Astrophysics of Krynn, part 3

 In the previous and previous posts, we figured:

• Earth: 6,371.00 km (mean radius)
• Krynn: 2,376.45 km (radius)
• Luna (Earth's moon): 1,737.4 km (mean radius)

... and:

• Solinari (36 days): 460,540 km radius orbit around Krynn
• Lunitari (28 days): 389,498 km radius orbit around Krynn
• Nuitari (8 days): 168,963 km radius orbit around Krynn

In this post, we'll try to deduce the radius of each of the moons of Krynn.

What's the data

There is no data. But we can use the following hints:

• Let's assume that Krynn's sun is of the same size, and same distance as Earth's sun.
• Let's assume that Lunitari is of the same size as Luna

This has interesting consequences:

• Lunitari is able to make a beautiful eclipse, hiding the sun (as Luna hides the sun)
• Solinari will hide more than the sun during an eclipse (i.e. probably no visible corona)
• Nuitari is unable to hide the sun, but will put a black point within the sun during an "eclipse"

Also, I want the following data to be true:

• When Solinari, Lunitari and Nuitari are in conjunction, I want a beautiful "eye" to form in the sky, as below:

In other words, the apparent radius of each moon need to have a specific size in relation to the others. For example, using the drawings I did:

• Solinari: 64 px
• Lunitari: 40 px
• Nuitari: 20 px

Two-moons conjunctions will result the following image in the sky:

For reference, you'll find below the image of each individual moon, with a green circle representing the size of Luna, for comparison's sake:

Also, all the moons are tidally locked with Krynn.

So? Lunitari

Disclaimer: Again, I'll be rounding a lot. In particular, the difference at sea altitude, between Krynn and Earth is negligible, and thus, neglected. Also, the difference between the radius of Krynn, and the radius of any moon's orbit can also be considered as negligible, so distances will be taken from the center of Krynn, not the surface. Last but not least, we'll assume angles are small enough to approximate A, tan(A) and sin(A) as the same values, in radians.

The good point about the data is that it simplifies some calculations: Luna and Lunitari have the same orbital period, and are at the same distance from their planet, and have the same apparent size, so, they have the same radius: 1,737.4 km

The angular diameter of Lunitari (and Luna) is thus (with rMoon being the radius of the moon, and D being the distance between the center of Krynn and the center of the moon):

δ = 2 arcsin(rMoon / D)

δ = 2 * rMoon / D

For Lunitari, this would be:

δ = 2 * 1,737.4 km / 389,498 km

δ = 0.00892122

Et voilà...

How to calculate the other moons?

The apparent diameter of the other moons is a factor of Lunitari's:

• Solinari = Lunitari * 1.6
• Nuitari = Lunitari / 2

This means the angular diameter of the moons are thus:

• δSolinari = 0.00892122 * 1.6 = 0.01427395
• δLunitari = 0.00892122
• δNuitari = 0.00892122 / 2 = 0.00446061

We can then, knowing their real distance, find their real radius:

δ = 2 * rMoon / D

rMoon = δ * D / 2

... which resolves into:

• rSolinari = 0.01427395 * 460,540 km / 2 = 3,287 km
• rLunitari = 1,737 km
• rNuitari = 0.00446061 * 168,963 km / 2 = 377 km

Et voilà...

Conclusion

We know have the following data:

• Earth: 6,371.00 km (mean radius)
• Krynn: 2,376.45 km (radius)
• Luna (Earth's moon): 1,737.4 km (mean radius)

... and:

• Solinari (36 days): 460,540 km radius orbit around Krynn
• Lunitari (28 days): 389,498 km radius orbit around Krynn
• Nuitari (8 days): 168,963 km radius orbit around Krynn

 ... and:

• Solinari (radius) = 3,287 km
• Lunitari (radius) = 1,737 km
• Nuitari (radius) = 377 km

So... yeah, Solinari is larger than Krynn... That is awkward, but unavoidable because our calculations our correct.

(Remember the cause of all of this is the small size of Krynn)

😁

The good thing is:

• No one will go there to verify (even if this can be deduced, the same way the ancient Greeks deduced the size of Luna when it passed through Earth's shadow)
• Even if one did, it's a fantasy world
• Also, Solinari (and the other moons) might have a very low density to counteract their size
• If we assume all the moons have the same albedo as Luna, then nights on Krynn can be much more luminous than on Earth
  

Astrophysics of Krynn, part 2

In the previous part, we found Krynn was a very small, but very dense planet, compared to Earth, both planets having the same mass, though.

Now, I will be using physics from high school for fun, to prove something that is very interesting, and then get an equation that will give the the orbit radius of all of Krynn's moons.

Disclaimer: I'm doing a lot of careless rounding, here, partly because I forgot half the tricks to do exact calculations with error margins, and partly because I don't really care about exact numbers.

Let's Physics!

Believe it or not, but the characteristics of a stable circular orbit of a moon around its planet depends only on the mass of the planet, and the distance between the moon and the planet.

