

v = 2πr/T = (2π × 1.5 × 10¹¹) ÷ 3.15 × 10⁷.
v = 9.42 × 10¹¹ ÷ 3.15 × 10⁷ = 2.99 × 10⁴ m/s.
About 30 km/s — the Earth covers its own diameter every seven minutes.

Answer≈ 3.0 × 10⁴ m/s (30 km/s)

MediumJupiter orbits at r = 7.8 × 10¹¹ m with a period of 11.9 years. Find its orbital speed and compare it with the Earth’s.

T = 11.9 × 3.15 × 10⁷ = 3.75 × 10⁸ s.
v = 2πr/T = (2π × 7.8 × 10¹¹) ÷ 3.75 × 10⁸ = 1.3 × 10⁴ m/s.
Five times farther out than the Earth, Jupiter moves at less than half the speed — v = 2πr/T rises with r far more slowly than T does, because gravity out there is 27 times weaker.

Answer≈ 13 km/s — farther is slower

HardUse the Earth’s orbit (r = 1.5 × 10¹¹ m, T = 3.15 × 10⁷ s) to find the mass of the Sun.

Gravity supplies the centripetal force: GMm/r² = mv²/r with v = 2πr/T, giving M = 4π²r³/(GT²).
M = (4π² × (1.5 × 10¹¹)³) ÷ (6.67 × 10⁻¹¹ × (3.15 × 10⁷)²).
M = 1.33 × 10³⁵ ÷ 6.62 × 10⁴ ≈ 2.0 × 10³⁰ kg — weighing the Sun with a calendar and a ruler.

AnswerM ≈ 2.0 × 10³⁰ kg

ChallengingHalley’s comet moves at 54 km/s at perihelion (0.59 AU) and climbs out to aphelion at 35 AU. Estimate its aphelion speed and explain the energy story of one full orbit.

For an orbit, angular momentum about the Sun is conserved: m v r is the same at the two extremes (velocity is perpendicular to the radius there), so vₐ = vₚ rₚ/rₐ.
vₐ = 54 × 0.59 ÷ 35 ≈ 0.91 km/s — sixty times slower than at perihelion.
Energy bookkeeping: falling inward, gravitational potential energy converts to kinetic energy, peaking at perihelion; climbing out, the trade reverses. Total energy never changes — the comet is a frictionless pendulum swinging in the Sun’s field, which is why its 76-year returns are so predictable.

Answer≈ 0.9 km/s at aphelion — fast and hot near the Sun, slow and frozen far away

Exam diagrams for this topic2 figures to inspect and practiseQuestions, hints and marking points in one compact subsection.

See it. Read it. Work it.

These figures come from GioPhysics practice papers. Open one, decode the drawing, work the guided questions, then follow its link to the full paper question.

01
InspectRead the figure comment.
02
TraceFollow labels, arrows and axes.
03
AnswerWork one part at a time.
04
CheckReveal hints and marking points.

IGCSE

01Fig. 3.1The Earth and the Solar SystemIGCSE

Figure comment

Fig. 3.1A planet moving on a nearly circular orbit around its star. The star is a disc at the centre of a dashed circle and the planet is a smaller disc on that circle, level with the star and to its right. The distance from the centre of the star out to the planet is marked 1.1 × 10¹¹ m, an arrow at the planet points along the orbit to show the direction in which it travels, and a note beneath states that one complete orbit takes 1.9 × 10⁷ s.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

6.1 The Earth and the Solar System Open IGCSE question · Space physics Q3

Guided questions 5 parts

Reading cue. The 1.1 × 10¹¹ m is measured from the centre of the star, so it is the orbit radius, not the distance travelled; 1.9 × 10⁷ s is the time for one full lap of the dashed circle.

aCalculate Calculate the distance the planet travels in going once round the dashed circle in Fig. 3.1.
recall2 marks
Check answer 2 marks
1. distance = 2πr with r = 1.1 × 10¹¹ m (1)
2. distance = 6.9 × 10¹¹ m (1)
bDetermine Light travels at 3.0 × 10⁸ m/s. Determine, in minutes, the time light takes to travel from the star to the planet at the position drawn in Fig. 3.1.
routine3 marks
Check answer 3 marks
1. t = d / v = 1.1 × 10¹¹ / 3.0 × 10⁸ (1)
2. t = 367 s (1)
3. t = 6.1 minutes (1)
cExplain The arrow at the planet in Fig. 3.1 shows the direction in which it is travelling. Explain how the direction of the star's gravitational pull on the planet is related to that arrow, and explain why this pull changes the planet's direction without changing its speed.
demanding4 marks
Check answer 4 marks
1. the gravitational force acts from the planet towards the centre of the star (1)
2. it is therefore at right angles to the arrow showing the direction of travel (1)
3. so the force has no part of it along the direction of motion, and does no work on the planet (1)
4. it changes only the direction of the velocity, so the planet keeps a constant speed while continually turning (1)
dExplain The orbit is drawn as a dashed circle, but the planet's path is described only as nearly circular. Explain how the planet's distance and speed would vary round a slightly squashed orbit, and explain what this means for any single speed calculated from Fig. 3.1.
top of the paper4 marks
Check answer 4 marks
1. the distance from the star would vary round the orbit rather than staying at 1.1 × 10¹¹ m (1)
2. the planet moves fastest at the point of the orbit closest to the star (1)
3. and slowest at the point furthest from the star (1)
4. a value found from circumference ÷ period is therefore only an average speed for the whole orbit (1)

