IGCSE Physics Paper 4 (Theory / Extended)

Topic 6: Space Physics -- Challenge Paper

Duration 1 hour 15 minutes

Total Marks 80

Questions 7

Time Remaining 75:00

Syllabus 0625 / 0972

⚡ Cambridge Challenge Level

These questions match real Cambridge IGCSE difficulty. Scoring 50%+ is a solid B, 60%+ is an A, 70%+ is A*. Don't worry if this feels harder -- that's the point!

Instructions

Answer all questions in the spaces provided.
Show all working for calculations. Answers without working may not receive full marks.
You may use a calculator.
Take g = 10 N/kg where needed.
Speed of light c = 3.0 × 10&sup8; m/s
1 light-year = 9.5 × 10¹⁵ m
Hubble constant H₀ = 2.2 × 10⁻¹⁸ s⁻¹
Answer all questions first. When finished, click "Submit Exam" at the bottom of Question 7.
Your answers will be automatically graded when you submit the exam. The model answers will be shown for review.

Question Navigation

Question 1 -- Orbital Mechanics and the Earth

Total: 12 marks

The European Space Agency (ESA) satellite Gaia orbits the Sun at the L2 Lagrange point, approximately 1.5 million km beyond Earth. Scientists at the Cambridge Institute of Astronomy use data from Gaia to map the positions and velocities of over a billion stars.

(a) [3 marks]

The Earth orbits the Sun at a mean distance of 1.5 × 10¹¹ m with an orbital period of 3.15 × 10⁷ s.

Calculate the orbital speed of the Earth using the equation v = 2πr / T. Show your working clearly.

Mark Scheme -- Part (a)

v = 2πr / T stated or substituted correctly

v = 2 × π × 1.5 × 10¹¹ / 3.15 × 10⁷ = 9.42 × 10¹¹ / 3.15 × 10⁷

v ≈ 29 900 m/s or 3.0 × 10⁴ m/s

Mark 1 -- v equals 2 pi r divided by T formula stated substituted Mark 2 -- correct substitution 9.42 times 10 to 11 divided 3.15 times 10 to 7 Mark 3 -- answer 29900 or 3.0 times 10 to 4 m/s

⚠ If you missed marks here: The classic slip is leaving out the 2π — 1.5 × 10¹¹ ÷ 3.15 × 10⁷ gives about 4 800 m/s, which loses the answer mark. The data is already in metres and seconds, so no unit conversion is needed; always write the substitution line to bank the method marks even if your arithmetic slips.

(b) [3 marks]

Mars has an orbital radius of 2.3 × 10¹¹ m and an orbital period of 5.94 × 10⁷ s.

Calculate the orbital speed of Mars and compare it with that of the Earth. Suggest why Mars moves more slowly than Earth.

Mark Scheme -- Part (b)

v = 2π × 2.3 × 10¹¹ / 5.94 × 10⁷ = 1.445 × 10¹² / 5.94 × 10⁷

v ≈ 24 300 m/s or 2.4 × 10⁴ m/s; Mars is slower than Earth

Mars is further from the Sun so the gravitational pull is weaker / less centripetal force needed, so orbital speed is lower

Mark 4 -- substitution 2 pi times 2.3 times 10 to 11 divided 5.94 times 10 to 7 Mark 5 -- answer 24300 or 2.4 times 10 to 4 m/s Mars slower Mark 6 -- further from Sun weaker gravitational pull less centripetal force lower speed

⚠ If you missed marks here: Two common losses: doing the calculation (24 300 m/s) but never actually stating that Mars is slower than Earth, and explaining the slowness as just "Mars is further away". The third mark needs the physics chain: further from the Sun → weaker gravitational pull → less centripetal force needed → lower orbital speed.

Comets orbit the Sun in highly elliptical orbits.

Explain, using the concept of conservation of energy, why a comet speeds up as it approaches the Sun and slows down as it moves away. Refer to kinetic energy and gravitational potential energy in your answer.

Mark Scheme -- Part (c)

Total energy (KE + GPE) is conserved throughout the orbit

As the comet approaches the Sun, gravitational potential energy decreases (becomes more negative) and kinetic energy increases, so the comet speeds up

As the comet moves away from the Sun, kinetic energy is converted back to gravitational potential energy, so the comet slows down

Mark 7 -- total energy conserved KE plus GPE constant throughout orbit Mark 8 -- approaching Sun GPE decreases KE increases comet speeds up Mark 9 -- moving away KE converted to GPE comet slows down

⚠ If you missed marks here: Saying "gravity pulls the comet so it speeds up" only earns partial credit. The mark scheme wants the energy story: total energy (KE + GPE) stays constant, GPE decreases and KE increases on the approach, then KE converts back to GPE moving away. Name both energy types explicitly, in both directions.

