Cambridge IGCSE Physics · unit 6
Space physics
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
A satellite raised to twice the orbital radius
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A spherical planet has radius R = 6.0×10⁶ m and surface gravitational field g₀ = 10 N kg⁻¹. A 100 kg satellite follows a circular orbit at radius 2R measured from the planet's centre. Ignore rotation and atmosphere. Use GM = g₀R² and the supplied inverse-square field data. Supplied model: Use g = GM/r², circular acceleration v²/r and orbital period T = 2πr/v. Supplied data (Radius / R; Field magnitude / N kg⁻¹): (1, 10), (2, 2.5), (3, 1.111111), (4, 0.625). (a) Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant. [3 marks] (b) The radius is increased to 4R. Predict ratios of field, speed and period, and explain why astronauts can feel weightless despite nonzero gravity. [3 marks]
Planetary gravitational field
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Field magnitude / N kg⁻¹
Horizontal axis (x): Radius / R
Read the exact data as a table
| Radius / R | Field magnitude / N kg⁻¹ |
|---|---|
| 1 | 10 |
| 2 | 2.5 |
| 3 | 1.111111 |
| 4 | 0.625 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the satellite's weight, orbital speed and orbital period. Explain what supplies the centripetal resultant.
Reveal the worked solution
- At 2R the field is g₀/4 = 2.5 N kg⁻¹, so the gravitational force is mg = 250 N. The satellite's mass remains 100 kg.
- Gravity alone supplies mv²/r: v = √(gr) = √(2.5×1.2×10⁷)≈5477 m s⁻¹.
- Period is circumference/speed, T = 2πr/v≈1.38×10⁴ s. 'Centripetal force' names the inward resultant; it is not an additional interaction alongside gravity.
- For a circular orbit g∝r⁻², v∝r⁻¹/² and T∝r³/². Doubling r therefore gives the stated ratios.
- Astronauts and spacecraft accelerate together in free fall. Little support force acts on the astronauts, so their apparent weight can be near zero.
- The gravitational field is not zero in orbit, and constant speed still requires acceleration because velocity direction changes.
Answer (a) Weight = 250 N; orbital speed ≈5.48 km s⁻¹; period ≈1.38×10⁴ s. (b) Field becomes 1/4, speed 1/√2 and period 2√2 times the initial orbital values.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A scale model that cannot show every scale
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. Use 1 AU = 1.50×10¹¹ m, solar radius = 7.0×10⁸ m and 1 parsec = 3.09×10¹⁶ m. A model puts Earth 15 m from the Sun. The supplied graph lists stellar parallax against distance; parallax p is the half-angle after Earth's six-month baseline shift, and d/pc = 1/(p/arcsec). A star has parallax 0.050 arcsec. Supplied data (Distance / pc; Parallax / arcsec): (5, 0.2), (10, 0.1), (20, 0.05), (40, 0.025). (a) Find the Sun's model radius and the star's true and model distances. Explain the practical scale limitation. [3 marks] (b) A parallax measurement has uncertainty ±0.005 arcsec. Estimate the fractional distance uncertainty and compare with a standard-candle method. [3 marks]
Parallax calibration
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Parallax / arcsec
Horizontal axis (x): Distance / pc
Read the exact data as a table
| Distance / pc | Parallax / arcsec |
|---|---|
| 5 | 0.2 |
| 10 | 0.1 |
| 20 | 0.05 |
| 40 | 0.025 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the Sun's model radius and the star's true and model distances. Explain the practical scale limitation.
Reveal the worked solution
- The model scale is 15/(1.50×10¹¹) = 1.0×10⁻¹⁰, so solar radius becomes 0.070 m.
- The parallax gives d = 1/0.050 = 20 pc, equal to 6.18×10¹⁷ m. Applying the same scale gives 6.18×10⁷ m, or 61800 km.
- A classroom model that fits planetary orbits cannot also put the nearest stellar distances nearby at the same scale; compressed diagrams must state their distortion.
- For d = 1/p, small fractional uncertainties satisfy Δd/d≈Δp/p = 0.005/0.050 = 0.10.
- The exact bounds are 1/0.055≈18.2 pc and 1/0.045≈22.2 pc, showing asymmetric inversion of the parallax limits.
- A standard candle uses known luminosity and measured flux with an inverse-square model. It reaches farther than practical parallax but requires a calibrated intrinsic luminosity and corrections for absorption.
