Revision Notes
O Level Physics: Space physics (Cambridge 5054) – Revision Notes
Condensed revision notes on space physics for Cambridge O Level Physics (5054): key facts, orbital speed method, star life cycle, redshift and a self-test.
- Subject
- Physics
- Level
- O LEVELS
- Topic
- Space physics
- Author
- Marlbridge Academic Team
- Updated
- Reviewed by
- Hassan (what this means)
Aligned to Cambridge O Level Physics (5054), 2026-2028. Official specification .
Syllabus page (what it covers and how it is assessed): Cambridge O Level Physics.
Syllabus points this page covers
5054
- 6.1 Earth and the Solar System
- 6.2 Stars and the Universe
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Need help with this topic? Request a free trial class for O Level Physics (5054).
These revision notes condense topic 6, Space physics, from the Cambridge O Level Physics (5054) syllabus for examination in 2026, 2027 and 2028. They cover sections 6.1 and 6.2. For full explanations and worked examples, read the study guide first. For exam-style questions, use the practice questions.
Related pages: Cambridge O Level Physics hub, the Physics checklist and the free 10-minute diagnostics. Paper 2 (Theory) has short-answer and structured questions and Paper 1 is multiple choice, so you need both quick recall and short calculations.
6.1.1 The Earth: facts to recall
| Fact | Value |
|---|---|
| Orbit of the Earth around the Sun | once in approximately 365 days |
| Shape of the orbit | an ellipse, approximately circular |
| Rotation on its axis (axis tilted) | once in approximately 24 hours |
| The Moon orbiting the Earth | approximately one month |
| Light from the Sun to the Earth | approximately 500 s |
Average orbital speed
| Quantity | Symbol | Note |
|---|---|---|
| Average orbital speed | v | km/s or m/s |
| Average radius of orbit | r | match the unit of v |
| Orbital period | T | in seconds |
Equation: v = 2πr / T. Rearranged: r = vT / 2π and T = 2πr / v.
Method in steps
- Convert the period into seconds (days × 24 × 3600).
- Keep r in km if you want km/s. Convert to metres only if you want m/s.
- Substitute into v = 2πr / T, or the rearranged form.
- Give the answer with a unit and to a sensible number of significant figures.
Quick reminder. A body orbits at r = 6.0 × 10⁷ km with a speed of 40 km/s. T = 2πr / v = 2π × 6.0 × 10⁷ / 40 = 9.4 × 10⁶ s, which is 109 days.
6.1.2 The Solar System
The Solar System contains
- one star, the Sun, which holds most of the mass;
- eight planets in order from the Sun: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune;
- minor planets: dwarf planets such as Pluto, and asteroids in the asteroid belt;
- moons, which orbit planets;
- smaller bodies: comets and natural satellites.
Gravitational field strength
| Idea | Statement |
|---|---|
| At the surface of a planet | depends on the mass of the planet |
| Around a planet | decreases as the distance from the planet increases |
| At the surface of the Sun | greater than at the surface of the planets |
Orbits
- The force that keeps an object in orbit around the Sun is the gravitational attraction of the Sun.
- As the distance from the Sun increases, the strength of the Sun’s gravitational field decreases and planetary orbital speeds decrease.
Reading planetary data. The syllabus lists orbital distance, orbital period, density, surface temperature and surface gravitational field strength. For a table question:
- Find the column and the rows you are asked about.
- State the trend as an increase or decrease with figures and units.
- If a speed is needed, use v = 2πr / T.
- Do not infer one property from another unless the question gives the link. Density does not tell you the surface field strength.
Quick reminder: comparing two planets. Planet M has orbital radius 1.0 × 10⁸ km and period 100 days. Planet N has radius 4.0 × 10⁸ km and period 800 days. M: v = 2π × 1.0 × 10⁸ / (100 × 24 × 3600) = 73 km/s. N: v = 2π × 4.0 × 10⁸ / (800 × 24 × 3600) = 36 km/s. The planet further from the Sun is slower, as the trend says it should be, and it also has the much longer path, so its period is eight times longer.
Quick reminder: the Sun’s light. The Sun is about 1.5 × 10⁸ km away. At 3.0 × 10⁵ km/s, light takes 1.5 × 10⁸ / 3.0 × 10⁵ = 500 s to arrive, which matches the recall fact.
6.2.1 The Sun as a star
- Medium-sized star, mostly hydrogen and helium.
- Radiates most of its energy in the infrared, visible and ultraviolet regions of the electromagnetic spectrum.
- Powered by nuclear reactions that release energy.
- In stable stars, the reaction is the fusion of hydrogen into helium.
6.2.2 Stars
| Term | Meaning |
|---|---|
| Galaxy | made up of many billions of stars |
| Milky Way | the galaxy that contains the Sun |
| Light-year | distance travelled by light in a vacuum in one year |
Other stars in the Milky Way are much further from the Earth than the Sun is.