(d) [3 marks]

A student at the Royal Observatory Greenwich observes that on 21 June, the Sun is above the horizon for approximately 16 hours 38 minutes, whereas on 21 December it is above the horizon for only 7 hours 50 minutes.

Explain this observation in terms of the Earth's axial tilt and orbital position.

Mark Scheme -- Part (d)

The Earth's axis is tilted at approximately 23.5° to the perpendicular of the orbital plane

On 21 June (summer solstice), the Northern Hemisphere is tilted towards the Sun, so the Sun follows a longer/higher arc across the sky, giving more hours of daylight

On 21 December (winter solstice), the Northern Hemisphere is tilted away from the Sun, so the Sun follows a shorter/lower arc, giving fewer hours of daylight

Mark 10 -- Earth axis tilted 23.5 degrees to perpendicular orbital plane Mark 11 -- June summer solstice Northern Hemisphere tilted towards Sun longer higher arc more daylight Mark 12 -- December winter solstice Northern Hemisphere tilted away shorter lower arc fewer daylight hours

⚠ If you missed marks here: The number one misconception is "Earth is closer to the Sun in summer" — that scores zero, because the Earth–Sun distance barely changes. You need the 23.5° axial tilt, plus the Northern Hemisphere tilted TOWARDS the Sun in June (longer, higher arc = more daylight) and AWAY from it in December (shorter, lower arc = less daylight).

Question 2 -- Solar System Formation and Planetary Data

Total: 12 marks

Scientists at the Jodrell Bank Centre for Astrophysics use radio telescopes to study planets and other objects in our Solar System. The table below shows data for the eight planets.

(a) [4 marks]

Study the data table below carefully.

Planet	Orbital radius / m	Type	Surface gravitational field strength / N/kg
Mercury	5.8 × 10¹⁰	Rocky	3.7
Venus	1.1 × 10¹¹	Rocky	8.9
Earth	1.5 × 10¹¹	Rocky	9.8
Mars	2.3 × 10¹¹	Rocky	3.7
Jupiter	7.8 × 10¹¹	Gas giant	24.8
Saturn	1.4 × 10¹²	Gas giant	10.4
Uranus	2.9 × 10¹²	Gas giant	8.9
Neptune	4.5 × 10¹²	Gas giant	11.2

Identify two patterns shown by the data in the table. For each pattern, use specific data from the table to support your answer.

Mark Scheme -- Part (a)

Pattern 1: The inner four planets (Mercury to Mars) are rocky, while the outer four (Jupiter to Neptune) are gas giants

Supporting data: e.g. Mercury, Venus, Earth, Mars are all labelled "Rocky" with orbital radii up to 2.3 × 10¹¹ m; Jupiter onwards are "Gas giant" with radii from 7.8 × 10¹¹ m

Pattern 2: The outer gas giant planets generally have larger gravitational field strengths than the inner rocky planets, suggesting they are more massive

Supporting data: e.g. Jupiter has g = 24.8 N/kg compared to Earth at 9.8 N/kg; Neptune has g = 11.2 N/kg compared to Mars at 3.7 N/kg

Mark 13 -- pattern 1 inner planets rocky outer planets gas giant Mark 14 -- supporting data with specific values from table for pattern 1 Mark 15 -- pattern 2 outer planets generally larger gravitational field strength more massive Mark 16 -- supporting data with specific values from table for pattern 2

⚠ If you missed marks here: Most lost marks come from stating a pattern without quoting numbers. "Outer planets have stronger gravity" alone earns only half; you must cite specific values, e.g. Jupiter g = 24.8 N/kg vs Earth 9.8 N/kg, or the rocky/gas split between 2.3 × 10¹¹ m and 7.8 × 10¹¹ m. Two patterns plus data for each = four separate marks.

(b) [4 marks]

Describe the accretion model of the formation of the Solar System. In your answer, explain why the inner planets are rocky while the outer planets are gaseous.

