High School Physics · Grades 10–12

23The Doppler Effect

An ambulance closes in, siren blazing, and sweeps past — and at that instant the note slides audibly downward, though the driver touched nothing. The siren is honest; the wave is not: motion squeezes it ahead and stretches it behind. This chapter measures the squeeze, then rides the same idea from sirens to radar guns, to blood in an artery, and out to a planet betrayed by a wobble in starlight.

23.1 The effect

Definition 23.1 (Doppler effect)

The Doppler effect is the change of the received frequency of a wave when source and receiver move relative to each other: the received frequency $f'$ is higher than the emitted frequency $f$ while they approach, lower while they recede, and equal to $f$ only while their distance momentarily stops changing.

Remark 23.2 (Pitch, not loudness)

Two things change as the ambulance passes. The loudness swells and fades because the wave spreads with distance (Chapter 21); the pitch shifts because of the speed of approach or recession, and drops abruptly at the passing. Loudness says “how far”; pitch says “how fast”.

23.2 A moving source: crowded wavefronts

The key picture: once a crest leaves the source, the medium owns it. Each crest expands as a circle at the wave speed $v$ (Chapter 20), centered on the point where it was emitted — the medium neither knows nor cares that the source has moved on.

Proposition 23.3 (Moving source)

A source emits crests with period $T$ (frequency $f = 1/T$ ) while moving in a straight line at speed $v_s < v$ through a medium in which the wave travels at speed $v$ . An observer at rest in the medium receives

\lambda' = (v \mp v_s)\,T, \qquad f' = f\,\frac{v}{v \mp v_s},

the upper signs for a source approaching, the lower for one receding.

Proof. Between two crests the first advances $vT$ while the source advances $v_s T$ . Ahead, the second crest is therefore emitted $v_s T$ closer to the first, $\lambda' = (v - v_s)T$ ; behind, the same $v_s T$ is added. The crests then sweep past a resting observer at the medium’s speed $v$ , one every $\lambda'/v$ seconds, so $f' = v/\lambda' = f v/(v \mp v_s)$ . ∎

Wavefronts of a source moving right: each circle is centered on the point where it was emitted (gray dots), so the crests bunch ahead, ' = (v - v_s)T, and stretch behind, ' = (v + v_s)T. — Wavefronts of a source moving right: each circle is centered on the point where it was emitted (gray dots), so the crests bunch ahead, $\lambda' = (v - v_s)T$ , and stretch behind, $\lambda' = (v + v_s)T$ .

Example 23.4 (The ambulance, in numbers)

A siren at $f = 700\,\mathrm{Hz}$ approaches at $v_s = 25\,\mathrm{m}/\mathrm{s}$ ; take $v = 340\,\mathrm{m}/\mathrm{s}$ in air (Chapter 21). Approaching: $f' = 700 \times 340/315 \approx 756\,\mathrm{Hz}$ ; receding: $f' = 700 \times 340/365 \approx 652\,\mathrm{Hz}$ . The pass drops the pitch by $104\,\mathrm{Hz}$ — a ratio $\approx 1.16$ , about two and a half semitones, unmistakable to any ear.

Remark 23.5 (What motion does not change)

The wave still travels at $v$ : the medium alone sets the speed (Chapter 20). Motion of the source squeezes wavelengths, never speeds. And the squeeze has a limit: as $v_s \to v$ the crests ahead pile onto each other and $f' \to \infty$ — the traffic jam of wavefronts a supersonic aircraft drags as a shock.

23.3 A moving observer, and the slow-motion shortcut

Proposition 23.6 (Moving observer)

An observer moves at speed $v_o$ straight toward (or away from) a source at rest in the medium. The received frequency is

f' = f\left(1 \pm \frac{v_o}{v}\right) \quad (+\ \text{approaching},\ -\ \text{receding}).

