Rosharan System maths

[Firesong she/her]

I did some maths on the Rosharan system, I am not going to share all the formulae, there are a lot. But I will show assumptions and the facts we know. 

So, we know that the Rosharan sun is a bit bigger than our sun, but not by much. I assumed 1.18 solar masses due to a few reasons that would come up later. We also know the Orbital Period is 1.1x the length of Earth's, these both let us make a lot of assumptions. 

It would make the sun only around 1.00004685954 Solar Radii, and make it 1.12559551306  AU away from Roshar (another calculation got 1.1261, which is still very close and I think one was just a mistake. This calculator is saying that the mass of the star and planet, with that distance, would be an orbital period of ~1.099, close, but less than 1.1. This is troublesome, but the difference from my own calculations are less than the change in distance from the sun that Earth goes over its orbit, which varies by 0.03341, so it shouldn't be too big an issue, this difference (between the 1.12559 and 1.1261 calculations, that is) is by 0.00050448694).

 

The habitable area is between 1.12488603236 and 1.62056677971 AU. (This is why I assumed 1.18, as it is around the closer end of the Habitable Zone, and still is "only a bit larger" than our own sun. 

Surface temperature of the star would be around 6276.65381902 Kelvin, with a Luminosity of around 1.39190544437 Solar Luminosity or 5.353268339×10^26 watts. Thus an Apparent Magnitude of around -26.946, which is only a bit brighter than the sun in our own sky (−26.832) Every unit of 1 is 2.512x brighter, thus it is around 0.2512x brighter (my brain is blanking out about how to write this better for some reason, but it is brighter, despite how it looks, think 1.2515x would be correct). 

So, despite being further away, their day would actually be brighter than our own. 

We know they see their star as white, and if you look up a star colour chart, 6,000 to 7,500 K stars are yellowish-white, and have an apparent colour of white. And lookie there, 6276.65 K under my calculations. So that works out just fine. 

It would have an angular size of around 0.473 degrees, around 94.6% of the 0.5 of our own sun. This should still look around the same size as our own sun, maybe a bit smaller, but it would be hard to notice due to the glow distorting the size. Around what the Empire State building would look like 46200 metres away. Or around 90.96% the size of the moon. It would be noticeable, but not too dramatic. The differences in distance to the sun and the greater size of the sun actually roughly cancel out to make the apparent difference rather miniscule. 

Roshar:

Mass: 3.296 * 10^24 kg

Distance from Star: ~1.1261094201 AU

Gravity: 6.86 m/s2 (0.7 Earth)

Density: 4.3327028902 g/cm3 (0.78505216347 Earth Density)

Surface Area: 4.029980895×10^11 m2 (~0.78833439464 Earth)

Land Area: ~4×10^10 m2 (9.92560536692% total area) (

Standard Surface Temperature: 261.815853715 K (ignoring greenhouse effects and assuming albedo of 0.29)

Orbital Period: 34,689,600 seconds (1.1 Earth Years)

Rotational Period: 69425.2373 seconds, 0.80353283912 Earth Days

Rotational Velocity: 1,845.131535 km/h (1.15320720938 that of Earth)

Angular Velocity: 0.0000905028999761 rad/s 

Standard Gravitational Parameter: 156,642,212,401,043,768,235 m3/s2 

Gravitational Time Dilation: 0.9999999996 seconds per 1 second without gravity (compared to the 0.9999999993 of Earth)

Axial Tilt: 0°

First Cosmic Velocity: 6,194 m/s2 (0.783059418458 Earth)

Escape Velocity: 8,759 m/s2 (0.782053571429 Earth)

 

Edited September 18, 2023 by Firesong

[alder24]

56 minutes ago, Firesong said:

I did some maths on the Rosharan system, I am not going to share all the formulae, there are a lot. But I will show assumptions and the facts we know. 

You had to be really bored, but I appreciate the math.

Rosharan's sun angular size wouldn't be noticeably smaller than our Sun's - the Moon's angular size changes by ~10% from 0,491° on its closest to 0,548° when it's the furthest away from the Earth - this isn't something that you can see with the naked eye. Sun's angular size changes by 0.02° too, from 0,52° to 0,54°. The difference between Rosharan's sun size and Earth's Sun size is the same as between Moon's sizes and that's ~0.05° - you won't see it unless you compare two really good photos with no sun flares.

