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DSE Physics — Astronomy and Cosmology Guide for Tutors: Parallax, Luminosity Distance and Four Hubble Law Errors
Students substitute into d equals one over p but leave the answer in parsecs when the question asks for light years, and they miss that the slope of a Hubble plot is the constant itself. This lesson lines the three distance tools up in order.
Objectives — Students can: (1) turn a parallax of 0.05 arcsecond into 20 parsecs and convert it to 65.2 light years; (2) use the distance modulus to find that a star at 100 parsecs with absolute magnitude 1 has apparent magnitude 6.0; (3) apply the Hubble law to a galaxy at 100 Mpc, getting 7000 km per second, and estimate the age of the universe from 70 km per second per Mpc as about 14.0 billion years.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Angle units | How many arcseconds are in one degree? | Three thousand six hundred |
| Logarithms | What is log of one hundred? | Two, because ten squared is one hundred |
| Speed of light | What is the speed of light? | About three hundred thousand kilometres per second |
| Kepler law | How do orbital radius and period relate? | Period squared is proportional to radius cubed |
02 Lesson flow (50 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-6 min | Prior check | Ask arcseconds, logarithms, light speed, Kepler | Answer orally, errors on the board |
| 6-20 min | Act 1: parallax | Model p of 0.05 arcsecond with unit conversion | Compute one case and give units |
| 20-34 min | Act 2: luminosity and distance | Model the distance modulus | Reverse the formula for distance |
| 34-44 min | Act 3: redshift and Hubble | Model redshift, recession speed and cosmic age | Sketch the speed against distance line |
| 44-50 min | Practice and exit | One question of each type, then three oral checks | Closed-book answers |
03 Teaching script (read it out)
Scene 1: Act 1: parallax is the only distance we measure directly
Build the geometry first. As the Earth orbits the Sun, the same star is photographed six months apart and shifts against the background, and half of that shift is the parallax p. Because p is tiny, the relation is d equals one over p, with p in arcseconds and d in parsecs.
Substitute: p is 0.05 arcsecond, so d is one over 0.05, which is 20 parsecs. Each parsec is 3.26 light years, so 20 times 3.26 gives 65.2 light years. This step is the most commonly dropped, because the question asks for light years and the student stops at parsecs.
Give the definition: one parsec is the distance at which the parallax is one arcsecond, so a smaller p means a larger d. That inverse relation is also why the method only works for nearby stars.
Misconception box: the angle in the data is usually the total shift over six months. Halve it before inverting.
Substitute: p is 0.05 arcsecond, so d is one over 0.05, which is 20 parsecs. Each parsec is 3.26 light years, so 20 times 3.26 gives 65.2 light years. This step is the most commonly dropped, because the question asks for light years and the student stops at parsecs.
Give the definition: one parsec is the distance at which the parallax is one arcsecond, so a smaller p means a larger d. That inverse relation is also why the method only works for nearby stars.
Misconception box: the angle in the data is usually the total shift over six months. Halve it before inverting.
Students restate: half the shift, invert for parsecs, multiply by 3.26 for light years.
Scene 2: Act 2: luminosity is true brightness, apparent magnitude is what we see
Separate the two terms. Absolute magnitude is the brightness at ten parsecs, apparent magnitude is the brightness now. The same object looks dimmer as it moves away, and the distance modulus links them: apparent minus absolute equals five log of distance over ten.
Model the forward case: at 100 parsecs the log is one, so the modulus is 5, and a star of absolute magnitude 1 has apparent magnitude 6.0. The reverse case is just as common: given both magnitudes, divide the difference by five and raise ten to that power.
Add the physical intuition: brightness falls as the square of distance, so doubling the distance cuts brightness to a quarter, about one and a half magnitudes fainter. That check catches sign errors instantly, because a larger distance must give a larger apparent magnitude number.
Misconception box: magnitudes run backwards. The smaller the number, the brighter the object.
Model the forward case: at 100 parsecs the log is one, so the modulus is 5, and a star of absolute magnitude 1 has apparent magnitude 6.0. The reverse case is just as common: given both magnitudes, divide the difference by five and raise ten to that power.
Add the physical intuition: brightness falls as the square of distance, so doubling the distance cuts brightness to a quarter, about one and a half magnitudes fainter. That check catches sign errors instantly, because a larger distance must give a larger apparent magnitude number.
Misconception box: magnitudes run backwards. The smaller the number, the brighter the object.
Students restate: subtract the magnitudes, divide by five, raise ten to the power for distance.
Scene 3: Act 3: redshift ties speed to distance
Start with what redshift is. Spectral lines from a receding galaxy shift towards the red because the wavelength is stretched. A redshift z of 0.05 at low speed gives a recession speed of roughly light speed times z, so 300000 times 0.05 is 15000 kilometres per second.
