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S6 Physics: Astronomy and Space Science – Practice Paper 1 (Foundation, 10 Questions)
S6 Physics: Astronomy and Space Science – Practice Paper 1 (Foundation, 10 Questions)
Paper one builds the three measuring sticks of astronomy: converting between the astronomical unit and the light year, light travel time, and parallax in parsecs, then comparing brightness with the inverse square law and the magnitude scale. Wien law converts temperature into peak wavelength, and the paper closes with the Earth orbital and escape speeds and the basic error of treating the light year as a unit of time.
Quick answer Light takes 500 s, about 8.3 minutes, to cover 1 AU; 4.24 ly is 4.01 times 10 to the 16 m; a parallax of 0.1 arcsecond gives 10 pc, or 32.6 ly; three times the distance leaves one ninth of the brightness; five magnitudes is 100 times and 2.5 magnitudes about 10 times; 5800 K peaks at 500 nm and 2900 K at 1000 nm in the infrared; the orbital speed is about 29.9 km/s and the escape speed 11.2 km/s; a light year is a unit of length.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | about 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S6 Physics | Astronomy and Space Science: Light Year, Luminosity, Redshift and Cosmic Expansion |
| Level | S6 |
02 Questions (10 items)
- FoundationHow long does light take to travel from the Sun to the Earth? Use 1 AU = 1.50 times 10 to the 11 m.
- FoundationProxima Centauri is 4.24 light years away. Express this in metres, using 1 ly = 9.46 times 10 to the 15 m.
- FoundationA star has a parallax angle of 0.1 arcsecond. Find its distance in parsecs and in light years.
- FoundationIf a star is moved to three times its original distance, what fraction of its original brightness do we receive?
- IntermediateTwo stars differ by 5 magnitudes. What is the brightness ratio? What if they differ by 2.5 magnitudes?
- IntermediateA star has a surface temperature of 5800 K. Find its peak wavelength using Wien law, peak wavelength = 2.90 times 10 to the minus 3 over T.
- IntermediateA red dwarf has a surface temperature of 2900 K. Find its peak wavelength and the band it belongs to.
- ChallengeThe Earth orbits at 1.50 times 10 to the 11 m with a period of 3.156 times 10 to the 7 s. Find its orbital speed.
- ChallengeThe Earth has radius 6.37 times 10 to the 6 m and mass 5.97 times 10 to the 24 kg. Find the escape speed at its surface.
- Error redoA student writes that a star 4 light years away is 4 years of light, treating the light year as a unit of time. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | t = 1.50 times 10 to the 11 divided by 3.00 times 10 to the 8, which is 500 s, about 8.3 minutes | A distance divided by the speed of light gives the light travel time; 1 AU is about 8.3 light minutes. |
| 2 | 4.24 times 9.46 times 10 to the 15, which is 4.01 times 10 to the 16 m | A light year is a unit of length, the distance light travels in one year. |
| 3 | d = 1 over p = 1 over 0.1, which is 10 pc, and 10 times 3.26 is 32.6 light years | The parallax method gives distance in parsecs as the reciprocal of the parallax angle in arcseconds, and works only for nearby stars. |
| 4 | Brightness falls as the square of distance, so one third squared is one ninth of the original value | The inverse square law is the basic tool of astronomical photometry and distance measurement. |
| 5 | Five magnitudes means 100 times; 2.5 magnitudes means the square root of 100, about 10 times | A five magnitude difference is a hundredfold brightness ratio, so one magnitude is about 2.512 times. |
| 6 | The peak wavelength is 2.90 times 10 to the minus 3 divided by 5800, which is 5.0 times 10 to the minus 7 m, or 500 nm, in the visible band | A hotter star peaks at a shorter wavelength, which is why blue stars are hotter than red ones. |
| 7 | The peak wavelength is 2.90 times 10 to the minus 3 divided by 2900, which is 1.0 times 10 to the minus 6 m, or 1000 nm, in the infrared | A peak in the infrared means the star is invisible to the eye and needs infrared telescopes. |
| 8 | v = 2 pi r over T, about 2.99 times 10 to the 4 m/s, or 29.9 km/s | The orbital speed is about 30 km/s, which is one over root two of the escape speed from the Sun at that radius. |
| 9 | v is the square root of 2 G M over R, about 1.12 times 10 to the 4 m/s, or 11.2 km/s | Escape speed grows as the square root of mass and falls as the square root of radius. |
| 10 | A light year is a unit of length; the distance is about 3.8 times 10 to the 16 m, and we see the star as it was 4 years ago | Seeing light that left four years ago describes a light travel time, not the unit of the distance itself. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
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