[Firesong she/her]

5 hours ago, alder24 said:

You had to be really bored, but I appreciate the math.

Rosharan's sun angular size wouldn't be noticeably smaller than our Sun's - the Moon's angular size changes by ~10% from 0,491° on its closest to 0,548° when it's the furthest away from the Earth - this isn't something that you can see with the naked eye. Sun's angular size changes by 0.02° too, from 0,52° to 0,54°. The difference between Rosharan's sun size and Earth's Sun size is the same as between Moon's sizes and that's ~0.05° - you won't see it unless you compare two really good photos with no sun flares.

I have added a lot if you want to check again (and comment on the discrepancy in semi-major axis I ran into)

[alder24]

13 hours ago, Firesong said:

It would make the sun only around 1.00004685954 Solar Radii, and make it 1.12559551306  AU away from Roshar (another calculation got 1.1261, which is still very close and I think one was just a mistake. This calculator is saying that the mass of the star and planet, with that distance, would be an orbital period of ~1.099, close, but less than 1.1. This is troublesome, but the difference from my own calculations are less than the change in distance from the sun that Earth goes over its orbit, which varies by 0.03341, so it shouldn't be too big an issue, this difference (between the 1.12559 and 1.1261 calculations, that is) is by 0.00050448694).

You're using a calculator? I use my own messy excel, where I barely can find anything now. I’ll check this for you. Roshar year is 500 days long, a day is 20h long, an hour is 50 minutes long, and assuming seconds are unchanging 1 min is 60 seconds, that means Rosharan year is 30 mil seconds long. Therefore Rosharan orbit is 1,021564378 AU away from its sun. If 1 min is 50 seconds, then it's 0,9046451172 AU.

But going with "Rosharan year is only 10% longer than the cosmere standard" gives 34 712 647,2 seconds in a year  (I've included more precise year measurement as 1 year = 365,2425 days, because of leap years, you can't ignore them, but there are even more precise values than that) - that means their callender isn't accurate with their orbital period (as Rosharan calendar year is shorter, not longer, than Earth's orbital year, which is ~31.6 mil seconds, or I did something wrong which is very, very likely (I did, going in reverse a bit later, it looks like Rosharan hour is ~57,85 Earth's minutes long and then that gives the correct value of seconds in a year (34.7 mil sec) - and this roughly checks out with this WoB (this was with a year = 365 days, not accurate enough) - so their calendar is aligned with planet's orbit, but their hour is ~57.85 Earth minutes long, and that's 50 Rosharan minutes (Rosharan second is ~ 1,157 Earth's seconds long) - I don't even want to go into how confusing that is...)). In this case Rosharn orbit is at 1,125926551 AU. And I think that's the correct number. You would get this if you accounted for leap years. I can't believe you were off by so much .

13 hours ago, Firesong said:

The habitable area is between 1.12488603236 and 1.62056677971 AU.

In my case it was 0.004 AU less than what you've got. I wonder why as I've used your luminosity value.

13 hours ago, Firesong said:

It would have an angular size of around 0.473 degrees, around 94.6% of the 0.5 of our own sun. Thus it would appear slightly smaller. Around what the Empire State building would look like 46200 metres away. Or around 90.96% the size of the moon. It would be noticeable, but not too dramatic. The differences in distance to the sun and the greater size of the sun actually roughly cancel out to make the apparent difference rather miniscule. 

I still disagree  

13 hours ago, Firesong said:

Rotational Velocity: 

Rotational Period: 63113.85125 Seconds, 0.730484389468 Earth Day

I calculated it for you. Equator is 35583 km long (or at least the average circumference) and I used 3 values for a day (2 are the same but with a different method just to check as it was done before I've calculated real value for Rosharan hour in Earth's minutes) - one is 20h*50 Rosharan min (57.8 Earth's minutes)*60sec and that gave rotational velocity at the equator equal to 1845,131535 km/h. Your length of day (where did you get that number? I got 69425,2944 sec, ~0.804) and that gave me 2029,646384 km/h, and my calculated earlier length of year divided by 500 ~ 69425,2944 sec per day gave me 1845,131535 km/h (the same as the first one). 