Proof. The wave pattern itself is undisturbed: crests of spacing $\lambda = vT$ travel at $v$ . An observer running at them meets crests at the relative speed $v + v_o$ , one every $\lambda/(v + v_o)$ seconds: $f' = (v + v_o)/\lambda = f(1 + v_o/v)$ . Running away, $v - v_o$ . ∎

Definition 23.7 (Radial velocity)

The radial velocity $v_r$ of a source relative to an observer is the component of their relative velocity along the line joining them — the rate at which the distance shrinks or grows. Only this component shifts the frequency; motion across the line of sight gives no shift at this order.

Proposition 23.8 (Slow motion: one formula for everybody)

When the radial speed is small, $v_r \ll v$ , it no longer matters who moves:

\frac{\Delta f}{f} = \frac{f' - f}{f} \approx +\frac{v_r}{v} \ \text{(approach)}, \qquad \frac{\Delta f}{f} \approx -\frac{v_r}{v} \ \text{(recession)}.

Proof. The observer formulas are already exactly $1 \pm v_o/v$ . For the source, write $x = v_s/v$ : the identity $\frac{1}{1 - x} = 1 + x + \frac{x^2}{1 - x}$ shows that $f'/f$ differs from $1 + x$ by a term of order $x^2$ — for $x = 0.1$ , a $1\%$ correction on a $10\%$ shift. To first order all four formulas collapse onto $1 \pm v_r/v$ . ∎

Frequency heard as a 440\, Hz siren passes at 15\, m/ s, 8\, m from the microphone: plateaus at fv/(v v_s), and the true f (dotted) crossed essentially at closest approach, where the motion is purely transverse. — Frequency heard as a $440\,\mathrm{Hz}$ siren passes at $15\,\mathrm{m}/\mathrm{s}$ , $8\,\mathrm{m}$ from the microphone: plateaus at $fv/(v \mp v_s)$ , and the true $f$ (dotted) crossed essentially at closest approach, where the motion is purely transverse.

Method 23.9 (Reading a pass)

A recording of a passing source shows two plateaus: $f_1$ (approach) and $f_2$ (recession). Then

f = \frac{2 f_1 f_2}{f_1 + f_2}, \qquad v_s = v\,\frac{f_1 - f_2}{f_1 + f_2}.

Indeed $f_1 + f_2 = \frac{2fv^2}{v^2 - v_s^2}$ and $f_1 - f_2 = \frac{2fv\,v_s}{v^2 - v_s^2}$ : divide for $v_s$ , and compute $2f_1 f_2/(f_1 + f_2)$ for $f$ . Two frequency readings yield both the true pitch and the speed — no stopwatch, no ruler.

23.4 Echoes that measure speed

Bounce a wave off a moving target and the Doppler effect strikes twice.

Proposition 23.10 (Reflection doubles the shift)

A wave of frequency $f$ is sent at a reflector approaching head-on at speed $u \ll v$ . The echo returns with

f' = f\,\frac{v + u}{v - u}, \qquad \text{hence} \qquad \Delta f \approx \frac{2u}{v}\,f .

Proof. Two shifts in series. As a moving observer the reflector receives $f(1 + u/v)$ ; re-emitting what it receives, it is now a moving source, so the echo arrives at $f' = f(1 + u/v)/(1 - u/v) \approx f(1 + 2u/v)$ for $u \ll v$ . ∎

Example 23.11 (The radar gun)

A traffic radar emits microwaves at $f = 24.15\,\mathrm{GHz}$ ( $v = c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ ) and superposes the echo on the outgoing wave; the beat (Chapter 21) directly sounds out $\Delta f = 2uf/c \approx 161\,\mathrm{Hz}$ per $\mathrm{m}/\mathrm{s}$ of car speed — a trivially measurable audio frequency carved out of a $24\,\mathrm{GHz}$ carrier. The factor $2$ is not optional: forgetting it flatters every driver by half.