Answer (a) Model Sun radius = 0.070 m; star distance = 20 pc = 6.18×10¹⁷ m; model star distance = 6.18×10⁷ m. (b) Approximate distance uncertainty = 10%, or about ±2 pc; exact bounds are about 18.2–22.2 pc.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Two stars at the same temperature with very different luminosities
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A main-sequence reference star and a red giant both have surface temperature 4000 K. The giant has luminosity 400 times the reference. A third star is a white dwarf with temperature 12000 K and luminosity 0.01 times the reference. Use L = 4πR²σT⁴. The supplied data place the three stars on temperature–luminosity axes; luminosities are relative to the reference star. Supplied data (Surface temperature / K; Relative luminosity): (4000, 1), (4000, 400), (12000, 0.01). (a) Determine the radius of the giant and dwarf relative to the reference. Explain why colour alone cannot determine luminosity. [3 marks] (b) Relate these observations to an H–R diagram and distinguish low-mass from high-mass stellar endpoints. [3 marks]
Three stellar observations
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Relative luminosity
Horizontal axis (x): Surface temperature / K
Read the exact data as a table
| Surface temperature / K | Relative luminosity |
|---|---|
| 4000 | 1 |
| 4000 | 400 |
| 12000 | 0.01 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Determine the radius of the giant and dwarf relative to the reference. Explain why colour alone cannot determine luminosity.
Reveal the worked solution
- At equal temperature, L∝R², so the giant radius ratio is √400 = 20.
- For the dwarf, Rw/Rref = √(Lw/Lref)(Tref/Tw)² = √0.01×(1/3)² = 1/90.
- Colour indicates a characteristic surface temperature, while luminosity also depends on emitting area. A hot small star may be less luminous than a cool large one.
- A standard H–R diagram places hotter temperatures toward the left and often uses logarithmic luminosity. The supplied linear data plot must be relabelled or transformed before being read as a standard H–R diagram.
- A Sun-like star can evolve into a red giant and leave a white dwarf after shedding outer layers. A white dwarf is not a newly formed small main-sequence star.
- Sufficiently massive stars can undergo core-collapse supernovae and leave neutron stars or black holes. Initial mass, mass loss and interactions matter; a single universal life-cycle chain is inadequate.
Answer (a) Giant radius = 20 times reference; dwarf radius = 1/90 of reference. (b) An H–R diagram compares intrinsic stellar properties, not spatial locations; stellar mass strongly influences evolution and endpoint.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A Hubble estimate with a local-motion warning
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A known spectral line has laboratory wavelength 500.0 nm. In one galaxy it appears at 510.0 nm. Independent distance measurements for other galaxies and their low-redshift recession speeds are supplied. Use c = 3.0×10⁵ km s⁻¹, v≈cz and 1 Mpc = 3.09×10¹⁹ km. Ignore curvature and detailed cosmological evolution in this local estimate. Supplied model: Use z = (λobserved−λrest)/λrest and v = H₀d in the stated low-redshift approximation. Supplied data (Distance / Mpc; Recession speed / km s⁻¹): (20, 1400), (40, 2800), (60, 4200), (80, 5600). (a) Find the redshift, approximate recession speed, Hubble parameter and inferred galaxy distance. [3 marks] (b) Estimate 1/H₀ in years using 1 yr = 3.16×10⁷ s, and explain two limits of treating it as an exact cosmic age. [3 marks]
Local distance–recession evidence
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Recession speed / km s⁻¹
Horizontal axis (x): Distance / Mpc
Read the exact data as a table
| Distance / Mpc | Recession speed / km s⁻¹ |
|---|---|
| 20 | 1400 |
| 40 | 2800 |
| 60 | 4200 |
| 80 | 5600 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Find the redshift, approximate recession speed, Hubble parameter and inferred galaxy distance.
Reveal the worked solution
- z = (λobserved−λrest)/λrest = (510−500)/500 = 0.0200.
- The low-z approximation gives v≈cz = 6000 km s⁻¹. The graph gradient gives H₀ = 1400/20 = 70 km s⁻¹ Mpc⁻¹.
- Distance estimated from v = H₀d is 6000/70≈85.7 Mpc. This inference assumes recession dominates the observed line shift.
- Convert H₀ to s⁻¹: 70/(3.09×10¹⁹)≈2.27×10⁻¹⁸ s⁻¹. Its reciprocal is 4.41×10¹⁷ s≈1.40×10¹⁰ yr.
- The cosmic expansion rate evolves, so a current rate extrapolated backward as constant need not give the true age.
- Galaxies also have peculiar motions. A nearby galaxy may be blueshifted despite the large-scale expansion, so one local observation does not overturn the population trend.
Answer (a) z = 0.0200; v≈6000 km s⁻¹; H₀≈70 km s⁻¹ Mpc⁻¹; distance≈85.7 Mpc. (b) Hubble time≈1.40×10¹⁰ years; it is an indicative timescale rather than an exact age.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.