 

You should add more info from Coppermind like radius or circumference.

[Firesong she/her]

2 hours ago, alder24 said:

You're using a calculator? I use my own messy excel, where I barely can find anything now. I’ll check this for you. Roshar year is 500 days long, a day is 20h long, an hour is 50 minutes long, and assuming seconds are unchanging 1 min is 60 seconds, that means Rosharan year is 30 mil seconds long. Therefore Rosharan orbit is 1,021564378 AU away from its sun. If 1 min is 50 seconds, then it's 0,9046451172 AU.

But going with "Rosharan year is only 10% longer than the cosmere standard" gives 34 712 647,2 seconds in a year  (I've included more precise year measurement as 1 year = 365,2425 days, because of leap years, you can't ignore them, but there are even more precise values than that) - that means their callender isn't accurate with their orbital period (as Rosharan calendar year is shorter, not longer, than Earth's orbital year, which is ~31.6 mil seconds, or I did something wrong which is very, very likely (I did, going in reverse a bit later, it looks like Rosharan hour is ~57,85 Earth's minutes long and then that gives the correct value of seconds in a year (34.7 mil sec) - and this roughly checks out with this WoB (this was with a year = 365 days, not accurate enough) - so their calendar is aligned with planet's orbit, but their hour is ~57.85 Earth minutes long, and that's 50 Rosharan minutes (Rosharan second is ~ 1,157 Earth's seconds long) - I don't even want to go into how confusing that is...)). In this case Rosharn orbit is at 1,125926551 AU. And I think that's the correct number. You would get this if you accounted for leap years. I can't believe you were off by so much .

In my case it was 0.004 AU less than what you've got. I wonder why as I've used your luminosity value.

I still disagree  

I calculated it for you. Equator is 35583 km long (or at least the average circumference) and I used 3 values for a day (2 are the same but with a different method just to check as it was done before I've calculated real value for Rosharan hour in Earth's minutes) - one is 20h*50 Rosharan min (57.8 Earth's minutes)*60sec and that gave rotational velocity at the equator equal to 1845,131535 km/h. Your length of day (where did you get that number? I got 69425,2944 sec, ~0.804) and that gave me 2029,646384 km/h, and my calculated earlier length of year divided by 500 ~ 69425,2944 sec per day gave me 1845,131535 km/h (the same as the first one). 

 

You should add more info from Coppermind like radius or circumference.

I did account for leap years on several steps. And on being different, this is what happens when you have so many equations which are put into equations which are put into equations. With a lot of exponents. Very small deviances grow over time. 

I got day length with the equation 500a= 365.2422 * 86400. With 365.2422 being the number on NASA's website. So I trust it is more accurate than 365.2425.

I didn't really try on the Rotational Velocity at the equator because I don't fully get the equation for Equatorial Bulge magnitude. I tried it, but the equivalences didn't work out. 5/4 * (w2 * a3)/GM was dramatically different from (15π/4 * 1/(GT^2 p)), when they should be approximately equal. So I am doing something wrong there. But this would dramatically change what the velocity at the equator is, as perfect hydrostatic equilibrium of a massive rotating body is a silly myth. 

1.12559551306 isn't too far from 1.125926551, not as much as you imply. They are the same to 3 decimal places, only off by around ~0.00031, not as much as you imply. 

Edited August 18, 2023 by Firesong

[alder24]

13 minutes ago, Firesong said:

I got day length with the equation 500a= 365.2422 * 86400. With 365.2422 being the number on NASA's website. So I trust it is more accurate than 265.2425.

That's wrong. You're equating Rosharan year to Earth's year, without accounting for difference in the value of a second. Rosharan second is  1,15708824 Earth's second. 500 Rosharan days isn't equal to Earth's year, but that's what your equasion means. You need to multiply the right side by 1.1 (because Rosharan year is 10% longer) to get the accurate value - for your 365.2422 days year it's still 69425,23738 seconds per day. The differences are really small depending on the value of Earth's year we used (there are some even more accurate).

