Astronomy & Celestial Mechanics
the sky, computed
Concept

Transit probability — where it appears

The chance that a planet's orbit is tilted enough to carry it across its star as seen from here, which is the ratio of the star's radius to the orbital distance. It is about half a per cent for an Earth-like orbit around a solar-type star, which is why a transit survey must watch a hundred thousand stars to find a few.

Named by 5 essays across one field — each of them below, with the objects they name alongside it.

A transit of a planet 0.103 of its star's radius. The star's brightness through one transit, computed by integrating the uniform stellar disc over the region the planet covers. The depth is 1.05%, which is exactly (Rp/R⋆)² = 0.01055. The four contact points are where the two discs are externally and internally tangent, at separations 1 ± 0.103 stellar radii.

A planet measured by the light it removes

A transit gives a depth, and the depth is a ratio of two radii rather than a size. Everything a transit says about a planet is said in units of a star nobody has visited either.

exoplanets · Transits A gap where planets should be. The number of planets per star per interval of log radius, for orbital periods under a hundred days, corrected for detection efficiency. There are two peaks — super-Earths near 1.3 R⊕ and sub-Neptunes near 2.4 R⊕ — and a deficit between them at 1.89 R⊕, where the occurrence falls to 33% of the peak. The gap is not a gap in what can be detected: detection efficiency rises smoothly through it, so a smooth underlying distribution could not produce a dip there.

The planets that were not seen

An occurrence rate is a count divided by a probability, and the probability can be a five-hundredth. Everything difficult about saying how common planets are lives in that denominator.

exoplanets · Occurrence rates 5 perturbers that draw one 58-day timing signal. Every perturbing planet that gives a 3-day transiting planet the same timing signal — a sinusoid with a 57.8-day super-period and an amplitude of 1.27 minutes — placed wide of the nearest first-order commensurabilities inside and outside its orbit, with its mass found by integrating until the amplitude matched. outside, near 4:3 at 4.070 days needs 7.6 Earth masses and would move the star by K = 3.1 m/s; outside, near 3:2 at 4.620 days needs 12.0 Earth masses and would move the star by K = 4.8 m/s; outside, near 2:1 at 6.329 days needs 25.6 Earth masses and would move the star by K = 9.2 m/s; inside, near 3:2 at 1.966 days needs 6.9 Earth masses and would move the star by K = 3.7 m/s; inside, near 2:1 at 1.462 days needs 45.8 Earth masses and would move the star by K = 26.7 m/s. The period ratio is along the bottom on a logarithmic axis and the required mass up the side. A super-period fixes the distance from some resonance and not which resonance it is, and the amplitude then fixes a mass for each guess — so the timing alone returns a list rather than a planet. The velocity semi-amplitudes differ by a factor of 8.5 across the list, which is one of the two ways the list is shortened.

One timing curve and five planets that could draw it

A transiting planet whose times wander at a 58-day period, by just over a minute, is being pulled by something — but the period says only how far from some resonance the pull comes, not from which. Perturbers inside and outside the orbit, near four different commensurabilities, each with its own mass, reproduce the same curve to a fraction of a per cent. Timing alone returns a list, and even the detail that shortens it hides a coincidence of its own.

exoplanets · Transit-timing Three biases against eccentricity, and they do not agree. Four quantities against orbital eccentricity, each relative to a circular orbit of the same semi-major axis, averaged over the argument of periastron. The transit probability rises as (1 − e²)⁻¹, because an eccentric planet spends part of its orbit inside its own semi-major axis: at e = 0.5 a transit is 1.33 times as likely. The transit duration falls as √(1 − e²), so the event carries less signal-to-noise, and the two together — probability times the square root of the time in transit — come to 1.24 at the same eccentricity. They very nearly cancel, and that is the surprise: a transit survey has almost no eccentricity bias at all. The radial-velocity curve is the one that does. A Keplerian of eccentricity e puts less of its variance in the fundamental and more into harmonics no sinusoidal search is looking at — 68 per cent remains at e = 0.6 and 47 per cent at e = 0.8 — so a velocity survey loses amplitude exactly where a transit survey does not. What no figure here can show is which of these the measured eccentricity distribution is made of, because the correction depends on a detection pipeline rather than on geometry, and the two surveys have to be corrected separately before their answers can be compared.

Every method prefers a circle, and not for the same reason

A transit is more likely on an eccentric orbit and shorter when it happens, and the two very nearly cancel. A velocity curve loses amplitude to harmonics no sinusoidal search is looking at, and that one does not cancel at all.

exoplanets · Detection bias How many planets a star has is the hardest thing a catalogue measures. The multiplicity distribution a transit catalogue would contain, for systems that all truly hold 5 planets, at four mutual inclination dispersions. 40,000 systems are drawn per dispersion with an isotropic viewing direction and Rayleigh-distributed inclinations about a common plane, at semi-major axes of 12, 16, 21, 27, 34 stellar radii; the bars are conditioned on at least one planet transiting, which is what makes a system appear in a catalogue at all. At 0.5° of dispersion 33 per cent of the detected systems show all 5 planets and the mean apparent multiplicity is 3.13; at 10° it is 1.39, with 68 per cent of them showing exactly one. Every one of those systems has 5 planets. The entire difference between a catalogue of singles and a catalogue of compact multiples is one number that nothing in the light curve measures. And the two effects run in opposite directions: the fraction of stars showing any planet RISES with the dispersion — 8%, 9%, 12%, 19% across the four — because scattering the orbits gives more of them a chance to cross the line of sight, while the number seen per detected star falls by a factor of 2.2. A survey that scatters its systems finds more stars with planets and fewer planets per star, and neither number on its own says which has happened. What no figure here can show is the true dispersion, because the observable is the ratio of those two and a system with fewer planets and a tighter plane reproduces it exactly.

How many planets a star has is not a measurement

Draw five thousand identical five-planet systems, scatter their orbital planes by half a degree, and a third of the detections show all five. Scatter them by ten degrees and two thirds show exactly one. Every system has five.

exoplanets · Detection bias

Named alongside it

The objects these essays reach for when they reach for this one.

Radial velocitySurvey completenessOccurrence rateSelection effectArgument of periastronChopping signalCompletenessCoplanarityDegeneracyDetection limitDetection thresholdEclipse

All concepts