6. Space physics

Syllabus
0625–2026–2027
Section
6
Level
—

6.1.1 The Earth

Syllabus
0625–2026–2027
Topic
6.1.1
Level
—

Explain Earth's rotation, day and night

Earth rotates once on its tilted axis in approximately 24 hours. The axis is the imaginary line through the North and South Poles.

At any moment, the hemisphere facing the Sun is illuminated and has daytime; the hemisphere facing away is in darkness and has night. As Earth rotates, each location moves into and then out of the illuminated half.

Earth rotates from west to east, so the Sun appears to move across the sky in the opposite direction, from east to west, rising in the east and setting in the west.

Because one rotation takes about 24 hours, the sequence of daylight and darkness and the Sun's apparent daily path repeat approximately every day.

Earth's rotation causes day and night. The tilt of the axis is important for seasons, but tilt is not the cause of the daily day/night cycle.

Explain Earth's year and the seasons

Earth completes one orbit of the Sun in approximately 365 days. This orbital period defines one year.

Earth's axis remains tilted in a nearly fixed direction as Earth moves around the Sun. During part of the orbit, one hemisphere is tilted towards the Sun; about half an orbit later it is tilted away.

Hemisphere orientation Sunlight received Season tendency
tilted towards Sun longer days and more direct rays summer
tilted away from Sun shorter days and less direct rays winter
neither strongly towards nor away intermediate day length and ray angle spring or autumn

The same orientations recur each orbit, so the seasons form a periodic cycle that repeats approximately every 365 days. The two hemispheres experience opposite seasons.

Seasons are not caused mainly by Earth being nearer to or farther from the Sun. They arise from axial tilt changing ray angle and daylight duration through the year.

Explain the Moon's cycle of phases

The Moon takes approximately one month to complete one orbit of Earth.

The Sun always illuminates half of the Moon. As the Moon moves around Earth, the viewing angle changes, so observers see different fractions of the illuminated half.

The visible sequence repeats: new Moon, waxing crescent, first quarter, waxing gibbous, full Moon, waning gibbous, last quarter and waning crescent, then new Moon again.

Because the Moon returns to approximately the same orbital arrangement after about one month, its cycle of phases is periodic with approximately a monthly timescale.

Ordinary Moon phases are not caused by Earth's shadow. Earth's shadow on the Moon produces a lunar eclipse, which is a different event.

Calculate average orbital speed

Average orbital speed is the total distance travelled around an orbit divided by the orbital period.

For an approximately circular orbit, the distance travelled in one orbit is 2πr, so v = 2πr/T, where r is the average orbital radius measured from the centre of the orbit and T is the time for one complete orbit.

Step Action
1 identify r and the full orbital period T
2 convert distance and time to units that match the required speed
3 calculate the circumference 2πr
4 divide by T and state the speed unit

For Earth, r ≈ 1.5 × 10¹¹ m and T ≈ 365 × 24 × 60 × 60 s. Therefore v ≈ 2π(1.5 × 10¹¹)/(3.15 × 10⁷) ≈ 3.0 × 10⁴ m/s.

The same relation can be rearranged: T = 2πr/v or r = vT/(2π).

Use the orbital radius, not the radius of the planet, unless the orbit is at the planet's surface. Average speed is not average velocity: velocity direction changes continuously around an orbit.

6.1.2 The Solar System

Syllabus
0625–2026–2027
Topic
6.1.2
Level
—

Describe the contents of the Solar System

The Solar System is the Sun and all natural objects held in orbit around it by gravity. The Sun is the Solar System's one star.

The eight planets in order of increasing distance from the Sun are Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus and Neptune.

Type of object Place in the Solar System
minor planets orbit the Sun; include dwarf planets such as Pluto and asteroids in the asteroid belt
moons or natural satellites orbit planets
comets and other smaller bodies orbit the Sun, often on very elongated paths

The Solar System has one star, not many. A galaxy contains many stars and their systems; the Milky Way is the galaxy that contains our Solar System.

Explain how the inner and outer planets formed

Mercury, Venus, Earth and Mars are the four inner planets: they are comparatively small and rocky. Jupiter, Saturn, Uranus and Neptune are the four outer planets: they are comparatively large and gaseous.

The accretion model begins with an interstellar cloud containing many elements in gas and dust. Gravity pulls the cloud material together.