Example 23.12 (Doppler ultrasound)

A medical probe sends $f = 5.0\,\mathrm{MHz}$ ultrasound into an artery ( $v \approx 1540\,\mathrm{m}/\mathrm{s}$ in tissue); red blood cells reflect it. Blood at $u = 0.50\,\mathrm{m}/\mathrm{s}$ returns $\Delta f = 2 \times 0.50 \times 5.0 \times 10^{6}/1540 \approx 3.2\,\mathrm{kHz}$ — an audible beat: the cardiologist literally hears the blood accelerate at each heartbeat, then measures $u$ from $\Delta f$ .

23.5 Doppler in starlight

Light is a wave (Chapter 22), its spectrum striped with sharp lines at wavelengths fixed by each chemical element (Chapter 2) — a laboratory-calibrated ruler printed on every star.

Definition 23.13 (Redshift and blueshift)

When the spectral lines of a source appear at wavelengths longer than in the laboratory, the spectrum is redshifted; when they appear shorter, it is blueshifted — shifted toward the red or the blue end of the visible spectrum.

Proposition 23.14 (Doppler shift of light)

For a source receding at radial speed $v_r \ll c$ , every wavelength is observed stretched by

\frac{\Delta \lambda}{\lambda} = \frac{\lambda' - \lambda}{\lambda} \approx \frac{v_r}{c} \quad \text{(redshift)},

and compressed by the same amount, $\Delta\lambda/\lambda \approx -v_r/c$ , for an approaching source (blueshift).

Proof. Admitted at this level. ∎

Remark 23.15 (Where the honest formula lives)

Light needs no medium, so the sound derivation cannot be copied — there is no “at rest in the medium”. The exact formula comes from special relativity, later this year (Chapter 35); the corrections are of order $(v_r/c)^2$ , negligible for planes, stars and nearby galaxies. Below we use the slow-motion formula with a clear conscience.

$The same hydrogen line in three spectra: at 656.30\, nm in the laboratory, blueshifted for an approaching star, redshifted for a receding one. The shift, read against the dashed reference, gives the radial velocity.$

The same hydrogen line in three spectra: at

656.30\,\mathrm{nm}

in the laboratory, blueshifted for an approaching star, redshifted for a receding one. The shift, read against the dashed reference, gives the radial velocity.

Example 23.16 (Weighing a star’s escape)

In the spectrum of the bright star Vega, the hydrogen line of laboratory wavelength $656.30\,\mathrm{nm}$ is measured at $656.27\,\mathrm{nm}$ : $\Delta\lambda = -0.03\,\mathrm{nm}$ , a blueshift, so Vega approaches at $v_r = c\,|\Delta\lambda|/\lambda \approx 1.4 \times 10^{4}\,\mathrm{m}/\mathrm{s}$ — $14\,\mathrm{km}/\mathrm{s}$ , read off a shift of one part in twenty thousand.

Example 23.17 (Wobbling stars: binaries and exoplanets)

When two stars orbit each other, their lines swing periodically blue and red — a spectroscopic binary, orbits measured without ever resolving the pair. A planet does the same to its star, faintly: both circle their common center of mass, so the star’s radial velocity oscillates by a few tens of $\mathrm{m}/\mathrm{s}$ — a shift $\Delta\lambda/\lambda \sim 10^{-7}$ . Detecting it is how the first planet around a Sun-like star was found; the weekend problem reruns that discovery, numbers and all.

Remark 23.18 (Galaxy redshifts)

The spectra of distant galaxies are all redshifted, and the more distant, the more shifted: the universe’s distances are stretching. For nearby galaxies $v_r = c\,\Delta\lambda/\lambda$ reads off the recession speed; for the farthest, shifts outgrow what the slow-motion formula can honestly handle, and the accounting is taken up again from Chapter 35 onward, and in the university volumes.

23.6 Exercises

Exercise 23.1 ★

A fire-engine siren emits at $700\,\mathrm{Hz}$ and drives at $25\,\mathrm{m}/\mathrm{s}$ ( $v = 340\,\mathrm{m}/\mathrm{s}$ ). Compute the frequency heard by a pedestrian (a) ahead of it; (b) behind it. (c) What does the driver hear?

