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.