Physics › Astrophysics › Quasars and exoplanets
Quasars and exoplanets
The two ends of the observable universe: objects so bright they outshine whole galaxies from billions of parsecs away, and planets so faint we have never really seen most of them. Both are found the same way, by reading starlight for what it refuses to say directly.
Builds on The Doppler effect and Hubble's law and The HR diagram and stellar evolution.
IN THIS TOPIC
- Describe the discovery and nature of quasars, and estimate their distances and power outputs from red shift.
- Explain why exoplanets are hard to detect directly.
- Interpret the radial velocity method and the transit light curve.
WHAT YOU PROBABLY THINK
We find planets around other stars by photographing them beside their suns.
Quasars: small, distant, absurdly bright
In the 1950s radio surveys turned up strong sources that matched nothing obvious in the visible sky. When optical counterparts were finally pinned down they looked like faint blue stars, hence the name quasar, from quasi-stellar radio source. Then their spectra came in, and the puzzle deepened: the lines showed enormous red shifts, far beyond any star's.
Take the red shift at face value and Hubble's law makes quasars the most distant objects we can measure, and that has a savage consequence. To appear even faintly visible from billions of parsecs away, a quasar must outshine an entire galaxy of stars. Yet their brightness flickers over days, and nothing can coordinate a flicker faster than light can cross the source: quasars are roughly solar-system sized. A galaxy's power from a light-day of space.
One engine fits: an active supermassive black hole, millions to billions of solar masses, at the centre of a young galaxy. Matter spiralling in heats by friction to millions of kelvin and blazes across the spectrum before it crosses the horizon. The supermassive black holes of the last lesson are the quiet survivors of that era; a quasar is one still feeding.
WORKED EXAMPLE
Distance and power from a spectrum
A quasar's 656.3 nm hydrogen line arrives at 984.5 nm. Estimate its distance (H = 65 km s−1 Mpc−1), and its absolute magnitude given an apparent magnitude of +17.0.
z = (984.5 − 656.3) / 656.3 = 0.50, so the naive speed is v = zc = 1.5 × 108 m s−1, and d = v/H = 150 000 / 65 ≈ 2300 Mpc. At half of c the small-shift formula is over-stretched, so this is an order-of-magnitude estimate.
M = m − 5 log(d/10 pc) = 17.0 − 5 log(2.3 × 108) = −24.8.
Calibrate that: around 33 times brighter than a whole bright galaxy at M ≈ −21, from a source the size of the solar system. The red shift did all the work: one measured wavelength gave speed, then distance, then power.
Exoplanets: why looking fails
A planet orbiting another star emits almost nothing of its own and reflects only a scrap of its star's light: the star typically outshines it by a factor of a billion, and at interstellar distances the angular separation between them is smaller than any telescope's diffraction limit can cleanly resolve. Photographing an exoplanet is like spotting a moth beside a lighthouse from across a sea, which is why direct images remain rare trophies, managed only for a few giant, young, far-flung planets. Nearly everything we know comes from two indirect methods.
The radial velocity method is the binary-star trick of the last lesson, pushed to extremes. The planet does not orbit a stationary star: both orbit their common centre of mass, so the star performs a miniature mirror-orbit, and its spectral lines swing to and fro with the planet's period. The period of the wobble gives the planet's year; the size of the wobble hints at its mass, since heavier planets drag their stars through faster counter-orbits.
The transit method watches for the tiny scheduled eclipse of a star whose planetary system happens to lie edge-on to us. Each orbit, the planet's disc blocks a fraction of the starlight equal to the ratio of the two discs' areas, so the fractional dip equals (rp/rs)2: the light curve's depth measures the planet's size, and its repeat interval gives the orbital period. A Jupiter crossing a Sun dims it by about one per cent; an Earth, by one part in ten thousand.
YOUR TURN
Sizing a planet from a dip
A star of radius 6.96 × 108 m dims by a fraction 4.0 × 10−4 during each transit. Find the planet's radius, and compare it with the Earth's 6.37 × 106 m, before opening the working.
Show the working
(rp/rs)2 = 4.0 × 10−4, so rp/rs = 0.020 and rp = 0.020 × 6.96 × 108 = 1.4 × 107 m.
That is about 2.2 Earth radii, a super-Earth. The square root is the step examiners watch: the dip compares areas, the answer wants a radius.
TRY IT UNSEEN
Why the wobble is hard
Jupiter makes the Sun orbit their shared centre of mass at about 13 m s−1. Find the fractional Doppler shift this produces, and the wavelength shift on a 550 nm line.
Show the working
Δλ/λ = v/c = 13 / 3.0 × 108 = 4.3 × 10−8.
Δλ = 4.3 × 10−8 × 550 nm = 2.4 × 10−5 nm: a shift a hundred-thousandth of a nanometre wide, which is why exoplanet spectrographs are among the most stable instruments ever built, and why the first planets found this way were heavy ones huddled close to their stars.
THE EXAM BIT
- The quasar story scores in sequence: strong radio source, star-like optical appearance, very large red shift, so very distant by Hubble's law, so galaxy-scale power from a solar-system-sized region, powered by an active supermassive black hole.
- Quasar distance estimates run z, then v = zc, then d = v/H; flag that large z strains the approximation. The examiner wants the honest caveat.
- Direct detection fails for two stated reasons: the star's overwhelming brightness, and an angular separation below the resolving limit. Give both.
- Radial velocity answers must mention the centre of mass: the star wobbles because both bodies orbit it. Period of wobble = period of orbit.
- Transit questions live on the light curve: flat, dip, flat, with depth (rp/rs)2 and the repeat time giving the orbital period. Sketch it with those labels and the marks follow.
CHECK YOURSELF
A quasar's brightness varies noticeably over about two days. Explain what this says about its size, and why that observation, combined with its red shift, forces an extraordinary power source.
Show a hint
Nothing can synchronise a change faster than light can cross the object.
Show the answer
A source cannot brighten as a whole faster than light can cross it, so its diameter is at most about two light-days, roughly 5 × 1013 m: solar-system sized, not galaxy sized.
The large red shift places the quasar billions of parsecs away by Hubble's law, and to be visible at all from there it must radiate more power than a whole galaxy.
Galaxy-scale power from a solar-system-scale volume rules out ordinary starlight; only matter heating violently as it falls toward a supermassive black hole fits both facts at once.
A quasar is an active supermassive black hole: galaxy power from a solar-system volume.
Exoplanets are found indirectly: the star's Doppler wobble, or the transit dip of (rp/rs) squared.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
CHECK YOUR PROGRESS
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- Describe the discovery and nature of quasars, and estimate their distances and power outputs from red shift.
- Explain why exoplanets are hard to detect directly.
- Interpret the radial velocity method and the transit light curve.
Open the full revision checklist to track your progress across the whole unit.
No animated video for this topic yet; these notes stand alone. InkPhysics on YouTube.