Solution

Solution of Exercise 23.1.

(a) $f' = 700 \times 340/315 \approx 756\,\mathrm{Hz}$ . (b) $f' = 700 \times 340/365 \approx 652\,\mathrm{Hz}$ . (c) $700\,\mathrm{Hz}$ : driver and siren move together, their distance never changes.

Exercise 23.2 ★

A friend claims: “the pitch drops as the ambulance leaves because the sound gets weaker with distance.” Untangle the two effects being confused, and say what each depends on.

Solution

Solution of Exercise 23.2.

The loudness fades because the wave spreads out — it depends on the distance. The pitch drops from approach to recession — the Doppler effect, which depends on how fast the distance changes. Two independent effects.

Exercise 23.3 ★

A cyclist rides at $6.0\,\mathrm{m}/\mathrm{s}$ straight toward a stationary siren emitting at $700\,\mathrm{Hz}$ . What frequency do they hear? And riding straight away? Which formula applies, and why not the moving-source one?

Solution

Solution of Exercise 23.3.

Moving observer: $f' = 700\,(1 + 6.0/340) \approx 712\,\mathrm{Hz}$ ; riding away, $700\,(1 - 6.0/340) \approx 688\,\mathrm{Hz}$ . The siren is at rest in the air, so the wave pattern is undisturbed; only the observer runs through it.

Exercise 23.4 ★

A train horn emits at $400\,\mathrm{Hz}$ while the train runs at $30\,\mathrm{m}/\mathrm{s}$ . Compute the wavelength ahead of the train, behind it, and at rest. Which observer receives which?

Solution

Solution of Exercise 23.4.

Ahead: $\lambda' = 310/400 = 0.775\,\mathrm{m}$ ; behind: $370/400 = 0.925\,\mathrm{m}$ ; at rest: $340/400 = 0.85\,\mathrm{m}$ . The short wavelength reaches whoever the train approaches, the long one whoever it leaves.

Exercise 23.5 ★

A phone playing a $440\,\mathrm{Hz}$ tone is carried by a runner at $5.0\,\mathrm{m}/\mathrm{s}$ . Use the slow-motion formula to estimate the shift heard by someone the runner approaches. Compare it to a semitone (about $6\%$ in frequency): would a musician notice?

Solution

Solution of Exercise 23.5.

$\Delta f/f \approx 5.0/340 \approx 1.5\%$ : $\Delta f \approx 6.5\,\mathrm{Hz}$ , so about $446\,\mathrm{Hz}$ — a quarter of a semitone. A musician would just notice.

Exercise 23.6 ★★

A bat flies at $6.0\,\mathrm{m}/\mathrm{s}$ straight at a wall, emitting at $50.0\,\mathrm{kHz}$ . Show that the echo it hears returns at $f' = f(v + v_b)/(v - v_b)$ , then compute $f'$ and the shift. Why is this the radar-gun formula in disguise?

Solution

Solution of Exercise 23.6.

The wall receives $f\,v/(v - v_b)$ (moving source); the bat, a moving observer closing on the echo, hears $(1 + v_b/v)$ times that: $f' = f(v + v_b)/(v - v_b) = 50.0 \times 346/334 \approx 51.8\,\mathrm{kHz}$ , a shift of $+1.8\,\mathrm{kHz}$ . It is the radar-gun double shift with emitter and reflector roles swapped: only the closing speed matters.

Exercise 23.7 ★★

A radar gun at $24.15\,\mathrm{GHz}$ measures a beat of $4.0\,\mathrm{kHz}$ from an approaching car. Find the car’s speed in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ . What beat would a car at exactly $50\,\mathrm{km}/\mathrm{h}$ give?

Solution

Solution of Exercise 23.7.