Then the Hubble law: speed equals the constant times distance. With 70 kilometres per second per megaparsec, a galaxy at 100 Mpc recedes at 7000 kilometres per second. In reverse, a speed of 15000 implies a distance of 15000 divided by 70, about 214 megaparsecs.
The age estimate is elegant: write distance as speed times time and the constant disappears, leaving time equal to one over the Hubble constant, which evaluates to about 14.0 billion years. That figure matches the oldest stars in order of magnitude and is the strongest evidence for the law itself. Warn about units: parsecs to kilometres, years to seconds, both conversions are compulsory.
Close with Kepler third law in astronomical units and years: period squared equals radius cubed, so a planet at 4 AU has a period of 8 years.
Misconception box: cosmological redshift is not simply a Doppler shift, because space itself expands between the galaxies.
Then the Hubble law: speed equals the constant times distance. With 70 kilometres per second per megaparsec, a galaxy at 100 Mpc recedes at 7000 kilometres per second. In reverse, a speed of 15000 implies a distance of 15000 divided by 70, about 214 megaparsecs.
The age estimate is elegant: write distance as speed times time and the constant disappears, leaving time equal to one over the Hubble constant, which evaluates to about 14.0 billion years. That figure matches the oldest stars in order of magnitude and is the strongest evidence for the law itself. Warn about units: parsecs to kilometres, years to seconds, both conversions are compulsory.
Close with Kepler third law in astronomical units and years: period squared equals radius cubed, so a planet at 4 AU has a period of 8 years.
Misconception box: cosmological redshift is not simply a Doppler shift, because space itself expands between the galaxies.
Students restate: redshift times light speed, divide by the constant, invert the constant for the age.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Parallax distance | p of 0.05 arcsecond | Invert, state parsecs, convert to light years | 20 parsecs, about 65.2 light years |
| Distance modulus | 100 parsecs, absolute magnitude 1 | Compute five log of distance over ten, then add | Apparent magnitude 6.0 |
| Hubble law | 100 Mpc | Multiply constant by distance, keep units | 7000 kilometres per second |
| Redshift distance | z of 0.05 | Convert to speed, divide by the constant | About 214 megaparsecs |
Sanity check (teach this): Sanity checks: a larger distance must give a larger apparent magnitude number, so any answer that gets brighter with distance has a sign error. A Hubble answer carries the unit kilometres per second, and dropping the per second is a unit error. The cosmic age must exceed the age of the oldest stars, around thirteen billion years, so 14.0 billion is plausible while a few million is a conversion slip. In Kepler law the period always exceeds the radius in number, because squaring loses to cubing, so four AU gives eight years.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| p is 0.05 arcsecond. Find the distance in light years. | 20 parsecs, about 65.2 light years | Was 3.26 applied |
| 100 parsecs, absolute magnitude 1. Find the apparent magnitude. | 6.0 | Logarithm used rather than intuition |
| z is 0.05. Find the speed and the distance. | 15000 kilometres per second, about 214 megaparsecs | Both formulas applied in order |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Forgetting to halve the shift | Inverting the full six-month displacement | The data give the total shift; parallax is half of it, so skipping the halving doubles the distance error. |
| Leaving the answer in parsecs | Writing twenty light years after inverting | The inverse gives parsecs; multiply by 3.26 when the question asks for light years. |
| Reversing the magnitude scale | Making the apparent magnitude smaller as distance grows | Smaller magnitude means brighter, so a distant star must have a larger number; a reversed answer means a sign error. |
| Wrong cosmic age unit | Reporting 13.97 years | One over the constant gives about 14.0 billion years, and skipping the conversion is a lost mark. |
07 Homework (paper, section and question numbers)
- Redo this chapter — Redo the three tasks showing formula, substitution and units
- Physics chapter index — Open the electromagnetic waves chapter and link redshift to the spectrum in three sentences
- Formula card — Put parallax, distance modulus, Hubble law and Kepler third law side by side with units
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| What is the distance when p is one arcsecond? | One parsec, about 3.26 light years | If wrong, repeat Act 1 |
| How does apparent magnitude change with distance? | It grows, so the star looks dimmer | If reversed, repeat Act 2 |
| How is the cosmic age estimated? | One over the Hubble constant, about fourteen billion years | If stuck, repeat Act 3 |
09 Tutor FAQ
Why does parallax only work nearby?
Distant stars shift by too small an angle for current instruments, so the error exceeds the answer.
Is redshift the same as the Doppler effect?
Only as a low-speed approximation; cosmological redshift comes from expanding space.
Is the Hubble constant really constant?
It changes with cosmic time, and current measurements still disagree by a few kilometres per second.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.