As the cloud contracts, its rotation becomes more important and the material forms a rotating accretion disc around the developing Sun. Dust and other particles collide and stick, building larger bodies by accretion.

Near the hot young Sun, mainly heat-resistant rocky material could condense and accrete, producing smaller rocky planets. Farther out, cooler conditions allowed much more gas and icy material to collect, producing larger gaseous planets.

Accretion means gradual growth by collecting matter. It is not a single collision that instantly produced each planet, and 'gaseous' does not mean an outer planet has no dense interior.

Relate gravitational field strength to mass and distance

Gravitational field strength at a point is the gravitational force per unit mass on a small test mass at that point. Its unit is N/kg.

At a planet's surface, gravitational field strength depends on the planet's mass: a more massive planet generally produces a stronger gravitational field at its surface when making syllabus-level comparisons.

Around any one planet, gravitational field strength decreases as distance from the planet increases. A test mass farther away therefore experiences less gravitational force per kilogram.

Comparison Expected conclusion
same planet, greater distance weaker gravitational field
planetary surface data, greater planet mass generally stronger surface field

Do not treat a planet's field as uniform everywhere around it. The surface value cannot be used unchanged far from the planet.

Calculate light-travel time in the Solar System

Light travels through a vacuum at approximately c = 3.0 × 10⁸ m/s. Use speed = distance/time, so t = d/c for a one-way journey.

Step Action
1 identify whether the stated distance is one-way or a return path
2 convert the distance to metres
3 calculate t = d/(3.0 × 10⁸)
4 state the answer in seconds or convert to the requested unit

The mean Sun–Earth distance is about 1.5 × 10¹¹ m, so light takes t = (1.5 × 10¹¹)/(3.0 × 10⁸) = 5.0 × 10² s, about 8.3 minutes, to travel from the Sun to Earth.

For a signal sent to an object and reflected back, the light travels twice the one-way separation, so use d = 2 × separation.

Match distance and speed units before dividing. Kilometres used directly with a speed in m/s give an answer wrong by a factor of 1000.

Explain why planets orbit the Sun

The Sun contains most of the mass of the Solar System.

Because gravitational effects depend on mass, the Sun produces the dominant gravitational field across the Solar System. Each planet is gravitationally bound mainly to the Sun.

The planets therefore orbit the Sun rather than another planet or a small body. Their moons can still orbit them because the planet's gravity dominates close to that planet.

The Sun's large size alone is not the explanation: its dominant mass is the relevant property. The Sun does not need to touch or push a planet to affect it; gravity acts across space.

Identify the force that maintains a solar orbit

The force that keeps a planet, minor planet or comet in orbit around the Sun is the gravitational attraction of the Sun.

This gravitational force acts towards the Sun. It continually changes the direction of the object's velocity, providing the inward, or centripetal, force required for an orbit.

Without the inward gravitational force, the object would continue approximately along a straight-line tangent rather than follow its curved path.

There is no separate outward force that balances gravity during an orbit. The object is accelerating because its velocity direction changes, even if its speed is momentarily constant.

Describe elliptical Solar System orbits

Planets, minor planets and comets travel around the Sun in elliptical orbits. A circle is a special, perfectly symmetric ellipse.

In an ellipse, the Sun lies away from the geometric centre, at one focus. It is therefore not at the centre of the orbit.

When an orbit is approximately circular, its two focal positions are very close together, so the Sun is approximately at the centre.

The Sun–object distance changes around a visibly elliptical orbit. This is why a quoted orbital distance is often an average value.

Elliptical does not always mean visibly stretched. A nearly circular planetary orbit is still an ellipse, while many comet orbits are much more elongated.

Analyse and interpret planetary data

Begin by identifying each column's variable and unit. Compare like with like, then describe only the trend supported by the values before explaining it.

Data comparison Useful interpretation
greater orbital distance usually longer orbital duration and lower orbital speed
greater distance from the Sun generally lower surface temperature
high density often associated with rocky rather than gaseous composition
greater planet mass generally stronger gravitational field at the surface

To compare orbital positions, calculate the fraction of an orbit completed: elapsed time/orbital period. Multiply this fraction by 360° only when a circular-orbit angle model is appropriate.