$u = c\,\Delta f/(2f) = 3.00 \times 10^{8} \times 4000/(2 \times 24.15 \times 10^{9}) \approx 24.8\,\mathrm{m}/\mathrm{s} \approx 89\,\mathrm{km}/\mathrm{h}$ . At $50\,\mathrm{km}/\mathrm{h}$ ( $13.9\,\mathrm{m}/\mathrm{s}$ ): $\Delta f = 2 \times 13.9 \times 24.15 \times 10^{9}/3.00 \times 10^{8} \approx 2.2\,\mathrm{kHz}$ .

Exercise 23.8 ★★

A $5.0\,\mathrm{MHz}$ Doppler ultrasound probe aimed along an artery ( $v = 1540\,\mathrm{m}/\mathrm{s}$ in tissue) hears a beat of $2.6\,\mathrm{kHz}$ . Find the blood speed. Why must the probe be angled along the flow rather than perpendicular to it?

Solution

Solution of Exercise 23.8.

$u = v\,\Delta f/(2f) = 1540 \times 2600/1.0 \times 10^{7} \approx 0.40\,\mathrm{m}/\mathrm{s}$ . Perpendicular flow has zero radial velocity, hence no shift at all.

Exercise 23.9 ★★

Take $f = 1000\,\mathrm{Hz}$ , $v = 340\,\mathrm{m}/\mathrm{s}$ and a closing speed of $34\,\mathrm{m}/\mathrm{s}$ . Compute $f'$ exactly when (a) the source moves, (b) the observer moves. Compare both to the slow-motion prediction $f(1 + v_r/v)$ and explain the size of the disagreement.

Solution

Solution of Exercise 23.9.

(a) Source: $f' = 1000 \times 340/306 \approx 1111\,\mathrm{Hz}$ . (b) Observer: $f' = 1000 \times 1.1 = 1100\,\mathrm{Hz}$ . Slow motion predicts $1100\,\mathrm{Hz}$ : exact for the observer, $11\,\mathrm{Hz}$ short for the source — the order- $x^2$ correction, $1\%$ of $f$ for $x = 0.1$ .

Exercise 23.10 ★★

The recording of a scooter passing a microphone shows plateaus at $465\,\mathrm{Hz}$ and $415\,\mathrm{Hz}$ . Using Method 23.9, find the horn’s true frequency and the scooter’s speed in $\mathrm{km}/\mathrm{h}$ . Why is the true frequency not their average?

Solution

Solution of Exercise 23.10.

$f = 2 \times 465 \times 415/880 \approx 439\,\mathrm{Hz}$ ; $v_s = 340 \times 50/880 \approx 19.3\,\mathrm{m}/\mathrm{s} \approx 70\,\mathrm{km}/\mathrm{h}$ . The true $f$ is the harmonic mean: approach raises the frequency more than recession lowers it, so $f$ sits below the arithmetic average ( $440\,\mathrm{Hz}$ ).

Exercise 23.11 ★★

In a star’s spectrum, the hydrogen line of laboratory wavelength $656.3\,\mathrm{nm}$ is observed at $656.5\,\mathrm{nm}$ . Redshift or blueshift? Compute the star’s radial velocity and direction.

Solution

Solution of Exercise 23.11.

Longer wavelength: redshift, the star recedes. $v_r = c\,\Delta\lambda/\lambda = 3.00 \times 10^{8} \times 0.2/656.3 \approx 9.1 \times 10^{4}\,\mathrm{m}/\mathrm{s} \approx 91\,\mathrm{km}/\mathrm{s}$ .

Exercise 23.12 ★★★

A star wobbles at $55\,\mathrm{m}/\mathrm{s}$ because of an unseen planet. Compute the amplitude of the wavelength swing of a $500\,\mathrm{nm}$ line, as a length and as a fraction of $\lambda$ . Order of magnitude: why did planet hunting have to wait for spectrographs stable to one part in $10^{7}$ ?

Solution

Solution of Exercise 23.12.