Mark Scheme -- Part (b)

A large cloud of gas and dust (nebula) collapsed under gravity, forming a spinning disc with the Sun at the centre

Particles in the disc collided and stuck together (accreted), gradually forming larger bodies called planetesimals, which grew into planets

Close to the Sun, temperatures were too high for gases to condense, so only rocky/metallic materials could form solid grains -- hence the inner planets are rocky

Further from the Sun, temperatures were low enough for gases such as hydrogen, helium, methane, and ammonia to be retained, forming the gas giant planets

Mark 17 -- cloud gas dust nebula collapsed gravity spinning disc Sun centre Mark 18 -- particles collided stuck accreted planetesimals grew planets Mark 19 -- close to Sun too hot gases condense only rocky metallic inner planets rocky Mark 20 -- further from Sun cooler gases retained hydrogen helium gas giants

⚠ If you missed marks here: Students usually describe the collapsing nebula and accretion but forget the temperature argument the question explicitly asks for: near the Sun it was too HOT for gases to condense (only rocky/metallic grains formed), while further out it was cool enough to retain hydrogen, helium, methane and ammonia. That explanation is worth two whole marks on its own.

In 1989, NASA's Voyager 2 spacecraft flew past Neptune, sending data back to Earth via radio signals travelling at the speed of light. The distance from the Sun to Neptune is 4.5 × 10¹² m.

(i) Calculate the time for light to travel from the Sun to Neptune. Give your answer in seconds and then convert to minutes.

(ii) Calculate how many times further Neptune is from the Sun compared to Earth (orbital radius 1.5 × 10¹¹ m).

(iii) Use your answers to explain why signals from Voyager 2 took over 4 hours to reach Earth when it was near Neptune.

Mark Scheme -- Part (c)

(i) t = d / c = 4.5 × 10¹² / 3.0 × 10⁸ = 15 000 s

(i) Convert: 15 000 / 60 = 250 minutes ≈ 4.2 hours

(ii) Ratio = 4.5 × 10¹² / 1.5 × 10¹¹ = 30; Neptune is 30 times further from the Sun than Earth

(iii) Radio signals travel at the speed of light (3.0 × 10⁸ m/s), which is finite; Neptune is 30 times further away than Earth, so signals take approximately 30 times longer than the 8 minutes from the Sun to Earth, giving about 250 minutes (over 4 hours)

Mark 21 -- time equals distance divided speed 4.5 times 10 to 12 divided 3.0 times 10 to 8 equals 15000 seconds Mark 22 -- converted 15000 divided 60 equals 250 minutes 4.2 hours Mark 23 -- ratio 4.5 times 10 to 12 divided 1.5 times 10 to 11 equals 30 times further Mark 24 -- signals travel speed light finite 30 times further 250 minutes over 4 hours

⚠ If you missed marks here: Watch the powers of ten: 4.5 × 10¹² ÷ 3.0 × 10⁸ = 15 000 s (not 1 500 or 150 000), then divide by 60 ONCE to get 250 minutes. For (iii), "Neptune is far away" is not enough — you must say radio signals travel at the finite speed of light AND link the 30× distance ratio to a ~30× longer travel time.

Question 3 -- Nuclear Fusion and the Sun

Total: 12 marks

Scientists at the Rutherford Appleton Laboratory in Oxfordshire use satellite data to study the Sun's emissions. The Sun is classified as a medium-sized star, composed mainly of hydrogen and helium.

(a) [3 marks]

In the Sun's core, hydrogen nuclei undergo nuclear fusion to form helium. Write a description of this process, including:
• the particles involved
• why extremely high temperatures are needed
• what is produced besides helium

Mark Scheme -- Part (a)

Hydrogen nuclei (protons) fuse / join together to form helium nuclei

Extremely high temperatures (millions of degrees) are needed to give the nuclei enough kinetic energy to overcome the electrostatic repulsion between the positive charges

A large amount of energy is released during fusion (as electromagnetic radiation / light and heat)

Mark 25 -- hydrogen nuclei protons fuse join together form helium Mark 26 -- high temperatures kinetic energy overcome electrostatic repulsion positive charges Mark 27 -- large amount energy released electromagnetic radiation light heat

⚠ If you missed marks here: Say hydrogen NUCLEI (protons) fuse — "atoms" or "molecules" joining is not fusion language and can cost the mark. The most-missed point is WHY high temperature is needed: both nuclei are positive, so they need enormous kinetic energy to overcome the electrostatic repulsion. "High temperature starts the reaction" without the repulsion reason does not score.

(b) [3 marks]

The Sun emits electromagnetic radiation including infrared, visible light, and ultraviolet.

Explain why the proportions of each type of radiation reaching the Earth's surface differ from those emitted by the Sun.

