Luminosity for cold Sudarskies

The source, I won’t say blame, for the equilibrium-temperature equation is Montalto 2021, arxiv version 2110.00489. This concerned the hot-Jupiter transiting TIC 257060897b.

Teq = T*Math.Sqrt(R/a)*Math.Pow(0.25,0.25). Or so it looks on the page; if so this Math.Pow(0.25,0.25) could have been printed 1/√2, although Montalto may have reasons why not.

T is the star’s effective temperature; R is stellar radius; a is semimajor. I don’t know if they integrated across the period over the variant values of r; eccentricity is simply not taken into account. R/a is the reciprocal of the distance in stellar-radius units, as opposed to AU. When a=R we are on the star’s surface. Yow! and note the reciprocal square roots again…

Also note absolutely nothing about the planet itself: no volume, no composition, and perfect absorption of light. It is a black dot in space.

I have little idea whence Montalto’s equation so I must guess. The stellar disc is not a point of light from a Hot Jupiter distance away. It makes sense, there, to have insolation as a function of stellar-radius units. Also with hot-Jupiters, as Sudarsky predicted for the fourth 900-1400 K range, albedo tends dark. Back in 2007 HD 189733b turned out Neptune-blue with 0.14 Bond albedo, half Neptune’s; this high, caused by its hot-spot, which might run over 1400 K into the class V range. Other planets in this heat-range – and distance from our scopes – are constrained even below this. I don’t blame Montalto whose team’s equations work fine for that paper. I blame those who’d cite this paper for the wrong systems.

For astrometric-calculated systems, past 2 AU from a main-sequence dwarf, Sudarsky is class II or I. Out here I prefer to estimate the star as a point with some fraction of our Sun’s luminosity. Because it’s easier, and because we’re actually told the maths.

The [Point] Luminosity is 4*Area*T*T*T*T*sigma, sigma being Stefan-Boltzmann. The stellar disc’s area = Math.PI*R*R obviously. I’d do this in Sol’s units so, when sigma = 1/4pi then L=1. Thus L=R*R*T*T*T*T. This T may be whence Montalto’s quartic-root… of one quarter. But who knows, nobody told me. I’ve sanity-checked this against (Wikipedia’s pre-DR2) Ursa Majoris 47, getting 1.487 against their 1.48 which I’ll call a win – for the equation. Actual luminosity can await actual data which isn’t Wiki data.

So first we make this Point Luminosity into an insolation-factor at 1 AU. Then multiply by Math.Pow(0.5, (a-1)). So at a=1, Sol-as-point delivers Earthlike insolation; a=2, half that, a=3, a quarter that and so forth.