OniLink

The Other Girl with the Gall

I'm Violet/OniLink. Trans and autistic and just kinda doing my best.

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I do informative Let's Plays on YouTube.

You can find me on FFXIV on Leviathan as Satora Lahnsi.

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OniLink
@OniLink

Ok this question has haunted my brain for a bit and it is very annoying so let's see if I can figure this out

For some reason I am unable to conceive of the proper way to add angular momenta in a complex system of bodies which both rotate and orbit each other but I think I figured it out also I think better when I think "out loud"

Ok so the moon is a sphere (ish) so I_moon approx = 2/5 m_moon r_moon^2 (from here when i say = i mean approx =)

With rotation speed w_moon, its angular momentum in its own frame is I_moon w_moon

But from the center of the earth, we need to use the parallel axis theorem to get I_moon orbit = I_moon + m_moon d_moon^2. Calculating its angular momentum from this perspective we get L_moon orbit = I_moon w_month + m_moon d_moon^2 w_month. But then this implicitly assumes that the moon is at the end of a "rod" effectively attaching it to the center of the earth, which then necessitates that the moon is rotating with w_moon = w_month.

Substitute w_moon for w_month on the "moon is rotating about its own axis" part of that angular momentum we get L_moon orbit = I_moon w_moon + m_moon d_moon^2 w_month. That makes sense. That also implies angular momentum adds in a very logical way. Which is something I could not find any confirmation on. It felt too simple, therefore it felt wrong.

Since angular momentum adds up nice and linearly, we can now add the Earth's angular momentum to the earth-moon system. Earth has I_earth = 2/5 m_earth r_earth^2, and contributes I_earth w_day angular momentum for a total of L_earth moon system = I_earth w_day + I_moon w_moon + m_moon d_moon^2 w_month.

Due to tidal forces, the earth and moon are slowly becoming tidally locked. Once that happens, w_day = w_moon = w_month = w_final, and L_earth moon system = (I_earth + I_moon + m_moon d_moon final^2) w_final.

Angular momentum is conserved, so:
(I_earth + I_moon + m_moon d_moon final^2) w_final = I_earth w_day + I_moon w_moon + m_moon d_moon^2 w_moon

We have one equation and two unknowns (d_moon final and w_final). Tidal forces do result in a loss of some of that energy to heat, but I'm not certain how to estimate that currently so we will assume no heat is produced and all energy remains in the motion of the system.

No heat is produced:
E_initial = E_final
E_initial = 1/2 (I_earth w_day^2 + I_moon w_moon^2 + m_moon d_moon^2 w_month^2)
E_final = 1/2 (I_earth + I_moon + m_moon d_moon final^2) w_final^2

Now we have two equations with two unknowns. Now for some hard numbers:
m_earth = 5.972168 e24 kg
r_earth = 6.3710 e6 m
m_moon = 7.342 e22 kg
r_moon = 1.7374 e6 m
d_moon = 3.84399 e8 m
w_earth = 2 pi / (23hr 56min 4.100sec) = 7.292 e-5 Hz
w_moon = 2 pi / (27.321661day) = 2.6616995 e-6 Hz
w_month = 2 pi / (27.321661day) = 2.6616995 e-6 Hz
I_earth = 0.3307 m_earth r_earth^2 = 8.016 e37 kg m^2 [from Wikipedia MOI factor]
I_moon = 0.3929 m_moon r_moon^2 = 8.708 e34 kg m^2 [from Wikipedia MOI factor]
m_moon d_moon^2 = 1.08487 e40 kg m^2

And now we can find initial angular momentum and kinetic energy
L_initial = I_earth w_day + I_moon w_moon + m_moon d_moon^2 w_moon = 5.845 e33 Js + 2.318 e29 Js + 2.8876 e34 Js = 3.4721 e34 Js
E_initial = 1/2 (I_earth w_day^2 + I_moon w_moon^2 + m_moon d_moon^2 w_month^2) = 2.131 e29 J + 3.084654 e23 J + 3.843 e28 J = 2.5153 e29 J

Moon's rotation is kinda negligible but that's fair. Earth dominates kinetic energy but moon orbit dominates angular momentum. Now to solve the pair of equations:

(I_earth + I_moon + m_moon d_moon final^2) w_final = 3.4721 e34 Js
1/2 (I_earth + I_moon + m_moon d_moon final^2) w_final^2 = 2.5153 e29 J

This isn't too bad. We recognize I_earth + I_moon + m_moon d_moon final^2 = I, consistent across both equations. Then
I w_final = 3.4721 e34 Js
1/2 I w_final^2 = 2.5153 e29 J

w_final = 3.4721 e34 Js / I -> w_final^2 = 1.2055 e69 J^2 s^2 / I^2
w_final^2 = 5.0306 e29 J / I
1.2055 e69 J^2 s^2 / I^2 = 5.0306 e29 J / I
I 1.2055 e69 J^2 s^2 = I^2 5.0306 e29 J
I = 1.2055 e69 J^2 s^2 / (5.0306 e29 J) = 2.3963 e39 kg m^2
We can solve this out for d_moon final to get
m_moon d_moon final^2 = I - I_earth - I_moon = 2.3161 e39 kg m^2
d_moon final^2 = 3.1545 e16 m^2
d_moon final = 1.7761 e8 m, or actually significantly closer to the earth than it is presently. I believe I have done a mistake somewhere as it should be getting further away. The mistake has to be in the fact that the moon's orbital angular momentum is dominating the earth's rotational angular momentum. It's past midnight so maybe I've done something quite stupid in here.

