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### Astronomy workshop

Posted: **15 Apr 2019 05:36**

by **Khemehekis**

This is a workshop to help conworlders build galaxies, solar systems, planets, and moons that are scientifically accurate. Conworlders can receive help with equations here.

Things like AU, solar mass and luminosity, the length of a year, gravity, density, escape velocity, orbital eccentricity, the phases of moons, etc. are all fair game here.

### Re: Astronomy workshop

Posted: **15 Apr 2019 05:37**

by **Khemehekis**

All right. My planet, Bodus, is 2.0251 AU from its Wanos sun. A year on Bodus lasts as long as 3.6504892 years on Earth (1,333.313801631 Earth days). What would the solar mass of Wanos be?

### Re: Astronomy workshop

Posted: **15 Apr 2019 13:07**

by **Salmoneus**

I'm not sure what help you want here - you have the same equations and equivalent calculator to any of us. Just enter in the numbers.

If you didn't know the equation, you could have just opened up wikipedia. Which tells you that the orbital period is 2*pi*sqrt((a^3)/GM), where G is the gravitational constant and M is the mass of the star.

### Re: Astronomy workshop

Posted: **16 Apr 2019 00:23**

by **Khemehekis**

Salmoneus wrote: ↑15 Apr 2019 13:07

I'm not sure what help you want here - you have the same equations and equivalent calculator to any of us. Just enter in the numbers.

I was using the website

https://astrographer.wordpress.com/book ... eoff-eddy/ where I found the equation:

M × T

2 = R

3 (Kepler’s third law)

where T is the orbital period and R is the distance from the sun.

I did the calculation:

2.0251^3 / 1,333.313801631^2 = 0.00000467169

which doesn't seem correct, as the solar mass of a star with a planet orbiting at such a distance and year length should be somewhere in between 1.3 and 1.7. How'd I get 0.00000467169?

### Re: Astronomy workshop

Posted: **16 Apr 2019 13:35**

by **sangi39**

Khemehekis wrote: ↑16 Apr 2019 00:23

Salmoneus wrote: ↑15 Apr 2019 13:07

I'm not sure what help you want here - you have the same equations and equivalent calculator to any of us. Just enter in the numbers.

I was using the website

https://astrographer.wordpress.com/book ... eoff-eddy/ where I found the equation:

M × T

2 = R

3 (Kepler’s third law)

where T is the orbital period and R is the distance from the sun.

I did the calculation:

2.0251^3 / 1,333.313801631^2 = 0.00000467169

which doesn't seem correct, as the solar mass of a star with a planet orbiting at such a distance and year length should be somewhere in between 1.3 and 1.7. How'd I get 0.00000467169?

I seem to recall that

*R* needs to be in AUs,

*M* is in Solar Masses and

*T* is in Earth Years in years, not in days, so you should get a mass for your sun of 0.623 Solar Masses, which would be a... K5(?) type star (more or less), which seems to put your planet way outside what the guide suggests is the "habitable" zone for an Earth-like planet.

For that and the year you'd want, and roughly the same distance, you'd be looking at an F0 type star (1.7 Solar Masses), with a distance from the star of 2.83 AU (giving a year of 1,333.66 Earth Days).

### Re: Astronomy workshop

Posted: **16 Apr 2019 16:02**

by **gach**

sangi39 wrote: ↑16 Apr 2019 13:35

I seem to recall that

*R* needs to be in AUs,

*M* is in Solar Masses and

*T* is in Earth Years in years

Yep, these particular units are in common use by people working with Solar System dynamics since they simplify the mathematics quite a bit.

