# Are these traits for my planet mathematically accurate? Could it support life?

Mass of star: 0.75 M☉

Surface temperature of star: 4,620° K

Luminosity of star: 0.2 L☉

Satellites: 2

Place in solar system: 2nd

Mass: 2.57 M⊕

Mean density: 5.5 g/cm3

Surface gravity: 1.37 g (13.4 m/s2)

Axial tilt: 126.4972091°

Tropical zones: latitudes 53°30’10.04724” (53.5027909°)

Frigid zone: latitudes 36°29’49.95276” (36.4972091°)

Aphelion: 82,540,321 km

Perihelion: 78,239,492 km

Semi-major axis: 80,390,000 km (0.54 AU)

Orbit perimeter: 505,010,000 km

Foci distance: 4,300,800 km

Eccentricity: 0.02675

Orbital period: 227.7 d (19,671,600.6s)

Average orbital speed: 25.67 km/s

Rotation period: 42h 8m 3.619s (151,683.619s)

Equatorial rotation velocity: 0.36201 km/s

Rotations per orbit: 129.69

Bond albedo: 0.298

Could a planet with these traits support life? Are all of the measurements accurate?

• This sounds realistic, though double checking all math would be tedious. – Alexander May 23 '20 at 23:14
• Got about half way down the list before brain-freeze set in (just roughly in my head comparing it to our system). Seemed fine. Surface temp (and presumably its life-support potential) will depend an awefull lot on the atmospheric composition. We also think of gas giants as cosmic roombas, do you plan to have those? – Rottweiler on market-day. May 24 '20 at 0:15
• Not using significant figures and not using scientific notation - please try these for readibility. – StephenG May 24 '20 at 2:09
• Could a planet with these traits support life? We have no solid theory for defining life, or what conditions are required for it to survive. – StephenG May 24 '20 at 2:13
• @Tantalus' touch. The gas giant vacuum cleaner idea is outdated. While jupiter reduces the amount of comets reaching the inner system, it disturbs as many belt objects, causing them to go into the inner system. At best Jupiter's effect is net zero. – TheDyingOfLight May 24 '20 at 9:28

Your star is not going to work, so the planet won't either.

Mass of star: 1.04 M☉

Surface temperature of star: 4,620° K

Luminosity of star: 0.2 L☉

There is a well known set of relationships for mass-luminosity (and hence temperature) for stars. It's explained on this Wikpedia page.

For stars of around 1 solar mass that relationship says :

$$\frac L {L_\odot}= \left( \frac M {M_\odot}\right)^4$$

So your star with $$M=1.04 M_\odot$$ should have $$L=1.17L_\odot$$ and not $$0.2L_\odot$$ as you state.

The knock on from this is of course the temperature and conditions on your planet.

The star is basically about 16% more luminous than our own, but your planet is about half the distance we are from the Sun. It's going to be hot !

The (very rough) calculation for effective temperature of a planet will be :

$$T = T_\oplus \left( \frac {L(1-A)a_\oplus^2} {L_\odot(1-A_\oplus)a^2} \right)^{\frac 1 4}$$

With your number for the star this gives $$T \approx 1.4 T_\oplus$$. So $$40$$ % hotter than earth, or about $$110^\circ K$$ hotter. That puts it well into runaway greenhouse territory and it would likely be far worse.

Most of that is a result of your very close orbit - move it out past $$1\, AU$$ and you'll get better results.

The star's radius is actually a bit tricky - at around $$M=M_\odot$$ the typical mass-radius behavior has a distinct boundary - different behaviors on each side. Probably the simplest thing is to treat your star as being almost the same as the Sun in terms of mass and you get a radius of about the same as the Sun.

• My star is a K2V main sequence star, if that helps. – Praearcturus May 25 '20 at 5:29
• @Praearcturus My understanding is that K-type main sequence stars have masses below about 0.85 solar masses, and your star has a mass above one solar mass. Your star's mass is borderline G or F, our own Sun being G-type. – StephenG May 25 '20 at 16:36
• What mass would fit with the 0.2 solar luminosity? – Praearcturus May 25 '20 at 17:03
• Using the expression for K-type stars listed on Wikipedia's page (the Cuntz-Wang formula), I get about $M=0.755\,M_\odot$. Using the formulas for general main sequence stars I get about $M=0.68\,M_\odot$. Somewhere in that range would be OK. – StephenG May 25 '20 at 18:24

## Stellar Heating

(using Stefan-Boltzmann equation) $${P \over A} = \epsilon \sigma T^4$$. Power per unit area at the surface of your star (0.69 suns ~ 480,000 km) is 2.58$$\times 10^7$$ Watts per square-meter.

At perihelion, the planet is 72 million kilometers distant. $$R2 \over R1$$ (where R1 is the solar surface radius) is 163, and the power received per unit area is

$$P_1 \times {R2 \over R1}^2$$ = 970.44 Watts per meter squared. At aphelion, 871.94 $$W \over m^2$$.

Compared to Earth receiving around 1,374 $$W \over m^2$$ peak, this planet gets about 63% to 70% to same amount of radiant heat.

When you include the lower Bond albedo, 619 to 689 $$W \over m^2$$ peak make it to the surface.

Compared to Earth, this world holds on to 72% as much radiant heat.

$$P_{average}$$ = $$P_{peak} \over 4$$. Therefore, the average power received is 163 $$W \over m^2$$. The predicted surface temperature, then is 231 Kelvin / -41 Celsius; compared to Earth's predicted surface temperature of -18 C.

I think it should be pointed out that Earth's observed surface temperature is +14 C, and that Earth gets a +32 degree C boost from greenhouse effect.

This planet may support life, but there must be some sort of greenhouse or other heating effect raising the mean temperature above freezing.

## Gravity

$$M2 \over M1$$ = 2.57; $$R2 \over R1$$ = 1.37; $$G2 \over G1$$ = 1.36 gs. About what you calculated.

• Higher g will favor a thicker atmosphere. Depending on the composition, I suspect a higher amount of heat may be trapped inside atmosphere; but a longer day... I don't know, average temperature may end OK, but the day/night extremes may be farther apart than on Earth. – Adrian Colomitchi May 24 '20 at 12:30
• Will the temperature in tropical regions also be ridiculously cold? – Praearcturus May 24 '20 at 16:44
• @praearcturus, the peak power received by the planet is more than enough to be very hot. How quickly that received power is distributed depends on a lot things. – James McLellan May 25 '20 at 20:59