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DSE Physics | Astronomy: parallax, luminosity and distance — lesson-prep guide for tutors

DSE Physics | Astronomy: parallax, luminosity and distance — lesson-prep guide for tutors

  • 2026-11-02

Students can say the words light year, then multiply the wrong numbers when asked how far 20 parsecs is. This lesson separates the three distance tools and gives each one a formula that can be checked.

Objectives — Students can: (1) use d(pc) = 1/p(arcsec) to get a distance from parallax and say why parallax only reaches nearby stars; (2) compare the brightness of equal-luminosity stars with the inverse-square law; (3) use m – M = 5 log(d/10) and v = H0 d to find distance and recession speed, naming the assumption behind each.

01 Prerequisites (3-minute check)

If any of these fails, fix it before the lesson starts.

Skill Quick check Pass mark
Units How many metres and light years is one parsec? About 3.09 x 10^16 m, about 3.26 light years
Inverse square If a source moves three times further away, what happens to flux? It falls to one ninth, because flux goes as one over distance squared
Logarithms What is log 100, and what is 10 squared? 2, and 100: the same statement in two forms

02 Lesson flow (60 minutes)

Time Stage Tutor does Student does
0-5 min Opening question Write 4.2 light years for Proxima and 8 min 20 s for sunlight, ask why the units differ Vote and name what each number measures
5-15 min Prerequisite check Three quick items: parsec, inverse square, logarithms Written answers, marked live
15-37 min Three-part script Teach parallax, luminosity, then distance modulus and Hubble law Recap the assumption of each tool in one sentence
37-49 min Question walk-through Model four questions, writing units before substituting Copy and recompute one themselves
49-55 min Independent practice Circulate; ask only what the valid range of this formula is Do three items and self-mark
55-60 min Exit check Collect answers and count misses aloud Written, no talking

03 Teaching script (read it out)

Scene 1: Part 1: parallax and the parsec

Parallax is the thumb trick. Hold a thumb up, close one eye then the other, and the thumb jumps against the background. Earth moves around the Sun, so six months apart the near star jumps by a tiny angle against far background stars. The definition is fixed: a parallax of one arcsecond means a distance of one parsec, so d in parsecs is simply 1 over p in arcseconds. A parallax of 0.05 arcseconds gives 20 parsecs, which is 20 times 3.26, about 65.3 light years. Watch the denominator: a smaller angle means a larger distance, because the two are inversely related. The limit is angular resolution, so below roughly 0.01 arcseconds the method fails and only reaches a few hundred parsecs.

If parallax falls from 0.05 to 0.02 arcseconds, does distance rise or fall, and by what factor?

Scene 2: Part 2: luminosity and the inverse square

Luminosity L is the energy a star emits each second and does not depend on where you stand. Flux F is the energy per second arriving on each square metre of your telescope and does depend on distance. The same bulb twice as far gives a quarter of the flux; ten times as far gives one hundredth. So two stars of equal luminosity, one at 10 parsecs and one at 100, differ by a factor of one hundred in flux, which is five magnitudes. This is the most used piece of reasoning in astronomy: when a star looks faint, ask first whether it is dim or simply far away.

Does luminosity change when a star is moved ten times further? What about flux?

Scene 3: Part 3: distance modulus and Hubble law

Apparent magnitude m is how bright it looks; absolute magnitude M is how bright it would look at 10 parsecs, so the difference is a distance measure. m – M = 5 log(d/10) with d in parsecs. For m = 6 and M = 1 the difference is 5, so log(d/10) = 1 and d = 100 parsecs. Beyond that, use redshift: v = H0 d. With H0 at 70 km/s per megaparsec, a galaxy at 100 megaparsecs recedes at 7000 km/s. Keep the assumptions apart: the distance modulus assumes you know the true luminosity of the star, and Hubble law assumes uniform expansion. Neither is a tape measure.

What must you already know before using the distance modulus, and before using Hubble law?