Light-year in kilometres (extension; the syllabus asks only for the definition). 3.0 × 10⁵ km/s × 3.15 × 10⁷ s ≈ 9.5 × 10¹² km. So 4.0 light-years is 4.0 × 9.5 × 10¹² = 3.8 × 10¹³ km.
Life cycle of a star
| Stage | Key statement |
|---|---|
| Interstellar cloud | gas and dust containing hydrogen |
| Protostar | cloud collapsing and warming because of its internal gravitational attraction |
| Stable star | inward gravitational force balanced by an outward force from the high temperature in the centre |
| End of hydrogen | all stars eventually run out of hydrogen as fuel |
| Red giant | most stars, when most of the central hydrogen has become helium |
| Red supergiant | more massive stars at the same point |
| Planetary nebula and white dwarf | from the red giant of a less massive star |
| Supernova | explosion of a red supergiant; nebula with hydrogen and new heavier elements; neutron star or black hole left at the centre |
| New stars | the supernova nebula may form new stars with orbiting planets |
6.2.3 The Universe
- The Milky Way is one of many billions of galaxies. Its diameter is approximately 100 000 light-years.
- Redshift: an increase in the observed wavelength of electromagnetic radiation from receding stars and galaxies.
- Light from distant galaxies is redshifted. The further the galaxy, the greater the redshift and the faster it moves away from the Earth.
- Big Bang evidence (qualitative): galaxies are receding, so the Universe is expanding. Reversing the expansion, everything started close together.
Must-know distinctions
| Pair | Difference |
|---|---|
| Light-year and year | one is a distance, the other is a time |
| Galaxy and Universe | a galaxy holds billions of stars, the Universe holds many billions of galaxies |
| Protostar and stable star | the protostar is still collapsing, and the stable star has balanced forces |
| Red giant and red supergiant | less massive and more massive stars |
| White dwarf and neutron star or black hole | end points of less massive and more massive stars |
| Orbital speed and orbital period | speed is distance over time, the period is the time for one orbit |
| Moon and planet | a moon orbits a planet, a planet orbits the Sun |
Quick self-test
- Which planet comes between Mars and Saturn?
- How long does light take to travel from the Sun to the Earth, approximately?
- A body orbits the Sun at an average radius of 2.0 × 10⁸ km, taking 300 days. Calculate its average orbital speed in km/s.
- A body has an average orbital speed of 20 km/s and an orbital period of 1.2 × 10⁷ s. Calculate the average radius of its orbit.
- What is the force that keeps the planets in orbit around the Sun?
- State what happens to the orbital speeds of planets as their distance from the Sun increases.
- Define a light-year, and find 4.0 light-years in km.
- State the condition that makes a protostar into a stable star.
- Name the objects left at the centre after a supernova.
- Galaxy X shows a larger redshift than galaxy Y. Compare the two galaxies.
- Name the nuclear process that powers stable stars.
Answers
- Jupiter.
- About 500 s.
- T = 300 × 24 × 3600 = 2.59 × 10⁷ s. v = 2π × 2.0 × 10⁸ / 2.59 × 10⁷ = 48 km/s.
- r = vT / 2π = 20 × 1.2 × 10⁷ / 2π = 3.8 × 10⁷ km.
- The gravitational attraction of the Sun.
- They decrease.
- The distance travelled by light in a vacuum in one year. 4.0 × 9.5 × 10¹² = 3.8 × 10¹³ km.
- The inward force of gravitational attraction is balanced by an outward force due to the high temperature at the centre.
- A neutron star or a black hole.
- X is further away and moving away from the Earth faster than Y.
- Fusion of hydrogen into helium.
Where marks are usually lost
- Leaving the period in days or years when the speed is needed in km/s or m/s.
- Using the radius, not 2πr, as the distance for one orbit.
- Defining a light-year as a unit of time.
- Describing the force on a planet as acting along its path rather than towards the Sun.
- Saying a protostar is “burning” hydrogen. Fusion begins once it is a stable star.
- Missing the balance of two forces when explaining a stable star.
- Putting a white dwarf after a supernova.
- Writing that redshift is a change of colour, rather than an increase in wavelength.
- Comparing galaxies by redshift but not saying both distance and speed.
Frequently asked questions
How do you calculate average orbital speed in O Level Physics 5054? Use v = 2πr / T, where 2πr is the distance round one orbit and T is the orbital period. Convert the period into the time unit the answer needs, such as seconds for m/s.
What keeps a planet in orbit? The Sun’s gravitational attraction on the planet acts towards the Sun, not along the planet’s path. Describing the force as acting along the orbit is a common error.
Is a protostar burning hydrogen? No. Nuclear fusion of hydrogen begins once the star becomes stable; a protostar is a collapsing cloud of gas and dust heating up because of its own gravity.
How do you compare galaxies using redshift? A greater redshift means a galaxy is moving away faster and is further away. When comparing galaxies, mention both distance and speed.
Official syllabus
Cambridge O Level Physics 5054 syllabus for examination in 2026, 2027 and 2028, published by Cambridge International, topic 6 Space physics. See also the study guide and book a free trial class.
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