Specification: AQA A-level Physics (7408) 3.9.3.1 Astrophysics option | Paper: 3 Section B | Data and formulae booklet: given in the exam
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📌 KEY DEFINITIONS – learn them word for word
Red shift and blue shift: when a source of waves is moving away from (receding from) the observer, the received wavelength is longer and the frequency lower than emitted: a red shift. When the source is moving towards (approaching) the observer, the received wavelength is shorter: a blue shift. The red shift is defined as z = Δf ÷ f = v ÷ c, valid only when v ≪ c.
Spectroscopic binary: two stars orbiting their common centre of mass, viewed in the plane of the orbit. The spectral lines of each star shift periodically between red and blue; the maximum shift gives the orbital speed and the time for one complete cycle of the shift pattern gives the orbital period.
Radial velocity method (exoplanets): an orbiting planet makes its star wobble about their common centre of mass, so the star’s spectral lines show a tiny periodic Doppler shift. The period of the shift equals the planet’s orbital period, and the size of the shift gives the star’s speed. (The other main method is the transit method: a periodic dip in the star’s brightness as the planet passes in front of it.)
The equation
red shift = change in frequency ÷ frequency = speed of source ÷ speed of light z = Δf ÷ f = v ÷ c and Δλ ÷ λ = −v ÷ c (for v ≪ c)
| Quantity | Symbol | Unit |
|---|---|---|
| red shift | z | no unit |
| change in frequency | Δf | hertz (Hz) |
| emitted (laboratory) frequency | f | hertz (Hz) |
| change in wavelength (observed − emitted) | Δλ | metres (m) or nm, the same unit as λ |
| emitted (laboratory) wavelength | λ | metres (m) or nm |
| speed of the source along the line of sight | v | metres per second (m s⁻¹) |
| speed of light, 3.00 × 108 | c | metres per second (m s⁻¹) |
Rearranged: v = cΔλ ÷ λ = zc | Δλ = λv ÷ c | Δf = fv ÷ c | The minus sign shows Δf and Δλ always have opposite signs. In practice, use the size of the shift to find v, then state the direction: a shift to longer wavelength (red shift) means receding, a shift to shorter wavelength (blue shift) means approaching. Δλ and λ only need the same unit (nm is fine), but give v in m s⁻¹. Hydrogen lines: H-alpha 656.3 nm, H-beta 486.1 nm.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) A distant galaxy is receding from the Earth at 6.0 × 106 m s⁻¹. Calculate its red shift z. (c = 3.00 × 108 m s⁻¹)
Show solution
F: z = v ÷ c
I: z = 6.0 × 106 ÷ 3.00 × 108
F: z is already the subject
A: z = 0.020 (no unit)
Q2 (Core) The hydrogen-alpha line has a laboratory wavelength of 656.3 nm. In the spectrum of a galaxy it is observed at 663.5 nm. Calculate (a) the red shift, to 3 significant figures, and (b) the speed of the galaxy, to 3 significant figures, stating whether it is approaching or receding. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) red shift
Δλ = 663.5 − 656.3 = 7.2 nm
F: z = Δλ ÷ λ
I: z = 7.2 ÷ 656.3
A: 0.010970… = 0.0110 (3 s.f.)
Step 2 – (b) speed
F: v = zc
I: v = 0.010970… × 3.00 × 108
A: 3.291… × 106 m s⁻¹ = 3.29 × 106 m s⁻¹ (3 s.f.), receding (the wavelength has increased, a red shift)
Watch out: divide by the emitted wavelength 656.3 nm, not the observed 663.5 nm.
Q3 (Core) Neutral hydrogen gas emits radio waves of frequency 1420.4 MHz. A galaxy is moving away from the Earth at 45 km s⁻¹. Calculate (a) the change in frequency, in kHz, to 3 significant figures, and (b) the frequency received on Earth, in MHz, to 6 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 1420.4 MHz = 1420.4 × 106 Hz; 45 km s⁻¹ = 45 × 103 m s⁻¹
F: Δf ÷ f = v ÷ c
I: Δf ÷ 1420.4 × 106 = 45 × 103 ÷ 3.00 × 108
F: Δf = 1420.4 × 106 × 45 × 103 ÷ 3.00 × 108
A: (a) 213 060 Hz = 213 kHz (3 s.f.)
(b) the galaxy is receding, so the received frequency is lower: 1420.4 − 0.21306 = 1420.18694 MHz = 1420.19 MHz (6 s.f.)