On one hand, you have the Newtonian: ΣF = m.a

On the other, you have the Newtonian gravitation: F = G.m1.m2 / (r²)

Assuming that the mass of the moon is negligible when compared to the mass of the planet simplify the problems because then, the planet doesn't move, and only the moon moves.

So, for the moon, what you have is:

mMoon.aMoon = G.mPlanet.mMoon / (r²)

Which simplifies into:

aMoon = G.mPlanet / (r²)

So, the distance between the two bodies (here, r), is:

r² = G * mPlanet / aMoon

Fun fact, the centripete acceleration of the moon in a stable circular orbit is a function of its tengent speed, and thus, of its orbital period. Indeed, the equation is:

a = v² / r (with v being the tangential speed)

We also know:

• - v = ω.r (where ω is the angular velocity)
• - ω = 2π / T (where T is the orbital period, i.e. the time to complete ONE orbit)

Dropping the minus sign (because we already know the direction), we can deduce:

v = ω.r = 2π / T * r = 2π * r / T

So, now, we can deduce the acceleration a from the speed v and the orbit radius r:

a = v² / r

a = (2π * r / T)² / r

a = 4π²  * r² / T² / r

a = 4π² * r / T²

As we are studying the acceleration of the moon aMoon:

r² = G * mPlanet / aMoon

r² = aMoon / G / mPlanet

... we can then replace aMoon with 4π² * r / T²:

r² = G * mPlanet / (4π² * r / T²)

r² = G * mPlanet * T² / 4π² / r

r^3 = G * mPlanet * T² / 4π²

As you can see, the orbit's radius of the moon only depends on its orbital period, and on the mass of the planet.

Assuming G and mPlanet never change, we can replace them with their values:

• G = 6.674×10^-11 m^3⋅kg^-1⋅s^-2
• mEarth = 5.97237×10^24 kg

(We use the mass of Earth as we showed previously the mass of Krynn was the same), which means the equation becomes:

r^3 = 6.674×10^-11 m^3⋅kg^-1⋅s^-2 * 5.97237×10^24 kg * T² / 4π²

r^3 = 6.674×10^-11 m^3⋅s^-2 * 5.97237×10^24 * T² / 4π²

T is given in seconds, but I would be more interested in its "days" unit.

So we can deduce that:

T = D * 60 * 60 * 24 s

Where D is the number of days. Thus:

T = D * 60 * 60 * 24 s

T = D * 86400 s

T² = D² * 7,464,960,000 s^2

T² = D² * 7.464,960,000 × 10^9 s^2

Going back to the equation, we get:

r^3 = 6.674×10^-11 m^3.s^-2 * 5.972,37×10^24 * D² * 7.464,960,000 × 10^9 s^2 / 4π²

r^3 = 6.674 m^3 * 5.972,37×10^22 * D² * 7.464,960,000 / 4π²

r^3 = 6.674 * 5.972,37×10^22 * D² * 7.464,960,000 / 4π² m^3

r^3 = 6.674 * 5.972,37 * D² * 7.464,960,000 / 4π² × 10^22 m^3

r^3 = 6.674 * 5.972,37 * 7.464,960,000 / 4π² * 10^22 * D² m^3

r^3 = 297.550 / 4π² * 10^22 * D² m^3

r^3 = 7.537,04 * 10^22 * D² m^3

r^3 = 75.370,4 * 10^21 * D² m^3

r^3 = 4.224,096^3 * (10^7)^3 * D^(2*3/3) m^3

r = 4.224,096 * 10^7 * D^(2/3) m

r = 4.224,096 * 10^4 * D^(2/3) km

... which means we have the equation to calculate the radius r of the orbit of a moon, if we know the moon's period, in days D around a planet of the same mass as Earth, like Krynn.

Wait, are you sure?

To prove I'm not wrong in the details, let's put Luna (Earth's moon) in the equation.

For, the Luna, let's assume that the period is 28 days:

28 ^ (2/3) = 9.220,87

... which means the equation becomes:

r = 4.224,096 * 10^4 * 9.220,87 km

r = 38.949,8 * 10^4 km

r = 389,498 km 

... which is more or less the average between the apogee and perigee of Luna.

So, yeah, my calculations are right...

😉

Back to Krynn: Solinari, Lunitari & Nuitari ?

The periods of each of Krynn's moon is:

• Solinari: 36 days
• Lunitari: 28 days
• Nuitari: 8 days

So, to get the orbital radius of these moons, all we have is to feed the equation with the data:

• Solinari: 36 days => 4.224,096 * 10^4 * 36^(2/3) km
• Lunitari: 28 days => 4.224,096 * 10^4 * 28^(2/3) km
• Nuitari: 8 days => 4.224,096 * 10^4 * 8^(2/3) km

... which gives:

• Solinari: 36 days => 42,240.96 * 10.902,7 km
• Lunitari: 28 days => 42,240.96 * 9.220,87 km
• Nuitari: 8 days => 42,240.96 * 4 km

... which gives the orbital radius (i.e. the distance between the center of the moon, and Krynn):

• Solinari: 36 days => 460,540 km
• Lunitari: 28 days => 389,498 km
• Nuitari: 8 days => 168,963 km

Et voilà...