Use counterexamples in the table to test a claim. A general increase does not prove direct proportionality; direct proportionality requires a constant ratio and a straight line through the origin.

Planetary variables are linked but not interchangeable. For example, low density does not by itself imply weak surface gravity, because planet mass and size also matter.

Link Solar gravitational field strength and orbital speed to distance

The Sun's gravitational field strength decreases as distance from the Sun increases.

The orbital speeds of the planets also decrease as their distance from the Sun increases: nearer planets orbit faster and farther planets orbit more slowly.

A nearer planet experiences a stronger solar gravitational field and needs a higher speed for its tighter orbit. A farther planet experiences weaker solar gravity and follows a larger orbit at a lower orbital speed.

Planet position Solar field Orbital speed
nearer the Sun stronger higher
farther from the Sun weaker lower

Do not infer that a farther planet completes an orbit sooner because it has a larger path. It travels a larger path at a lower speed, so its orbital period is longer.

Explain speed changes in an elliptical orbit

An object in an elliptical orbit travels faster when it is closer to the Sun and slower when it is farther away.

As the object moves towards the Sun, its gravitational potential energy decreases. Gravity transfers energy to its kinetic store, so kinetic energy and speed increase.

As the object moves away from the Sun, kinetic energy is transferred to its gravitational potential store, so kinetic energy and speed decrease.

Ignoring resistive effects, the total of kinetic energy and gravitational potential energy remains constant. Energy changes store; it is not created near the Sun or destroyed farther away.

Orbital position Gravitational potential energy Kinetic energy and speed
closest to Sun minimum maximum
farthest from Sun maximum minimum

The changing speed does not mean gravity switches on and off. Gravity acts throughout the orbit, while the balance between kinetic and gravitational potential energy changes continuously.

6.2.1 The Sun as a star

Syllabus
0625–2026–2027
Topic
6.2.1
Level
—

Describe the Sun as a star

The Sun is a medium-sized star. It is the one star in our Solar System and the nearest star to Earth.

The Sun consists mostly of the elements hydrogen and helium. Other elements are present in much smaller amounts.

The Sun radiates most of its energy in three neighbouring regions of the electromagnetic spectrum: infrared, visible light and ultraviolet.

Region Position relative to visible light
infrared longer wavelength than visible red light
visible light detected by the human eye
ultraviolet shorter wavelength than visible violet light

The Sun does not emit only visible light, and most of its energy is not in radio waves, X-rays or gamma rays. 'Medium-sized star' does not mean the Sun is a planet or a galaxy.

Explain how fusion powers a stable star

Stars are powered by nuclear reactions in their cores. These reactions release energy that is eventually transferred from the star into space.

In a stable star, hydrogen nuclei join together to form helium nuclei. Joining small nuclei to make a larger nucleus is nuclear fusion.

Fusion releases energy. This energy moves through the star and is radiated from its surface as electromagnetic radiation.

Process What happens to nuclei Role in a stable star
fusion small nuclei join to form a larger nucleus hydrogen forms helium and energy is released
fission a large nucleus splits into smaller nuclei not the process that powers a stable star

Stellar fusion is not chemical burning: no oxygen is required. It is also not nuclear fission, and the direction is hydrogen to helium, not helium to hydrogen.

6.2.2 Stars

Syllabus
0625–2026–2027
Topic
6.2.2
Level
—

Describe galaxies and define the light-year

A galaxy is a vast collection containing many billions of stars. The Sun is one star in the galaxy called the Milky Way.

The Sun is much closer to Earth than every other star in the Milky Way. Even the nearest stars beyond the Sun are so far away that astronomical distance units are useful.

A light-year is the distance travelled by light through the vacuum of space in one year.

If a star is four light-years away, its light takes about four years to reach us. We therefore observe that star as it was about four years earlier.

A light-year is a unit of distance, not time. The Solar System is inside the Milky Way; the Milky Way contains many stars, while the Solar System contains only one star, the Sun.

Use the distance represented by one light-year

One light-year is approximately 9.5 × 10¹⁵ m, or 9.5 × 10¹² km.

Light travels at about 3.0 × 10⁸ m/s and one year is about 365 × 24 × 60 × 60 = 3.15 × 10⁷ s. Therefore d = ct ≈ (3.0 × 10⁸)(3.15 × 10⁷) ≈ 9.5 × 10¹⁵ m.