04 Question types and worked steps

Type Example Steps Answer
Parallax calculation Given a parallax angle, find distance. (1) Check the unit is arcseconds; (2) d = 1/p; (3) multiply by 3.26 for light years; (4) state the method limit. Example: p = 0.05 arcsec gives 20 pc, about 65.3 ly.
Inverse-square comparison Compare brightness of equal-luminosity stars. (1) Write F proportional to L over d squared; (2) take the distance ratio; (3) square and invert; (4) convert to magnitudes if asked. Example: 10 pc against 100 pc gives a flux ratio of 1 to 1/100.
Distance modulus Given m and M, find d. (1) Compute m – M; (2) divide by 5; (3) add 1 for the exponent; (4) d = 10^(1+(m-M)/5) pc. Example: m = 6, M = 1 gives d = 100 pc.
Hubble law Given redshift or speed, find distance. (1) Get v from redshift when v is well below c; (2) d = v/H0; (3) align units in Mpc; (4) state the uniform-expansion assumption. Example: v = 7000 km/s with H0 = 70 gives 100 Mpc.
Sanity check (teach this): Reasonableness check: sunlight takes 1.5 x 10^11 divided by 3.0 x 10^8, which is 500 s or 8 min 20 s. Proxima at 4.2 light years means light travelling 4.2 years, the same quantity in another unit. 20 parsecs times 3.09 x 10^16 m is 6.18 x 10^17 m, which divided by 9.46 x 10^15 gives 65.3 light years. One over H0 at 70 km/s per Mpc is about 14 billion years, the same order as the age of the universe; a student who gets 140 years has mixed units.

05 Independent practice (3 questions)

Question Answer What to watch
A star has parallax 0.05 arcseconds. Give its distance in parsecs and light years. 20 parsecs, and 20 times 3.26 is about 65.3 light years. Check that the reciprocal comes first; doing 0.05 times 3.26 is the slip to catch.
Two equal-luminosity stars sit at 10 and 100 parsecs. Give the flux ratio. 1 to 1/100, so the far star is 100 times fainter, a difference of 5 magnitudes. Check for the square; a factor of 10 means the square was dropped.
A star has m = 6 and M = 1. Find its distance. m – M = 5, log(d/10) = 1, so d = 100 parsecs. Check that 5 divided by 5 is then added to 1; answering 10 pc is the common error.

06 Common mistakes and what to say

Mistake Student writes Say this
Treating parallax as proportional to distance Writes that a larger parallax means a greater distance. It is inverse. A parallax of 0.05 is 20 parsecs and 0.02 is 50. Smaller denominator, larger answer.
Confusing luminosity with flux Writes that the star gets dimmer, so its luminosity falls. Luminosity is what the star emits. What changes is flux, because the same energy spreads over a larger sphere.
Dropping the plus one Gets m – M = 5 and writes d = 10 parsecs. Five over five is one, but you still add the original 10, so it is 10 squared. Writing 10 loses a power of ten.
Mixing units in Hubble law Divides km/s by a distance in parsecs. H0 is km/s per megaparsec, so the distance must be in Mpc. Otherwise the answer is wrong by a factor of a million.

07 Homework (paper, section and question numbers)

08 Exit check (last 5 minutes)

Question Answer Decision
What distance corresponds to a parallax of 0.1 arcseconds? 10 parsecs Cannot answer: reteach part 1
A star of equal luminosity moves four times further. What happens to flux? It falls to one sixteenth Cannot answer: reteach part 2
State the distance modulus formula. m – M = 5 log(d/10), with d in parsecs Cannot answer: reteach part 3

09 Tutor FAQ

Which year is this for?

S6 elective astronomy mainly; S5 classes that have finished waves and optics can start it earlier.

Must students memorise the Hubble constant?

The paper supplies it. What they must know is the unit, km/s per megaparsec, because the unit decides the distance unit.

Is there a matching paper?

The homework link goes to the secondary Physics index, which pairs with the radioactive decay chapter on exponentials.

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