$\Delta\lambda = \lambda K/c = 500 \times 55/3.00 \times 10^{8} \approx 9.2 \times 10^{-5}\,\mathrm{nm}$ ; $\Delta\lambda/\lambda = 55/3.00 \times 10^{8} \approx 1.8 \times 10^{-7}$ . The whole signal is two parts in ten million: a spectrograph drifting by more than $10^{-7}$ buries it.

Exercise 23.13 ★★★

Every line in a galaxy’s spectrum is found stretched by $2.4\%$ . Compute its recession speed. Astronomers find that such speeds grow in proportion to distance, for every direction of the sky: what picture does this suggest? Is the slow-motion formula still trustworthy here?

Solution

Solution of Exercise 23.13.

$v_r = 0.024\,c \approx 7.2 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ . Speeds proportional to distance, the same in every direction: all distances stretching uniformly — an expanding universe with no privileged center. At $2.4\%$ the corrections are of order $(v_r/c)^2 \approx 6 \times 10^{-4}$ : still trustworthy.

Exercise 23.14 ★★★

The Sun rotates: one edge of its disk approaches us at about $2.0\,\mathrm{km}/\mathrm{s}$ while the other recedes equally fast. What does this do to the $656.3\,\mathrm{nm}$ line in light from the whole disk — shift it or broaden it? Compute the effect in $\mathrm{nm}$ .

Solution

Solution of Exercise 23.14.

The two edges shift opposite ways, so the disk-averaged line is broadened, not shifted: full width $2\lambda v/c = 2 \times 656.3 \times 2.0 \times 10^{3}/3.00 \times 10^{8} \approx 8.8 \times 10^{-3}\,\mathrm{nm}$ (about $\pm4.4 \times 10^{-3}\,\mathrm{nm}$ ).

Exercise 23.15 ★★★

A police car at $40\,\mathrm{m}/\mathrm{s}$ , siren at $700\,\mathrm{Hz}$ , chases a truck driving at $30\,\mathrm{m}/\mathrm{s}$ in the same direction. (a) Justify $f' = f(v - v_o)/(v - v_s)$ for the truck driver. (b) Compute $f'$ . (c) Show that if the truck matched the car’s speed the shift would vanish, and say why that is as it should be.

Solution

Solution of Exercise 23.15.

(a) Ahead of the car the crests are spaced $(v - v_s)T$ ; the truck flees at $v_o$ and meets them at relative speed $v - v_o$ : $f' = (v - v_o)/\bigl((v - v_s)T\bigr) = f(v - v_o)/(v - v_s)$ . (b) $f' = 700 \times 310/300 \approx 723\,\mathrm{Hz}$ . (c) $v_o = v_s$ gives $f' = f$ : the separation is constant, and no change of distance means no Doppler shift, by definition.

23.7 Problem: The Hunt for an Exoplanet

Problem 23.1

Weekend problem — the hunt for an exoplanet: a train horn calibrates the ear, a radar gun rehearses the double shift, and by Sunday night a star’s spectrum, wobbling by one part in five million, has weighed a planet nobody has ever seen

An astronomy club spends a weekend on one idea — the Doppler shift — rehearsed on Earth, then aimed at a Sun-like star. Data: $v = 340\,\mathrm{m}/\mathrm{s}$ for sound in air, $c = 3.00 \times 10^{8}\,\mathrm{m}/\mathrm{s}$ , $G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}$ ; the star has mass $M = 2.0 \times 10^{30}\,\mathrm{kg}$ . Admitted, from the mechanics of circular orbits (Chapter 27): a planet of mass $m$ on a circular orbit of radius $r$ has period and speed given by $r^3 = \frac{G M P^2}{4\pi^2}$ and $V = \frac{2\pi r}{P}$ , and the star, circling the common center of mass, wobbles at speed $K = \frac{m}{M}\,V$ .

Part I — Friday: the level crossing. A train passes at constant speed, horn on; a phone records $f_1 = 471\,\mathrm{Hz}$ while it approaches, $f_2 = 411\,\mathrm{Hz}$ after it passes.