Just think of it; a difference in 4 place after the period won't make a change in a range of 8000 or something. My leap year is based on the Gregorian calendar - less accurate but we're basing Rosharan period on Rosharan calendar, so I think that's more comparable.

Another way you could get that is to take the "Orbital Period: 34,689,600 seconds (1.1 Earth Years)" and divide it by 500, which gives  69379,2 seconds per day. But your value of orbital period isn't accurate, as that's what I was referring to while pointing out leap years. This value was calculated by using 365 days per year, not accounting for a leap years (365*24*60*60*1,1=34 689 600, while 365,2522*24*60*60*1,1= 34 712 618,69). This is easy to spot because your value ends with 0 and it lacks numbers after the period - they should be there because we both want to use an accurate year so we won't get a full number of seconds.

36 minutes ago, Firesong said:

I didn't really try on the Rotational Velocity at the equator because I don't fully get the equation for Equatorial Bulge magnitude. I tried it, but the equivalences didn't work out. 5/4 * (w2 * a3)/GM was dramatically different from (15π/4 * 1/(GT^2 p)), when they should be approximately equal. So I am doing something wrong there. But this would dramatically change what the velocity at the equator is, as perfect hydrostatic equilibrium of a massive rotating body is a silly myth. 

Dramatically change? Not really. Earth's equatorial radius is just 21 km larger than polar, which is just 7 km larger than the mean radius. Even when I gave Roshar 20 km larger radius at the equator than its mean radius, it only changed its equatorial velocity by just 6.4 km/h. Not much at all. Barely noticeable and it's in the range of a good approximation. Giving it +7 km, it makes it just 2.2 km/h faster. You can just round the value to tenth digit at it will be just fine.

31 minutes ago, Firesong said:

1.12559551306 isn't too far from 1.125926551, not as much as you imply. They are the same to 3 decimal places, only off by around ~0.00031, not as much as you imply. 

That was a joke, sorry  

[Firesong she/her]

25 minutes ago, alder24 said:

That's wrong. You're equating Rosharan year to Earth's year, without accounting for difference in the value of a second. Rosharan second is  1,15708824 Earth's second. 500 Rosharan days isn't equal to Earth's year, but that's what your equasion means. You need to multiply the right side by 1.1 (because Rosharan year is 10% longer) to get the accurate value - for your 365.2422 days year it's still 69425,23738 seconds per day. The differences are really small depending on the value of Earth's year we used (there are some even more accurate).

Just think of it; a difference in 4 place after the period won't make a change in a range of 8000 or something. My leap year is based on the Gregorian calendar - less accurate but we're basing Rosharan period on Rosharan calendar, so I think that's more comparable.

Another way you could get that is to take the "Orbital Period: 34,689,600 seconds (1.1 Earth Years)" and divide it by 500, which gives  69379,2 seconds per day. But your value of orbital period isn't accurate, as that's what I was referring to while pointing out leap years. This value was calculated by using 365 days per year, not accounting for a leap years (365*24*60*60*1,1=34 689 600, while 365,2522*24*60*60*1,1= 34 712 618,69). This is easy to spot because your value ends with 0 and it lacks numbers after the period - they should be there because we both want to use an accurate year so we won't get a full number of seconds.

Dramatically change? Not really. Earth's equatorial radius is just 21 km larger than polar, which is just 7 km larger than the mean radius. Even when I gave Roshar 20 km larger radius at the equator than its mean radius, it only changed its equatorial velocity by just 6.4 km/h. Not much at all. Barely noticeable and it's in the range of a good approximation. Giving it +7 km, it makes it just 2.2 km/h faster. You can just round the value to tenth digit at it will be just fine.

That was a joke, sorry  

Oh wait, yeah, I forgot to add a 1.1x, you're right. That would turn out as 69425.2373... I made a mistake there. Pretty easy mistake when there are so many steps across every formula I did. Which would change it to around 1.12609339186... AU. I would go and fix some things

And don't be sorry, its fine.