AstronomyThe Solar System

Finding Planets Around Other Stars

A passing planet dims its star by (R_planet/R_star)², and a planet's tug makes its star wobble.

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For centuries, planets around other stars were a guess. The first one around a Sun-like star, 51 Pegasi b, was found in 1995, and today more than 5,000 are confirmed (NASA Exoplanet Archive). Hardly any were seen directly: a planet is billions of times fainter than its star and lost in its glare. Instead astronomers watch the star, using ideas you already know: Doppler shifts and orbits.

Why it works

Transits. If a planet's orbit is edge-on to us, the planet crosses its star's face once per orbit and blocks a small disk of light. The fraction blocked is the ratio of the two disks' areas: $\dfrac{\pi R_p^2}{\pi R_\star^2} = \left(\dfrac{R_p}{R_\star}\right)^2$. The depth gives the planet's size; the time between dips gives its period. Wobble. A star and its planet both orbit their common center of mass, so the star moves in a small circle, and its spectral lines shift back and forth. Balancing momenta, $M_\star v_\star = m_p v_p$, so the size of the wobble gives the planet's mass.

Key idea

Transit depth $= \left(\dfrac{R_p}{R_\star}\right)^2$ gives a planet's size; the star's Doppler wobble, $v_\star = v_p\,\dfrac{m_p}{M_\star}$, gives its mass. Together they give its density: rocky or gassy.
A planet one-tenth of its star's radius crossing the star. The light curve dips by $0.1^2 = 1\%$.

Worked examples

Worked example: Jupiter and Earth from afar

Jupiter's radius is about $0.10$ of the Sun's; Earth's is about $\frac{1}{109}$. How much would each dim the Sun during a transit?
  1. Jupiter: $(0.10)^2 = 0.01 = 1\%$.
  2. Earth: $\left(\frac{1}{109}\right)^2 = 8.4 \times 10^{-5}$, about $0.008\%$.
  3. Finding Earth-sized planets needs measuring starlight to better than one part in 10,000, which is why the Kepler and TESS telescopes observe from space.

Answer: $1\%$ and about $0.008\%$.

Worked example: The Sun's wobble

Jupiter orbits at $13.1$ km/s and has $\frac{1}{1047}$ of the Sun's mass. How fast does Jupiter make the Sun wobble?
  1. $v_\odot = v_J \frac{m_J}{M_\odot} = 13.1\ \text{km/s} \times \frac{1}{1047} = 0.0125$ km/s.
  2. That is $12.5$ m/s, about the speed of a fast bicycle, measured from light-years away by its Doppler shift.

Answer: About $12.5$ m/s.

Common mistake

It is natural to imagine exoplanets being found in photographs, but only a small share have been imaged directly. Most were found by the transit method (thousands by the Kepler and TESS missions) or by the star's wobble. Check: a planet like Jupiter seen from a nearby star would be about a billion times fainter than the Sun beside it.
A star and a companion circling their center of mass, with the Doppler (radial velocity) curves. A planet makes its star wobble in the same way, only much less.

Practice

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Sources and further reading

  • OpenStax, Astronomy 2e, section 21.4, Planets beyond the Solar System: Search and Discovery
  • OpenStax, Astronomy 2e, section 21.5, Exoplanets Everywhere: What We Are Learning
  • NASA Exoplanet Archive (discovery counts by method)
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