Physics › Astrophysics › Detecting exoplanets
Detecting exoplanets
Nearly all known exoplanets were found indirectly. This lesson explains why direct imaging rarely works and how the two main indirect methods do. The radial velocity method measures the star's periodic Doppler shift as it moves about the common centre of mass. The transit method measures a small dip in brightness that repeats once per orbit.
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Quasars and exoplanets, part 2 of 2. Part 1 is Quasars.
Builds on Quasars and The Doppler effect and red shift and Orbits and satellites.
IN THIS TOPIC
- Explain why direct exoplanet detection fails, and interpret the radial velocity method and the transit light curve.
COMMON MISCONCEPTION
We find planets around other stars by photographing them beside their suns.
Direct photographs are rare because the star vastly outshines the planet: most exoplanets are found indirectly, from the star's periodic Doppler shift or the small dip in brightness during a transit.
Exoplanets: why looking fails
A planet orbiting another star emits almost nothing of its own and reflects only a tiny fraction of its star's light. This causes two problems, and an exam answer needs both. The star typically outshines the planet by a factor of a billion. At interstellar distances the angle between them is also below what any telescope's diffraction limit can cleanly resolve. Direct images are therefore rare, managed only for a few giant, young, widely separated planets, and nearly everything we know comes from two indirect methods.
The radial velocity method is the binary-star method of The Doppler effect and red shift, pushed to extremes. The planet does not orbit a stationary star. Both bodies 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 that variation gives the planet's orbital period. Its amplitude depends on the planet's mass, because a more massive planet gives the star a larger counter-orbital speed.
Describe the drawing
A planet's dark disc crosses the face of a drawn star, and below it the brightness curve runs flat, ramps down to a shallow flat floor during the transit, and recovers: the fractional depth of the dip is the squared ratio of the radii. The curve is idealised (a central, non-grazing crossing with limb darkening neglected) and the dip is exaggerated for visibility.
The transit method looks for the small regular 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 manages one part in ten thousand.
GUIDED PRACTICE
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.
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 a required step. The dip compares areas, and the answer wants a radius.
INDEPENDENT PRACTICE
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 of a few hundred-thousandths of a nanometre. Exoplanet spectrographs are consequently among the most stable instruments ever built, and the first planets found this way were heavy ones huddled close to their stars.
ASSESSMENT FOCUS
- Direct detection fails for two stated reasons, and one alone will not do. The star's overwhelming brightness, and an angular separation below the resolving limit.
- Transit questions live on the light curve, so sketch it. Flat, dip, flat, with depth (rp/rs)2 and the repeat time giving the orbital period. Radial velocity answers must mention the centre of mass: the star wobbles only because both bodies orbit it.
- The two methods answer different questions, so pair them when a question offers both. The dip's depth gives the planet's size, the wobble's size reflects its mass, either period is the planet's year, and mass with radius gives a density and a rocky-or-gaseous verdict.
CHECK YOURSELF
A star of radius 7.0 × 108 m dims by a fraction 1.6 × 10−4 once every 12 days. Find the planet's radius and orbital period, and state what a radial velocity measurement with the same period would add.
Show a hint
The dip compares disc areas, so a square root is compulsory before any radius appears.
Show the answer
(rp/rs)2 = 1.6 × 10−4, so rp/rs = 0.0126 and rp = 0.0126 × 7.0 × 108 ≈ 8.9 × 106 m, about 1.4 Earth radii.
The dips repeat once per orbit, so the orbital period is 12 days.
The amplitude of the star's radial-velocity variation depends on the planet's mass, because a more massive planet gives the star a larger counter-orbital speed about their shared centre of mass. Mass and radius together give a density, and with it a rocky or gaseous verdict.
Direct imaging fails for two reasons: the star is about a billion times brighter than the planet, and their angular separation is below the diffraction limit.
The transit dip is (rp/rs) squared and repeats once per orbit; the star's Doppler wobble shares that period, and its size scales with the planet's mass.
Or read them with their mark schemes on the quasars questions page.
Or revise the whole astrophysics deck of 74 cards.
CHECK YOUR PROGRESS
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- Explain why direct exoplanet detection fails, and interpret the radial velocity method and the transit light curve.
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