Mark Scheme -- Part (b)

The Earth's atmosphere absorbs, reflects, or scatters some of the radiation before it reaches the surface

The ozone layer absorbs most of the ultraviolet radiation, so less UV reaches the surface than is emitted

Greenhouse gases (e.g. water vapour, CO₂) absorb some infrared radiation; visible light passes through relatively unaffected, so the proportion of visible light at the surface is higher than emitted

Mark 28 -- atmosphere absorbs reflects scatters radiation before reaching surface Mark 29 -- ozone layer absorbs ultraviolet UV less reaches surface Mark 30 -- greenhouse gases absorb infrared visible light passes through proportions different

⚠ If you missed marks here: A vague "the atmosphere blocks some radiation" earns one mark at best. Be specific about which part absorbs which type: the ozone layer absorbs most of the UV, greenhouse gases (water vapour, CO₂) absorb some infrared, and visible light passes through largely unaffected — that mismatch is exactly why the proportions at the surface differ.

The Sun converts approximately 4 × 10⁹ kg of matter into energy every second. Despite this enormous rate of mass loss, the Sun has been stable for about 4.6 × 10⁹ years.

Suggest why the Sun can sustain fusion for such a long time.

Mark Scheme -- Part (c)

The Sun has an extremely large mass (approximately 2 × 10³⁰ kg), providing a vast supply of hydrogen fuel

Only a tiny fraction of the Sun's total mass is converted to energy each second, so the fuel lasts billions of years

The Sun is in a state of equilibrium (main sequence) -- the outward radiation pressure balances the inward gravitational collapse, keeping it stable

Mark 31 -- Sun extremely large mass vast supply hydrogen fuel Mark 32 -- tiny fraction total mass converted each second fuel lasts billions years Mark 33 -- equilibrium main sequence radiation pressure balances gravitational collapse stable

⚠ If you missed marks here: Writing only "the Sun has lots of hydrogen" misses two marks. Add the scale point — the 4 × 10⁹ kg lost per second is a tiny fraction of the Sun's ~2 × 10³⁰ kg mass — and the stability point: outward radiation pressure balances inward gravitational collapse (main sequence equilibrium).

(d) [3 marks]

Compare the Sun with a red dwarf star and a blue supergiant star. For each comparison, state one difference in terms of size, temperature, or luminosity.

Mark Scheme -- Part (d)

A red dwarf is smaller / cooler / less luminous than the Sun (e.g. lower surface temperature, much dimmer)

A blue supergiant is larger / hotter / more luminous than the Sun (e.g. much higher surface temperature, thousands of times brighter)

Clear comparative statement for each: e.g. the Sun has a surface temperature of about 5500°C; a red dwarf is about 3000°C; a blue supergiant exceeds 25 000°C

Mark 34 -- red dwarf smaller cooler less luminous dimmer than Sun Mark 35 -- blue supergiant larger hotter more luminous brighter than Sun Mark 36 -- clear comparative statement with specific property size temperature luminosity

⚠ If you missed marks here: Each answer must be a clear comparison AGAINST THE SUN, naming the property: a red dwarf is smaller/cooler/dimmer than the Sun, a blue supergiant is larger/hotter/brighter than the Sun. Writing "red dwarfs are small" without "than the Sun" is not a comparison and typically loses the third mark.

Question 4 -- Stellar Evolution

Total: 12 marks

Astronomers at the Scottish Dark Sky Observatory photograph a variety of nebulae and stellar objects as part of their public outreach programme.

(a) [4 marks]

A student at the Scottish Dark Sky Observatory photographs the Orion Nebula (a stellar nursery) and the Ring Nebula (a planetary nebula).

Explain the significance of each nebula in the context of stellar life cycles.

Mark Scheme -- Part (a)

The Orion Nebula is a stellar nursery -- a large cloud of gas and dust where new stars are being born / where gravity pulls gas together to form protostars

It represents the beginning of a star's life cycle -- protostars form and eventually become main sequence stars when nuclear fusion begins

The Ring Nebula is a planetary nebula -- the outer layers of a dying low/medium-mass star that have been ejected into space

It represents the end stage of a Sun-like star's life -- the core that remains becomes a white dwarf, which will eventually cool to a black dwarf

Mark 37 -- Orion Nebula stellar nursery cloud gas dust new stars born protostars Mark 38 -- beginning star life cycle protostars main sequence fusion begins Mark 39 -- Ring Nebula planetary nebula outer layers ejected dying star Mark 40 -- end stage Sun-like star core becomes white dwarf cools black dwarf

⚠ If you missed marks here: The trap: a "planetary nebula" has nothing to do with planets — it is the ejected outer layers of a dying low/medium-mass star, whose remaining core becomes a white dwarf (eventually a black dwarf). For Orion, do not stop at "stars form there": say gravity pulls the gas and dust into protostars, which become main sequence stars when fusion begins.

(b) [4 marks]

Describe the complete life cycle of a star that is approximately 20 times the mass of the Sun. Include all stages from nebula to final remnant.