Whatever, let's find that tidally locked day.

w_final = 3.4721 e34 Js / I = 1.4489 e-5 Hz = 2 pi / day length
day length = 5 days 27 min

See that makes sense. It makes sense that the earth's days will get longer. The earth is actively slowing its rotation due to tidal friction with the moon, which orbits the earth slower than the earth rotates. What doesn't make sense is the moon getting closer. Or maybe it does make sense? For the earth to slow down, the moon needs to speed up. If its orbit drifts further away... it would slow down. So it would need to get closer. But I critically assumed that energy would be conserved, no losses to heat. So that would quite possibly account for the discrepancy with observed reality that the moon is drifting away from us.

How account for energy losses due to friction 😭 why are my physics skills so rusty 😭


OniLink
@OniLink

waiting for roommate to get out of the shower so i can shower so let's do another calculation

continuing from above, i'm curious how the moon would look from earth in this hypothetical situation. currently the moon is (d_moon - r_earth) = 3.7803 e8 m from the surface of the earth. with a radius r_moon = 1.7374 e6 m, its angular diameter in the sky is 2 atan(r_moon / (d_moon - r_earth)) = 0.53 degrees. If it were to come closer to a distance of 1.7761 e8 m from the earth's center or 1.7124 e8 m from the earth's surface, its angular diameter would be 1.16 degrees. A little over twice as big. Big moon. Very big moon.


OniLink
@OniLink

I got like 4 hours of sleep and I have remembered a pretty fucking important aspect I forgot to include in my calculations: GRAVITATIONAL POTENTIAL ENERGY. YOU FOOL. I'LL FIX IT WHEN I GET UP


OniLink
@OniLink

Ok so my math is fine except for energy conservation. So let's start from there.

E_initial = 1/2 (I_earth w_day^2 + I_moon w_moon^2 + m_moon d_moon^2 w_month^2) - G m_earth m_moon / d_moon E_final = 1/2 (I_earth + I_moon + m_moon d_moon final^2) w_final^2 - G m_earth m_moon / d_moon

Adding in the gravitational potential energy to my original initial energy calculations gives

E_initial = 2.5153 e29 J - 9.359 e26 J = 2.5059 e29 J

Then for the energy conservation we have

E_final = 1/2 (I_earth + I_moon + m_moon d_moon final^2) w_final^2 - G m_earth m_moon / d_moon final = 2.5059 e29 J

Our equations are now
(I_earth + I_moon + m_moon d_moon final^2) w_final = 3.4721 e34 Js
1/2 I w_final^2 - 3.598 e35 J m / d_moon final = 2.5059 e29 J

This is unfortunately a bit more complicated as d_moon final is part of calculating I. But we'll see what happens! Isolating w_final...

1/2 I w_final^2 = 2.5059 e29 J + 3.598 e35 J m / d_moon final
w_final^2 = 5.0118 e29 J / I + 7.196 e35 J m / (I d_moon final)
I w_final = 3.4721 e34 Js
w_final = 3.4721 e34 J s / I -> w_final^2 = 1.2055 e69 J^2 s^2 / I^2
5.0118 e29 J / I + 7.196 e35 J m / (I d_moon final) = 1.2055 e69 J^2 s^2 / I^2
I [5.0118 e29 J + 7.196 e35 J m / d_moon final] = 1.2055 e69 J^2 s^2
I d_moon final 5.0118 e29 J + I 7.196 e35 J m = d_moon final 1.2055 e69 J^2 s^2

Expanding out I, we have

(I_earth d_moon final + I_moon d_moon final + m_moon d_moon final^3) 5.0118 e29 J + (I_earth + I_moon + m_moon d_moon final^2) 7.196 e35 J m = 0

It's a fucking cuuuubiiiiiiiiiic

d_moon final^3 [m_moon 5.0118 e29 J] + d_moon final^2 [m_moon 7.196 e35 J m] + d_moon final [(I_earth + I_moon) 5.0118 e29 J - 1.2055 e69 J^2 s^2] + [I_earth + I_moon] 7.196 e35 J m = 0
d_moon final^3 [3.6796 e52 J^2 s^2 / m^2] + d_moon final^2 [5.284 e58 J^2 s^2 / m] + d_moon final [4.0218 e67 J^2 s^2 - 1.2055 e69 J^2 s^2] + 5.775 e73 J^2 m = 0

deferring to wolfram alpha because cubics gives three solutions

d_moon final = -1.78701 e8 m; clearly incorrect, the moon cannot be "so far inside the earth it is out the other side and at... pretty close to the original height i calculated"
d_moon final = 49,558.9 m; wow that sure is close to the earth's surface! they would be literally intersecting!
d_moon final = 1.77216 e8 m; that is sounding more like the actual answer. This is also only a very marginal difference from the original calculation. GPE had a minimal effect on the problem. >:(


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