Watch out: for a receding source the frequency decreases even though it is called a red shift. Do not add Δf.
Q4 (Core) A star is moving towards the Earth at 18 km s⁻¹. Calculate the wavelength at which its hydrogen-beta line (laboratory wavelength 486.1 nm) is observed, to 5 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 18 km s⁻¹ = 1.8 × 104 m s⁻¹
F: Δλ ÷ λ = −v ÷ c
I: size of Δλ = 486.1 × 1.8 × 104 ÷ 3.00 × 108
F: size of Δλ = 0.029166 nm; the star is approaching, so this is a blue shift and the wavelength decreases
A: 486.1 − 0.029166 = 486.070834 nm = 486.07 nm (5 s.f.)
Q5 (Exam) A star in a spectroscopic binary is viewed in the plane of its orbit. Its hydrogen-beta line (laboratory wavelength 486.1 nm) varies smoothly between 485.95 nm and 486.25 nm. The longest wavelength is recorded on day 1.2, and the next time it is recorded is day 4.4. Calculate (a) the orbital speed of the star, to 2 significant figures, and (b) the orbital period in days. Assume the binary as a whole is not moving relative to the Earth. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) orbital speed
the line swings equally either side of 486.1 nm, so the maximum shift is Δλ = 486.25 − 486.1 = 0.15 nm
F: v = cΔλ ÷ λ
I: v = 3.00 × 108 × 0.15 ÷ 486.1
A: 9.257… × 104 m s⁻¹ = 9.3 × 104 m s⁻¹ (93 km s⁻¹, 2 s.f.)
Step 2 – (b) period
one complete cycle of the shift pattern runs from one maximum wavelength to the next, so T = 4.4 − 1.2 = 3.2 days
Watch out: using the full range 486.25 − 485.95 = 0.30 nm doubles the speed. The shift is measured from the laboratory (rest) wavelength.
Q6 (Exam) The quasar 3C 273 has a red shift of 0.158. (a) Calculate the wavelength at which its hydrogen-beta line (laboratory wavelength 486.1 nm) is observed, to 4 significant figures. (b) Calculate its speed of recession, to 3 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) observed wavelength
F: z = Δλ ÷ λ
I: 0.158 = Δλ ÷ 486.1
F: Δλ = 0.158 × 486.1 = 76.8038 nm
A: observed λ = 486.1 + 76.8038 = 562.9038 nm = 562.9 nm (4 s.f.), shifted from blue-green into the yellow-green
Step 2 – (b) speed
v = zc = 0.158 × 3.00 × 108 = 4.74 × 107 m s⁻¹ (3 s.f.)
Watch out: v is about 16% of c here, so v ≪ c is only roughly true. The AQA equation is an approximation, but it is the one you must use in the exam.
Q7 (Exam) The star 51 Pegasi wobbles because of an orbiting planet. The maximum speed of the star along the line of sight is 55 m s⁻¹. Calculate the maximum shift in the wavelength of its hydrogen-alpha line (laboratory wavelength 656.3 nm), in metres, to 2 significant figures, and explain why this makes the planet hard to detect. (c = 3.00 × 108 m s⁻¹)
Show solution
Convert first: 656.3 nm = 656.3 × 10−9 m
F: Δλ ÷ λ = −v ÷ c
I: size of Δλ = 656.3 × 10−9 × 55 ÷ 3.00 × 108
F: size of Δλ = λv ÷ c
A: 1.203… × 10−13 m = 1.2 × 10−13 m (2 s.f.), shifting alternately to the red and the blue over each orbit
This is about 2 parts in 107 of the wavelength, so a very high resolution, very stable spectrometer is needed to measure it.
Q8 (Exam) A binary system of two stars is viewed in the plane of the orbit. At one instant the hydrogen-alpha line (laboratory wavelength 656.3 nm) appears split into two lines at 656.18 nm and 656.62 nm. Calculate (a) the velocity of the binary system as a whole relative to the Earth, to 2 significant figures, and (b) the orbital speed of each star at this instant, to 2 significant figures. (c = 3.00 × 108 m s⁻¹)
Show solution
Step 1 – (a) velocity of the whole system
the centre of mass is at the mid-point of the two lines: (656.18 + 656.62) ÷ 2 = 656.40 nm
Δλ = 656.40 − 656.3 = +0.10 nm (a red shift)
F: v = cΔλ ÷ λ
I: v = 3.00 × 108 × 0.10 ÷ 656.3
A: 4.571… × 104 m s⁻¹ = 4.6 × 104 m s⁻¹ away from the Earth (receding, 2 s.f.)