Conversion Operation
light-years to metres multiply by 9.5 × 10¹⁵
metres to light-years divide by 9.5 × 10¹⁵
light-years to kilometres multiply by 9.5 × 10¹²

A distance of 6.6 × 10²⁰ m corresponds to (6.6 × 10²⁰)/(9.5 × 10¹⁵) ≈ 6.9 × 10⁴ light-years.

Do not attach seconds to 9.5 × 10¹⁵: it is a distance in metres. Converting metres to kilometres divides by 1000, so the numerical power changes from 10¹⁵ to 10¹².

Describe the life cycle of a star

A star begins in an interstellar cloud of gas and dust containing hydrogen. Internal gravitational attraction makes part of the cloud collapse; as it contracts, its temperature rises and it becomes a protostar.

A protostar becomes a stable star when the inward force of gravitational attraction is balanced by an outward force due to the high temperature at its centre. Hydrogen fusion supplies the energy that maintains this hot stable state.

Eventually every star runs short of hydrogen fuel in its centre because much of that hydrogen has been converted to helium. The star expands: most stars become red giants, while more massive stars become red supergiants.

Lower-mass route Higher-mass route
red giant red supergiant
outer layers form a planetary nebula explodes as a supernova
white dwarf remains at the centre nebula remains, with hydrogen and newly formed heavier elements
— collapsed remnant is a neutron star or, for a sufficiently massive core, a black hole

Material in the nebula from a supernova can later be pulled together by gravity to form new stars. Accretion discs around those forming stars may also produce orbiting planets, so stellar material is recycled into later systems.

Shared beginning: hydrogen-rich gas and dust → protostar → stable star → hydrogen fuel runs low. Initial mass then controls which end branch the star follows.

A red giant does not explode as a supernova in the lower-mass route, and a white dwarf does not normally become a neutron star or black hole. A planetary nebula is ejected stellar gas, not a nebula containing a newly formed planet.

6.2.3 The Universe

Syllabus
0625–2026–2027
Topic
6.2.3
Level
—

Place the Milky Way within the Universe

The Universe contains many billions of galaxies. The Milky Way is one of those galaxies.

The Milky Way itself contains many billions of stars and has an approximate diameter of 100 000 light-years.

Light therefore takes about 100 000 years to travel a distance equal to the Milky Way's diameter. This is a galactic scale, far larger than the Solar System.

Scale Contains
Solar System the Sun and objects orbiting it
Milky Way galaxy many billions of stars, including the Sun
Universe many billions of galaxies

The Milky Way is not the whole Universe, and 100 000 light-years is its approximate diameter, not its distance from Earth.

Describe redshift from a receding source

Redshift is an increase in the observed wavelength of electromagnetic radiation from a star or galaxy that is moving away from the observer.

Astronomers compare identifiable spectral lines in received starlight with the wavelengths of the same lines measured in a laboratory. A shift to longer wavelengths is a redshift.

For a receding source, successive wavefronts arrive more spread out, so the observed wavelength is longer than the emitted or laboratory wavelength.

Redshift does not mean that all received light is visibly red. The whole pattern of spectral wavelengths is shifted towards longer wavelengths.

Recognise redshift in light from distant galaxies

Light from distant galaxies generally appears redshifted when compared with light produced by the same elements on Earth.

If a hydrogen line has wavelength 656 nm in a laboratory but is observed at 667 nm from a galaxy, the observed wavelength has increased: 667 − 656 = 11 nm. The galaxy's light is redshifted.

The known laboratory spectrum acts as the reference. Matching the pattern of lines identifies the element; their displacement to longer wavelength identifies redshift.

A longer measured wavelength must be compared with the correct laboratory line. A single wavelength without a reference does not by itself establish redshift.

Use redshift as evidence for an expanding Universe

Redshift in the light from distant galaxies shows that those galaxies are receding from us.

When light from galaxies in all directions is generally redshifted, the large-scale separations between galaxies are increasing. This is evidence that the Universe is expanding.

Running the expansion model backwards implies that matter was closer together in the past. Redshift therefore supports the Big Bang Theory.

More distant galaxies generally show greater redshift and recession speed, strengthening the expansion interpretation.