Transfer challenge

A communications satellite moves in a circle of radius 4.2 × 10⁷ m about the centre of the Earth, taking 8.64 × 10⁴ s for one orbit. Calculate its orbital speed, and explain why it stays above the same point on the equator.

Check answer 5 marks

circumference = 2π × 4.2 × 10⁷ = 2.64 × 10⁸ m (1)
v = 2.64 × 10⁸ / 8.64 × 10⁴ (1)
v = 3.1 × 10³ m/s (1)
its orbital period is equal to the time the Earth takes to spin once on its axis (1)
so it keeps pace with the ground beneath it and stays above the same point (1)

02Fig. 7.1The Earth and the Solar SystemIGCSE

Figure comment

Fig. 7.1A not-to-scale plan view of the Sun, the Earth and the Moon. The Sun is a disc at the centre of a large dashed circle, and the Earth is a smaller disc sitting on that circle to the right of the Sun. A second, much smaller dashed circle is drawn around the Earth with the Moon on it, up and to the right of the Earth. A short arrow on each dashed circle shows the direction in which the Earth travels around the Sun and the Moon around the Earth.

Read the comment once, then trace every arrow, label, axis or component in the drawing before opening the questions.

6.1 The Earth and the Solar System Open IGCSE question · Space physics Q7

Guided questions 5 parts

Reading cue. This is a plan view and not to scale: check what each dashed circle is drawn around before you name it, and note that both arrows go the same way round.

aIdentify Identify what each of the two dashed circles in Fig. 7.1 represents, and state the body at the centre of each one.
recall2 marks
Check answer 2 marks
1. the large circle is the path of the Earth, centred on the Sun (1)
2. the small circle is the path of the Moon, centred on the Earth (1)
bExplain The Sun is the only source of light drawn in Fig. 7.1. Explain why an observer on Earth sees the whole of the Moon's lit face when the Moon reaches the point of the small circle furthest from the Sun, and identify where on that circle the Moon is almost invisible.
routine3 marks
Check answer 3 marks
1. the Moon is seen only by the sunlight it reflects (1)
2. at the far point the lit half of the Moon faces the Earth, so the whole lit disc is seen (1)
3. on the side of the small circle nearest the Sun, the lit half faces away from the Earth, so almost none of it can be seen (1)
cDetermine The Moon takes 27 days to travel once round the small dashed circle and the Earth spins once on its axis in 24 hours, both in the directions shown by the arrows. Determine the angle the Moon moves round its orbit in one day, and hence determine the extra time the Earth must spin each day before the observer faces the Moon again.
demanding4 marks
Show a hint
In one day the Earth must turn through 360° plus the extra angle the Moon has moved on round its own circle.
Check answer 4 marks
1. angle moved in one day = 360 / 27 = 13.3° (1)
2. after one complete spin the Earth must turn a further 13.3° to point at the Moon again (1)
3. extra time = 13.3 / 360 × 24 hours (1)
4. = 0.89 hours, about 53 minutes (1)
dSuggest Fig. 7.1 is drawn not to scale. Suggest two ways in which a true scale drawing would look different, and suggest why the diagram is drawn as it is.
top of the paper3 marks
Check answer 3 marks
1. the Moon's orbit would be very much smaller compared with the Earth's orbit (1)
2. the Sun would be drawn very much larger than the Earth, and both would be far too small to see beside orbits of this size (1)
3. the diagram is drawn out of scale so that the directions of travel and the relative positions of the three bodies can all be shown on one page (1)

Transfer challenge

Mars spins once on its axis in 24.6 hours, and its moon Phobos orbits Mars in only 7.7 hours, in the same direction as Mars spins. Explain what an observer standing on Mars sees Phobos do, and calculate how many times in one Martian day it does this.

Check answer 5 marks

Phobos goes once round its orbit in less time than Mars takes to spin once (1)
so it overtakes the observer, moving round faster than the ground turns, and appears to rise in the west and set in the east, the opposite way to our Moon (1)
Phobos moves round at 360 / 7.7 = 46.8° per hour while the ground turns at 360 / 24.6 = 14.6° per hour, so Phobos gains about 32° each hour (1)
it gains a whole turn on the observer every 360 / 32 = 11.2 hours (1)
24.6 / 11.2 = 2.2, so it crosses the sky about twice each Martian day (1)

Extra challenging questions

Take the ideas further with course-specific extension questions, supplied-data graphs and worked solutions.