Step 2 – (b) orbital speeds
each line is 656.62 − 656.40 = 0.22 nm from the mid-point, one red shifted and one blue shifted
v = 3.00 × 108 × 0.22 ÷ 656.3 = 1.005… × 105 m s⁻¹
A: 1.0 × 105 m s⁻¹ each (2 s.f.), one moving towards us and one away; equal speeds mean equal masses
Watch out: measuring the orbital shifts from 656.3 nm instead of from the mid-point mixes the motion of the whole system into the orbital speed.
Q9 (Stretch) Two identical stars orbit their common centre of mass in circles of the same radius, viewed in the plane of the orbit. The hydrogen-alpha line (656.3 nm) of one star shifts to a maximum of 0.18 nm either side of 656.3 nm. Its wavelength is a maximum at t = 0, returns to 656.3 nm at t = 1.25 days and is a minimum at t = 2.5 days. Calculate (a) the orbital speed, (b) the radius of each star’s orbit, to 2 significant figures, and (c) the mass of each star, to 2 significant figures. (c = 3.00 × 108 m s⁻¹, G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Step 1 – (a) orbital speed
v = cΔλ ÷ λ = 3.00 × 108 × 0.18 ÷ 656.3 = 8.227… × 104 m s⁻¹ (82 km s⁻¹)
Step 2 – period from the shift pattern
maximum to minimum is half a cycle, so T = 2 × 2.5 = 5.0 days = 5.0 × 86 400 = 432 000 s
Step 3 – (b) radius of orbit
v = 2πr ÷ T, so r = vT ÷ 2π = 8.227… × 104 × 432 000 ÷ 2π
A: 5.657… × 109 m = 5.7 × 109 m (2 s.f.)
Step 4 – (c) mass
the stars are 2r apart, so gravity provides the centripetal force: Gm² ÷ (2r)² = mv² ÷ r, giving m = 4rv² ÷ G
m = 4 × 5.657… × 109 × (8.227… × 104)² ÷ 6.67 × 10−11 = 2.296… × 1030 kg
A: 2.3 × 1030 kg (2 s.f.), about 1.2 solar masses
Watch out: the separation of the stars is 2r, not r. Using r in Newton’s law of gravitation gives a mass four times too small.
Q10 (Stretch) A Sun-like star of mass 1.99 × 1030 kg has a planet detected by the radial velocity method. The hydrogen-alpha line (656.3 nm) of the star shifts by a maximum of 1.23 × 10−13 m either side of its mean, with a period of 4.23 days. Assume the orbits are circular, viewed edge-on, and the planet’s mass is much smaller than the star’s. Calculate (a) the star’s orbital speed, (b) the radius of the planet’s orbit, (c) the planet’s orbital speed and (d) the mass of the planet, each to 3 significant figures. (c = 3.00 × 108 m s⁻¹, G = 6.67 × 10−11 N m² kg⁻²)
Show solution
Step 1 – (a) star’s speed
vs = cΔλ ÷ λ = 3.00 × 108 × 1.23 × 10−13 ÷ 656.3 × 10−9 = 56.22… m s⁻¹ = 56.2 m s⁻¹ (3 s.f.)
Step 2 – (b) radius of planet’s orbit
T = 4.23 × 86 400 = 365 472 s; GMm ÷ r² = m(2πr ÷ T)² ÷ r gives r³ = GMT² ÷ 4π²
r³ = 6.67 × 10−11 × 1.99 × 1030 × 365 472² ÷ 4π² = 4.490… × 1029 m³
A: r = 7.657… × 109 m = 7.66 × 109 m (3 s.f.), only 0.051 AU from the star
Step 3 – (c) planet’s speed
vp = 2πr ÷ T = 2π × 7.657… × 109 ÷ 365 472 = 1.316… × 105 m s⁻¹ = 1.32 × 105 m s⁻¹ (3 s.f.)
Step 4 – (d) planet’s mass
the star and planet have equal and opposite momenta about the centre of mass: Mvs = mvp
m = 1.99 × 1030 × 56.22… ÷ 1.316… × 105 = 8.498… × 1026 kg
A: 8.50 × 1026 kg (3 s.f.), roughly half the mass of Jupiter: a “hot Jupiter”
Watch out: if the orbit is not edge-on, the Doppler shift only measures the line-of-sight part of the star’s velocity, so this is a minimum mass.
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