Explain, with the crowded-wavefront picture, why $f_1 > f_2$ although the horn never changes.
Write the two equations linking $f_1$ , $f_2$ to the true frequency $f$ and train speed $u$ .
Show that $u = v\,\dfrac{f_1 - f_2}{f_1 + f_2}$ and compute it in $\mathrm{m}/\mathrm{s}$ and $\mathrm{km}/\mathrm{h}$ .
Show that $f = \dfrac{2 f_1 f_2}{f_1 + f_2}$ and compute it. Why is $f$ slightly below the average of $f_1$ and $f_2$ ?
Compute the wavelengths ahead of and behind the train, and the at-rest wavelength.

Part II — Saturday: the radar gun. A traffic patrol demonstrates its radar: $f = 24.125\,\mathrm{GHz}$ , and the gun reads the beat between the outgoing wave and the echo.

Explain why the echo of a moving car is shifted twice — name the role the car plays in each shift.
Show that for a car at speed $u \ll c$ the beat is $\Delta f \approx \dfrac{2u}{c}\,f$ .
A car returns $\Delta f = 3.55\,\mathrm{kHz}$ : compute $u$ .
Convert to $\mathrm{km}/\mathrm{h}$ and compare to the $80\,\mathrm{km}/\mathrm{h}$ limit posted there.
How many hertz of beat correspond to $1\,\mathrm{km}/\mathrm{h}$ ? Comment: what makes such a tiny relative shift of a $24\,\mathrm{GHz}$ wave easy to measure?

Part III — Sunday: the star that wobbles. The club downloads the measured radial velocity of a Sun-like star, night after night, plotted below.

The star’s radial velocity over twelve nights: a clean sinusoid — the signature of a body on a circular orbit, seen edge-on.

What is actually measured, night after night, in the star’s spectrum? State the relation used to turn it into $v_r$ .
Read the amplitude $K$ and the period $P$ of the wobble off the curve.
Compute the corresponding swing $\Delta\lambda$ of a $550\,\mathrm{nm}$ line, and $\Delta\lambda/\lambda$ . Compare with Part I: how much harder is this measurement than the train’s?
Why does an unseen planet make the star wobble at all? What does the sinusoidal shape indicate about the orbit?
Why does this method measure only the radial part of the star’s motion, and what would be seen if the orbit were face-on to us?

Part IV — Sunday night: weighing the invisible. Take the orbit circular and edge-on, so $K$ is the star’s full orbital speed.

From $P$ and the admitted relation $r^3 = G M P^2/(4\pi^2)$ , compute the planet’s orbital radius $r$ . Compare it to the Earth–Sun distance, $1.5 \times 10^{11}\,\mathrm{m}$ .
Compute the planet’s orbital speed $V = 2\pi r/P$ .
From $K = (m/M)\,V$ , compute the planet’s mass $m$ .
Compare $m$ to Jupiter ( $1.9 \times 10^{27}\,\mathrm{kg}$ ) and to Earth ( $6.0 \times 10^{24}\,\mathrm{kg}$ ), and $r$ to the Sun–Mercury distance ( $5.8 \times 10^{10}\,\mathrm{m}$ ). What sort of world is this?
If the orbit were tilted rather than edge-on, would the true mass be larger or smaller than your value? Conclude in one sentence: what did a periodic shift of one part in five million just deliver?

Solution

Solution of Problem 23.1.

1. Each crest expands from the point where it was emitted: crests bunch ahead of the moving horn and stretch behind it, so the approaching phone meets them faster ( $f_1$ ) than the receding one ( $f_2$ ).

2. $f_1 = f\,\dfrac{v}{v - u}$ , $f_2 = f\,\dfrac{v}{v + u}$ .