Explain what determines whether the remnant is a neutron star or a black hole.

Mark Scheme -- Part (b)

Nebula → protostar → main sequence star: gravity pulls gas and dust together; when the core is hot enough, hydrogen fusion begins and the star joins the main sequence

Main sequence → red supergiant: when hydrogen fuel in the core is exhausted, the star expands and cools to become a red supergiant (heavier elements fuse in the core)

Red supergiant → supernova: the core collapses suddenly and the outer layers are blown off in a massive explosion (supernova), creating and scattering heavy elements

The remnant depends on the remaining core mass: if the core is below about 3 solar masses, it becomes a neutron star; if the core exceeds about 3 solar masses, gravity is so strong that nothing can stop the collapse and it becomes a black hole

Mark 41 -- nebula protostar main sequence gravity gas dust fusion begins Mark 42 -- hydrogen exhausted expands cools red supergiant heavier elements fuse Mark 43 -- core collapses supernova outer layers blown off explosion heavy elements Mark 44 -- remnant core mass determines neutron star below 3 solar masses black hole above 3 solar masses

⚠ If you missed marks here: A 20-solar-mass star does NOT become a red giant then a white dwarf — that is the Sun-like path and it loses marks here. The sequence is nebula → protostar → main sequence → red SUPERGIANT → SUPERNOVA. Then the remnant rule is its own mark: core below about 3 solar masses → neutron star, above about 3 → black hole. State the mass condition explicitly.

The star Betelgeuse in the constellation Orion is a red supergiant approximately 700 light-years from Earth. Presenters on the BBC programme The Sky at Night have discussed the possibility that Betelgeuse could explode as a supernova.

(i) If Betelgeuse were to explode as a supernova tomorrow, explain when we would observe the explosion and why.

(ii) Calculate the distance to Betelgeuse in metres. (1 light-year = 9.5 × 10¹⁵ m)

Mark Scheme -- Part (c)

(i) We would not observe the explosion for 700 years after it happened

(i) Because light travels at a finite speed (3.0 × 10⁸ m/s) and the light from the explosion must travel 700 light-years to reach Earth

(ii) Distance = 700 × 9.5 × 10¹⁵ m

(ii) Distance = 6.65 × 10¹⁸ m ≈ 6.7 × 10¹⁸ m

Mark 45 -- not observe for 700 years after explosion Mark 46 -- light travels finite speed must travel 700 light-years reach Earth Mark 47 -- distance equals 700 times 9.5 times 10 to 15 Mark 48 -- answer 6.65 times 10 to 18 or 6.7 times 10 to 18 metres

⚠ If you missed marks here: For (i), "we would see it tomorrow" scores zero — the light needs 700 years to reach us because it travels at a finite speed. For (ii), MULTIPLY: 700 × 9.5 × 10¹⁵ = 6.65 × 10¹⁸ m. Dividing by 9.5 × 10¹⁵ instead (a common reflex) gives 7.4 × 10⁻¹⁴, which should ring alarm bells as absurdly small for a distance to a star.

Question 5 -- Distances in Space

Total: 10 marks

Researchers at the Royal Astronomical Society use data from space telescopes to measure the vast distances between galaxies. The Milky Way, our home galaxy, has a diameter of approximately 1.0 × 10⁵ light-years.

(a) [3 marks]

The Andromeda Galaxy is approximately 2.5 × 10⁶ light-years from Earth. The diameter of the Milky Way is approximately 1.0 × 10⁵ light-years.

(i) Calculate the distance to the Andromeda Galaxy in metres. (1 light-year = 9.5 × 10¹⁵ m)

(ii) Calculate the diameter of the Milky Way in metres.

(iii) Determine how many Milky Way diameters would fit between our galaxy and Andromeda.

Mark Scheme -- Part (a)

(i) Andromeda distance = 2.5 × 10⁶ × 9.5 × 10¹⁵ = 2.375 × 10²² m ≈ 2.4 × 10²² m

(ii) Milky Way diameter = 1.0 × 10⁵ × 9.5 × 10¹⁵ = 9.5 × 10²⁰ m

(iii) Ratio = 2.5 × 10⁶ / 1.0 × 10⁵ = 25 Milky Way diameters (or equivalently 2.4 × 10²² / 9.5 × 10²⁰ = 25)

Mark 49 -- Andromeda 2.5 times 10 to 6 times 9.5 times 10 to 15 equals 2.4 times 10 to 22 metres Mark 50 -- Milky Way diameter 1.0 times 10 to 5 times 9.5 times 10 to 15 equals 9.5 times 10 to 20 metres Mark 51 -- ratio 25 Milky Way diameters fit between galaxies

⚠ If you missed marks here: These are all multiply-by-9.5 × 10¹⁵ conversions — dividing instead is the classic error. Check the exponents carefully: 2.5 × 10⁶ × 9.5 × 10¹⁵ = 2.4 × 10²² m (23.75 × 10²¹ rewritten in standard form). For (iii) you can skip metres entirely: 2.5 × 10⁶ ÷ 1.0 × 10⁵ = 25 diameters.