😉

In the next post, we will calculate the radius of each of Krynn's moons.

Astrophysics of Krynn, part 1

One would assume that, by default, Krynn would be a Earth-like planet.

Not so fast...

The Sources

There are conflicting information around. One, from the Atlas of Dragonlance, gives a 4-hours offset between the Tower of the High Clerist and Silvanost. Another, using a globe of krynn, instead gives a 2-hours offset for the same locations.

I choose the last source, because it would roughly make Krynn twice as large as the first one would make.

Also, to measure distances, I used Sean Macdonald's map of post-Cataclysm Ansalon.

Using the Globe and the Map

One interesting detail in this globe is the following line:

Indeed, this line is 50° long over the globe, and is situated at the latitude -30° (which is the same, for all our current purposes, as +30°)

And if we look at Sean Macdonald's map, it does from Mount Nevermind, to the very eastern edge of the Misty Isle, which is a distance I measured to be as 1,796km (sorry for the imperialist, I will be using international units system, here).

And with all that information, and a bit of geometry, I can calculate the radius of Krynn.

Let's do it.

Using Geometry

Assuming the following...

From the Mount Nevermind to the right edge of the Misty Isle:

• On Sean Macdonald's map, I measure 1,796km. Let's call this length SMD.
• On the 3D globe, I count 5 longitudes (each longitude being 10°), at lattitude -30° (which is the same than at lattitude 30°)

This means, the perimeter of the globe at lattitude -30° is:

P(30°) = SMD * 36 / 5

P(30°) = SMD * 7.2

What I want, is the radius of the globe at lattitude 0°, because it will be the radius of the sphere itself.

So, first, I need to transform the perimeter into a radius, which is done with the formula:

R = P / 2π

So:

R(30°) = P(30°) / 2π

R(30°) = SMD * 7.2 / 2π

R(30°) = SMD * 3.6 / π

Now, I want R(0°), which can be deduced from R(30°) because:

R(30°) = R(0°) * cos(30°)

... with cos(30°) = (3^0.5) / 2

So we have:

R(0°) = R(30°) / cos(30°)

R(0°) = R(30°) / (3^0.5) * 2

R(0°) = SMD * 3.6 / π / (3^0.5) * 2

R(0°) = SMD * 7.2 / π / (3^0.5)

We then replace the variables/constants with their values:

R(0°) = 1,796km * 7.2 / 3.14159 / 1.73205

R(0°) = 12,931.2km / 3.14159 / 1.73205

R(0°) = 4,116.13km / 1.73205

R(0°) = 2,376.45km

So, the Radius of Krynn is 2,376.45km...

That's small...

For comparison:

• Earth: 6,371.00 km (mean radius)
• Krynn: 2,376.45 km (radius)
• Luna (Earth's moon): 1,737.4 km (mean radius)

I mean, Krynn is barely larger than Luna (Earth's moon).

In the next part, we'll calculate the radius of each of Krynn's moon orbits.

And  yet, it changes nothing...

Apart from less space to put continents and seas (Ansalon is the same size as Western Europe...

Yeah, it's centered on France. Deal with it... 😝

... or 4 times the surface of Ukraine)...

... it doesn't really change anything because it is clear in the novels our heroes are dealing on a 1G gravity planet, which means Krynn's mass is the same as Earth's mass.

In other words, to have exactly the same gravity as Earth, if Krynn is smaller, it is also much denser.

And for the next physics calculations, this detail will make things much simpler.

P.S. Does it really change nothing?

From the simple calculations I intend to do later, it changes nothing.

First, if Krynn is more dense than Earth, then it makes sense its core is much more dense than Earth's core. We can imagine exotic metals, instead of the mix of nickel and iron that is the inner and outer core of Earth. Dragonmetal? Starmetal? Denser material would mean heavier than iron, which means, probably, more radioactive material than on Earth. Fun fact, the radioactive material of Earth's core is what gives the planet its heat. Would Krynn be warmer?

Also, from storytelling perspective, this changes a lot, because there isn't a lot of room on Krynn. It becomes a "small world", and relatively, it makes Ansalon larger, and thus, more important. Remember that the gods of Krynn seem obsessed with Ansalon, up to the point of sucker-punching it with a meteor.

If you are building a campaign where only the 2 first trilogies, it means the rest of Krynn can be explained as you want, and you won't need to spend a lot of time on it, because the overall surface is small. And it might be easy to repurpose the other continents (e.g. Taladas and Adlatum) into very exotic continents.

For example, one might be a land of dinosaurs, where the dragons reign supreme. Another might be a land of chaos, ravaged by the Graygem, where inhabitants are monstrous mutant-like creatures (this would let us repurpose the fiends, and more particularly, the demons, to inhabit this continent).

Another might be a wasteland where life isn't possible anymore, having been ravaged eons ago, and possibly the destination of my players a few campaign by now, to discover the first apostasis. Who knows?

😉