3. $f_1 - f_2 = \frac{2fv\,u}{v^2 - u^2}$ and $f_1 + f_2 = \frac{2fv^2}{v^2 - u^2}$ ; dividing, $u = v\,\frac{f_1 - f_2}{f_1 + f_2} = 340 \times 60/882 \approx 23\,\mathrm{m}/\mathrm{s} \approx 83\,\mathrm{km}/\mathrm{h}$ .

4. $\dfrac{2f_1 f_2}{f_1 + f_2} = f$ (the common factor cancels): $f = 2 \times 471 \times 411/882 \approx 439\,\mathrm{Hz}$ — the harmonic mean, below the average $441\,\mathrm{Hz}$ because approach boosts $f$ more than recession cuts it.

5. Ahead: $(v - u)/f = 316.9/439 \approx 0.72\,\mathrm{m}$ ; behind: $363.1/439 \approx 0.83\,\mathrm{m}$ ; at rest: $340/439 \approx 0.77\,\mathrm{m}$ .

6. Once as a moving observer receiving the wave, once as a moving source re-emitting what it received.

7. $f' = f\,(1 + u/c)/(1 - u/c) \approx f(1 + 2u/c)$ , so $\Delta f = f' - f \approx \dfrac{2u}{c}\,f$ .

8. $u = c\,\Delta f/(2f) = 3.00 \times 10^{8} \times 3550/(2 \times 24.125 \times 10^{9}) \approx 22\,\mathrm{m}/\mathrm{s}$ .

9. $22.1 \times 3.6 \approx 79\,\mathrm{km}/\mathrm{h}$ : just under the $80\,\mathrm{km}/\mathrm{h}$ limit.

10. $\Delta f = 2f/(3.6\,c) \approx 45\,\mathrm{Hz}$ per $\mathrm{km}/\mathrm{h}$ . Beating the echo against the emitted wave leaves only the difference: an audio frequency, trivially counted, though it is a $10^{-7}$ fraction of the carrier.

11. The wavelengths of known spectral lines, night after night; then $v_r = c\,\Delta\lambda/\lambda$ .

12. $K = 55\,\mathrm{m}/\mathrm{s}$ , $P = 4.2\,\mathrm{days}$ .

13. $\Delta\lambda = 550 \times 55/3.00 \times 10^{8} \approx 1.0 \times 10^{-4}\,\mathrm{nm}$ , $\Delta\lambda/\lambda \approx 1.8 \times 10^{-7}$ — one part in five million, some $4 \times 10^{5}$ times smaller than the train’s $7\%$ shift.

14. Star and planet attract each other equally, so both orbit their common center of mass: the star cannot stand still. A sinusoidal $v_r$ is uniform circular motion seen edge-on — a circular orbit.

15. Only the line-of-sight component changes the star–Earth distance, hence the wavelength. Face-on, $v_r = 0$ at all times: no signal whatsoever.

16. $P = 3.63 \times 10^{5}\,\mathrm{s}$ : $r^3 = 6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times (3.63 \times 10^{5})^2/(4\pi^2) \approx 4.4 \times 10^{29}$ , so $r \approx 7.6 \times 10^{9}\,\mathrm{m}$ — about $1/20$ of the Earth–Sun distance.

17. $V = 2\pi r/P = 2\pi \times 7.6 \times 10^{9}/3.63 \times 10^{5} \approx 1.3 \times 10^{5}\,\mathrm{m}/\mathrm{s}$ .

18. $m = M K/V = 2.0 \times 10^{30} \times 55/1.3 \times 10^{5} \approx 8.3 \times 10^{26}\,\mathrm{kg}$ .

19. About $0.4$ Jupiter masses ( $\approx 140$ Earths), orbiting $8$ times closer than Mercury: a gas giant roasting against its star — a “hot Jupiter”.

20. A tilt hides part of the motion: the measured $K$ is only the radial share, so the true mass is larger — $8.3 \times 10^{26}\,\mathrm{kg}$ is a minimum. A periodic shift of one part in five million just delivered the orbit, speed and minimum mass of a planet nobody has ever seen.