(b) [4 marks]

A galaxy is observed to be receding from Earth at a velocity of 6.6 × 10⁶ m/s.

(i) Using the Hubble equation v = H₀d, where H₀ = 2.2 × 10⁻¹⁸ s⁻¹, calculate the distance to this galaxy in metres.

(ii) Convert your answer to light-years. (1 light-year = 9.5 × 10¹⁵ m)

(iii) State what is meant by the term "recession velocity" in this context.

Mark Scheme -- Part (b)

(i) d = v / H₀ = 6.6 × 10⁶ / 2.2 × 10⁻¹⁸

(i) d = 3.0 × 10²⁴ m

(ii) In light-years: 3.0 × 10²⁴ / 9.5 × 10¹⁵ = 3.16 × 10⁸ ly ≈ 3.2 × 10⁸ light-years (320 million light-years)

(iii) Recession velocity is the speed at which a galaxy is moving away from us / the speed at which the distance between the galaxy and Earth is increasing due to the expansion of the Universe

Mark 52 -- d equals v divided H0 equals 6.6 times 10 to 6 divided 2.2 times 10 to minus 18 Mark 53 -- distance 3.0 times 10 to 24 metres Mark 54 -- converted 3.0 times 10 to 24 divided 9.5 times 10 to 15 equals 3.2 times 10 to 8 light-years 320 million Mark 55 -- recession velocity speed galaxy moving away expansion Universe

⚠ If you missed marks here: The big one: rearrange v = H₀d to d = v ÷ H₀. Multiplying instead gives about 1.5 × 10⁻¹¹ m — smaller than an atom, so sanity-check your answer! Correctly, 6.6 × 10⁶ ÷ 2.2 × 10⁻¹⁸ = 3.0 × 10²⁴ m (dividing by 10⁻¹⁸ makes the number bigger). For (iii), say the galaxy is moving AWAY from us / the distance is increasing due to the expansion of the Universe — not just "its speed".

Explain why redshift measurements of very distant galaxies provide information about the early Universe, rather than the Universe as it is now.

Mark Scheme -- Part (c)

Light from very distant galaxies has taken billions of years to reach us, so we see them as they were in the past, not as they are now

The further away a galaxy is, the further back in time we are looking (because of the finite speed of light)

The redshift of these distant galaxies tells us how fast the Universe was expanding at that earlier time, giving us a "snapshot" of the early Universe

Mark 56 -- light taken billions years reach us see them as they were past Mark 57 -- further away further back time finite speed light Mark 58 -- redshift tells expansion rate earlier time snapshot early Universe

⚠ If you missed marks here: "They are very far away" alone does not score. You need the time link: light from these galaxies took billions of years to reach us, so we see them as they WERE, and the further the galaxy, the further back in time we look (finite speed of light). Then close the loop: their redshift reveals how fast the Universe was expanding at that earlier time.

Question 6 -- Cosmology: Big Bang, CMBR, and the Age of the Universe

Total: 12 marks

Data from the ESA Planck satellite and observations at the Lovell Telescope at Jodrell Bank have contributed to our understanding of the origin and evolution of the Universe.

(a) [3 marks]

Describe the Big Bang theory. Include what happened in the first few minutes after the Big Bang and how matter began to form.

Mark Scheme -- Part (a)

The Big Bang theory states that the Universe began from a single point / singularity of extremely high temperature and density, and has been expanding ever since

In the first few minutes, the Universe was incredibly hot; as it expanded and cooled, sub-atomic particles (protons, neutrons, electrons) formed

Simple nuclei (hydrogen and helium) formed through nuclear fusion in the first few minutes; atoms formed later as the Universe cooled enough for electrons to be captured by nuclei

Mark 59 -- Big Bang Universe began single point singularity high temperature density expanding Mark 60 -- first minutes incredibly hot expanded cooled protons neutrons electrons formed Mark 61 -- hydrogen helium nuclei fusion first minutes atoms formed cooled electrons captured

⚠ If you missed marks here: Do not stop at "the Universe began at a single point and is expanding" — that is only one of three marks. The other two are about the first few minutes: as the Universe expanded and COOLED, protons, neutrons and electrons formed, then hydrogen and helium NUCLEI formed by fusion; whole atoms came later, once it was cool enough for electrons to be captured.