Redshift is supporting evidence, not a claim that Earth is at a special central point. Expansion is observed on the large scale between galaxies.

Identify cosmic microwave background radiation

Cosmic microwave background radiation, abbreviated CMBR, is microwave electromagnetic radiation of a specific frequency range observed at all points in space around us.

Detectors receive this faint microwave background from every direction rather than from one local star, planet or galaxy.

Its all-sky presence is why it is described as a cosmic background: it is a property observed throughout the surrounding Universe.

CMBR is not ordinary microwave transmission from communication equipment and is not concentrated in one direction or one visible object.

Explain the origin of the CMBR

The radiation that became the CMBR was produced shortly after the Universe formed.

As the Universe expanded, the wavelength of this radiation was stretched to longer wavelengths.

That expansion shifted the radiation into the microwave region of the electromagnetic spectrum, where it is observed today from all directions.

The CMBR's ancient origin, stretched wavelength and all-sky distribution support the model of a hot early Universe that has expanded.

The CMBR was not newly produced by modern galaxies. Its wavelength changed because the Universe expanded after the radiation was released.

Determine galaxy recession speed from redshift

The recession speed v of a galaxy can be found from the redshift of its starlight: the measured change in wavelength relative to laboratory wavelengths.

Step Observation or inference
1 identify a known spectral line in the galaxy's light
2 compare its observed wavelength with the laboratory wavelength
3 find the increase in wavelength, or redshift
4 use the redshift relation supplied for the question to determine recession speed

A larger wavelength increase means a larger redshift and, under the syllabus model, a greater speed away from Earth.

Redshift determines recession speed, not distance directly. Galaxy distance is obtained using a separate observation such as supernova brightness.

Determine galaxy distance from a supernova

The distance d to a far galaxy can be determined using the observed brightness of a suitable supernova in that galaxy.

A supernova of known intrinsic brightness appears dimmer when it is farther away. Comparing its observed brightness with the expected intrinsic brightness provides the distance.

Astronomers can therefore obtain the two quantities needed for Hubble analysis in different ways: supernova brightness gives distance, while redshift gives recession speed.

The overall brightness of an unknown galaxy is not the same standard measurement. The syllabus relation uses the brightness of a supernova in that galaxy.

Define and use the Hubble constant

The Hubble constant H₀ is the ratio of a galaxy's recession speed v to its distance d from Earth: H₀ = v/d.

If H₀ is treated as constant, v = H₀d, so recession speed is directly proportional to distance. A graph of v against d has gradient H₀ when units are consistent.

Required quantity Relation
Hubble constant H₀ = v/d
recession speed v = H₀d
galaxy distance d = v/H₀

If v is in m/s and d is in m, the metres cancel and H₀ has unit s⁻¹. Convert kilometres and light-years before substitution when required.

H₀ is not the speed of light and it is not itself the age of the Universe. It is a speed-per-distance ratio.

Recall the current estimate of the Hubble constant

The current syllabus estimate of the Hubble constant is H₀ = 2.2 × 10⁻¹⁸ s⁻¹, read as 2.2 × 10⁻¹⁸ per second.

The s⁻¹ unit follows from H₀ = v/d: (m/s)/m = 1/s.

The very small value is expected because cosmic expansion is measured over extremely large distances and timescales.

Do not confuse 2.2 × 10⁻¹⁸ s⁻¹ with the speed of light, 3.0 × 10⁸ m/s. Their quantities, units and powers of ten are different.

Estimate the age of the Universe from H₀

From H₀ = v/d, rearranging gives d/v = 1/H₀. The ratio distance/speed has units of time and provides an estimate of the age of the Universe.

Using H₀ = 2.2 × 10⁻¹⁸ s⁻¹ gives age ≈ 1/H₀ ≈ 4.5 × 10¹⁷ s. Dividing by about 3.2 × 10⁷ s/year gives approximately 1.4 × 10¹⁰ years.

Galaxies are now receding, and more distant galaxies generally recede faster. Extrapolating their separations backwards implies that matter was closer together and supports the idea that all matter was once concentrated at a single point.

This is an estimate based on using the current expansion relation over cosmic history; it is not a direct stopwatch measurement.

Use the reciprocal 1/H₀, not H₀ itself. The reciprocal has units of seconds because H₀ has units s⁻¹.