I will say that deciding the planet's orbit first and then trying to fit everything else to it is a pretty backwards way to do things. In this case it leaves us with a planet with so low energy influx from the star that it feels safe to say no realistic greenhouse effect could turn it habitable.

sangi39's suggestion of replacing the central star with a F0 type star comes with its own problems since more massive stars exhaust their fuel and become giants much faster than low mass stars. A rough order of magnitude estimate for a star's main sequence lifetime is given by the formula

*t* ≈

*M/L* × 10

10 yr,

where both the mass

*M* and luminosity

*L* of the star is given in solar units (

*M*Sun = 1 and

*L*Sun = 1 =>

*t*Sun ≈ 10 billion years). Observationally we know that for the main sequence stars luminosity scales with the mass roughly as

*L*/

*L*Sun = (

*M*/

*M*Sun)

3.8.

If you plug this relation into the main sequence lifetime formula, you get an estimate

*t* ≈ (

*M*/

*M*Sun)

-2.8 × 10

10 yr.

For a 1.7 Solar mass star this would give a roughly 2 billion year main sequence lifetime. That could potentially give enough time for complex life to evolve on a planet before the star heats up too much, but the example of life's evolution on Earth suggests that it's most likely way too little.

### Re: Astronomy workshop

Posted: **16 Apr 2019 17:06**

by **sangi39**

gach wrote: ↑16 Apr 2019 16:02

sangi39 wrote: ↑16 Apr 2019 13:35

I seem to recall that

*R* needs to be in AUs,

*M* is in Solar Masses and

*T* is in Earth Years in years

Yep, these particular units are in common use by people working with Solar System dynamics since they simplify the mathematics quite a bit.

I will say that deciding the planet's orbit first and then trying to fit everything else to it is a pretty backwards way to do things. In this case it leaves us with a planet with so low energy influx from the star that it feels safe to say no realistic greenhouse effect could turn it habitable.

sangi39's suggestion of replacing the central star with a F0 type star comes with its own problems since more massive stars exhaust their fuel and become giants much faster than low mass stars. A rough order of magnitude estimate for a star's main sequence lifetime is given by the formula

*t* ≈

*M/L* × 10

10 yr,

where both the mass

*M* and luminosity

*L* of the star is given in solar units (

*M*Sun = 1 and

*L*Sun = 1 =>

*t*Sun ≈ 10 billion years). Observationally we know that for the main sequence stars luminosity scales with the mass roughly as

*L*/

*L*Sun = (

*M*/

*M*Sun)

3.8.

If you plug this relation into the main sequence lifetime formula, you get an estimate

*t* ≈ (

*M*/

*M*Sun)

-2.8 × 10

10 yr.

For a 1.7 Solar mass star this would give a roughly 2 billion year main sequence lifetime. That could potentially give enough time for complex life to evolve on a planet before the star heats up too much, but the example of life's evolution on Earth suggests that it's most likely way too little.

I hadn't thought about that! I was more concerned with the higher UV output

But yeah, it's a bit of a problem, and it doesn't look like you can have it all:

A) If you have a year 3.6504892 Earth-years long, for it to be habitable it would have to orbit at 2.83AU around a star with a mass around 1.7 Solar Masses.

B) If you have a distance from the star of 2.0251AU, for it to be habitable it would have to orbit around a star of mass around 1.3 Solar Masses once every 923 days.

It does look like you can fiddle with the mass of the star around those values, but as gach mentioned above, the result will be a in a star outputting a relatively high amount of UV light over a relatively shorter lifespan, which might then affect the sort of atmosphere your planet might have or the biology of the species on the planet.

### Re: Astronomy workshop

Posted: **17 Apr 2019 05:56**

by **Khemehekis**

Thanks for the help, **Sangi39** and **Gach**!

I guess I'll keep my year length and change the number of AU to 2.83.

And Sangi, thanks for explaining how I was doing the math wrong! Now I'll know that one is supposed to plug in Earth-years, not Earth-days.

I'll post the new planet data for Bodus when I've got this done.

**EDIT**: My Bodus description now reads: Bodus is a planet in the solar system Wanos, the fifth planet from the sun. Bodus is 2.830244 AU from its Wanos sun (with a solar mass of 1.701255). A year on Bodus lasts as long as 3.6504892 years on Earth (1,333.66 Earth days).