(b) [3 marks]

Explain what cosmic microwave background radiation (CMBR) is and why its near-uniformity across the sky supports the Big Bang theory. State the approximate temperature of the CMBR.

Mark Scheme -- Part (b)

CMBR is low-energy microwave radiation that fills the entire Universe; it is the remnant / "afterglow" of the extremely hot, dense early Universe that has cooled as the Universe expanded

Its near-uniformity (almost the same temperature in every direction) supports the Big Bang because it shows the early Universe was very uniform / homogeneous, consistent with everything originating from one event

The approximate temperature of the CMBR is about 2.7 K (about −270°C)

Mark 62 -- CMBR microwave radiation fills Universe remnant afterglow early Universe cooled expanded Mark 63 -- near-uniformity same temperature every direction early Universe uniform homogeneous one event Mark 64 -- temperature approximately 2.7 K or minus 270 degrees Celsius

⚠ If you missed marks here: Three marks, three commonly missed pieces: (1) CMBR is the cooled remnant / "afterglow" of the hot early Universe — not just "radiation from space"; (2) its near-uniformity in EVERY direction shows the early Universe was homogeneous, consistent with a single origin event; (3) the temperature, about 2.7 K (−270°C) — easy to forget even though the question asks for it directly.

Two galaxies are observed: Galaxy A has a recession velocity of 3.3 × 10⁶ m/s and Galaxy B has a recession velocity of 8.8 × 10⁶ m/s.

Using H₀ = 2.2 × 10⁻¹⁸ s⁻¹, calculate the distance to each galaxy. Which galaxy shows a greater redshift? Explain your reasoning.

Mark Scheme -- Part (c)

Galaxy A: d = v / H₀ = 3.3 × 10⁶ / 2.2 × 10⁻¹⁸ = 1.5 × 10²⁴ m

Galaxy B: d = v / H₀ = 8.8 × 10⁶ / 2.2 × 10⁻¹⁸ = 4.0 × 10²⁴ m

Galaxy B shows a greater redshift because it is further away and has a higher recession velocity; the greater the recession velocity, the more the light is stretched (shifted to longer wavelengths / redshifted)

Mark 65 -- Galaxy A distance 3.3 times 10 to 6 divided 2.2 times 10 to minus 18 equals 1.5 times 10 to 24 metres Mark 66 -- Galaxy B distance 8.8 times 10 to 6 divided 2.2 times 10 to minus 18 equals 4.0 times 10 to 24 metres Mark 67 -- Galaxy B greater redshift further away higher recession velocity light stretched longer wavelengths

⚠ If you missed marks here: Same d = v ÷ H₀ rearrangement twice: A = 1.5 × 10²⁴ m, B = 4.0 × 10²⁴ m (multiplying by H₀ gives tiny nonsense answers). The explain mark is where most slip: "B is further so more redshift" is incomplete — link the higher recession velocity to the light being stretched to LONGER wavelengths. Redshift means the wavelength is INCREASED, not just "moving away".

(d) [3 marks]

Using H₀ = 2.2 × 10⁻¹⁸ s⁻¹, estimate the age of the Universe in years. (1 year = 3.15 × 10⁷ s)

Discuss one limitation of using 1/H₀ as an estimate for the age of the Universe.

Mark Scheme -- Part (d)

Age = 1 / H₀ = 1 / (2.2 × 10⁻¹⁸) = 4.55 × 10¹⁷ s; converting: 4.55 × 10¹⁷ / 3.15 × 10⁷ = 1.44 × 10¹⁰ years ≈ 14.4 billion years

Correct final answer of approximately 14 billion years (accept 13-15 billion years range)

Limitation: 1/H₀ assumes the rate of expansion has been constant throughout the history of the Universe, but in reality the expansion rate has changed (it was slowing due to gravity, and is now accelerating due to dark energy), so 1/H₀ is only an approximation

Mark 68 -- 1 divided 2.2 times 10 to minus 18 equals 4.55 times 10 to 17 seconds converted 3.15 times 10 to 7 Mark 69 -- answer approximately 14.4 billion years 1.44 times 10 to 10 Mark 70 -- limitation assumes constant expansion rate actually changed slowing gravity accelerating dark energy approximation

⚠ If you missed marks here: Two traps: stopping at 4.55 × 10¹⁷ s without converting to years (divide by 3.15 × 10⁷ to get 1.44 × 10¹⁰ ≈ 14.4 billion years), and giving "H₀ might be measured inaccurately" as the limitation. The scheme wants the assumption: 1/H₀ assumes the expansion rate has been CONSTANT for all time, but it has actually changed (slowed by gravity, now accelerating).

Question 7 -- Gravity, Orbits, and Space Exploration

Total: 10 marks

In 2015, British ESA astronaut Tim Peake spent 186 days aboard the International Space Station (ISS) as part of the Principia mission. The ISS orbits at approximately 408 km above the Earth's surface.

(a) [4 marks]

The radius of the Earth is 6370 km. The ISS completes one orbit in approximately 92 minutes.

Calculate:
(i) the orbital radius of the ISS in metres
(ii) the orbital speed of the ISS using v = 2πr / T
(iii) the total distance Tim Peake travelled during his 186-day mission

Mark Scheme -- Part (a)

(i) Orbital radius = Earth radius + altitude = 6370 + 408 = 6778 km = 6.778 × 10⁶ m

(ii) v = 2π × 6.778 × 10⁶ / (92 × 60) = 4.259 × 10⁷ / 5520 = 7714 m/s ≈ 7700 m/s

(iii) Time = 186 days = 186 × 24 × 3600 = 1.607 × 10⁷ s

(iii) Distance = speed × time = 7700 × 1.607 × 10⁷ = 1.237 × 10¹¹ m ≈ 1.2 × 10¹¹ m

Mark 71 -- orbital radius 6370 plus 408 equals 6778 km or 6.778 times 10 to 6 metres Mark 72 -- v equals 2 pi times 6.778 times 10 to 6 divided 5520 equals 7700 m/s Mark 73 -- time 186 days converted 1.607 times 10 to 7 seconds Mark 74 -- distance 7700 times 1.607 times 10 to 7 equals 1.2 times 10 to 11 metres

⚠ If you missed marks here: Two classic slips: using just the 408 km altitude as the orbital radius instead of adding the Earth's radius (6370 + 408 = 6778 km = 6.778 × 10⁶ m), and forgetting to convert 92 minutes to 5520 s — either one wrecks the speed. For (iii), 186 days must become seconds (186 × 24 × 3600 = 1.6 × 10⁷ s) before multiplying by 7700 m/s.

(b) [3 marks]

Explain why astronauts aboard the ISS experience apparent weightlessness even though they are still within Earth's gravitational field. Use the concept of free fall in your answer.

Mark Scheme -- Part (b)

The ISS and the astronauts inside it are both in free fall towards the Earth at the same rate (due to gravity)

Because they are falling at the same rate, there is no contact force / normal reaction between the astronaut and the floor of the ISS, so they feel weightless

They are still within Earth's gravitational field (g is about 8.7 N/kg at ISS altitude), but the gravitational force provides the centripetal force needed for their circular orbit, so they do not fall to the ground

Mark 75 -- ISS astronauts both free fall towards Earth same rate gravity Mark 76 -- falling same rate no contact force normal reaction feel weightless Mark 77 -- still within gravitational field gravity provides centripetal force circular orbit

⚠ If you missed marks here: "There is no gravity in space" scores zero — g is still about 8.7 N/kg at ISS altitude. The wanted answer: the astronaut AND the ISS are both in free fall towards Earth at the same rate, so there is no contact/normal force between astronaut and floor — that is why they FEEL weightless. Gravity is still acting; it provides the centripetal force for the orbit.

The ESA Rosetta spacecraft took 10 years to reach Comet 67P/Churyumov-Gerasimenko. The comet orbits the Sun in a highly elliptical orbit.

Explain why the comet's speed varies significantly during its orbit and at which point in its orbit the speed is greatest.

Mark Scheme -- Part (c)

The comet's orbit is highly elliptical, so its distance from the Sun varies greatly during one orbit

By conservation of energy, as the comet gets closer to the Sun, gravitational potential energy decreases and kinetic energy increases, so the comet speeds up; the speed is greatest at perihelion (the closest point to the Sun)

As the comet moves away from the Sun towards aphelion (the furthest point), kinetic energy is converted to gravitational potential energy and the comet slows down; the speed is lowest at aphelion

Mark 78 -- elliptical orbit distance from Sun varies greatly Mark 79 -- conservation energy closer Sun GPE decreases KE increases speeds up greatest perihelion closest point Mark 80 -- moves away KE converted GPE slows down slowest aphelion furthest point

⚠ If you missed marks here: Name the points: speed is greatest at PERIHELION (closest to the Sun) and lowest at aphelion (furthest). "It goes faster near the Sun because gravity is stronger" misses the energy marks — state that GPE converts to KE on the approach and KE back to GPE moving away, and start by noting the elliptical orbit means the comet's distance from the Sun varies greatly.