Wikipedia:Reference desk/Archives/Science/2023 July 4

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July 4

Lunar distance and Earth's rotation

Lunar distance (astronomy)#Orbital history

Laser measurements show that the average lunar distance is increasing, which implies that the Moon was closer in the past, and that Earth's days were shorter.

I understand why this implies that the Moon was closer in the past, but what does this have to do with the speed of Earth's rotation? The increasing distance means that the Earth-Moon centre of gravity is moving slightly farther from Earth's centre; does this affect the rotation speed? Nyttend (talk) 21:50, 4 July 2023 (UTC)[reply]

One explanation may be related to tidal locking and the fact that the Moon’s rate of rotation is locked to the Earth. As the Moon moves further from Earth, the Moon’s orbital speed and angular velocity about the Earth reduce in order to conserve mechanical energy. As the Moon’s angular velocity about the Earth reduces, tidal locking causes the Moon’s rate of rotation to decrease but also applies a brake to the Earth’s rate of rotation so the length of one day on Earth increases. Dolphin (t) 22:07, 4 July 2023 (UTC)[reply]

Mechanical energy isn't conserved; it's consumed by tidal heating. What is conserved is angular momentum. As the Moon moves away from the Earth, its orbital speed drops, but slower than the orbital radius increases, so its orbital angular momentum increases. This comes from the Earth's spin angular momentum (also a tiny bit from the Moon's spin angular momentum). The orbital period and both spin periods increase. PiusImpavidus (talk) 23:12, 4 July 2023 (UTC)[reply]

Given that the earth's rotation is gradually slowing down, is there a predictable point in time where it will be tidally locked to the moon just as the moon is tidally locked to the earth? ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:47, 5 July 2023 (UTC)[reply]

No. The simple theory predicts that the rate of rotation will get closer and closer to the Moon’s rate of revolution but will never actually reach it. One will approach the other asymptotically. Dolphin (t) 01:07, 5 July 2023 (UTC)[reply]

It would be asymptotic if the force depended in the speed of rotation, but it doesn't. These sources agree (oh, two of them are the same writer; but that's still two sources) that the predictable time Bugs asked for is about 50 billion years from now. However, by that time the Sun will have had its red giant phase and I don't think it's known for sure whether the Earth and Moon will still exist. --142.112.221.43 (talk) 07:06, 5 July 2023 (UTC)[reply]

Many years ago I read a piece which said that the day and synodic month would eventually stabilise at a length of 47 days as the day is now. Does that seem in the right ballpark? 2A02:C7B:301:3D00:10CE:42BE:967B:27F0 (talk) 11:10, 5 July 2023 (UTC)[reply]

"47 days as the day is now", i.e. one day then would be 47×24 hours long? Nyttend (talk) 23:27, 5 July 2023 (UTC)[reply]

Yes, I've also seen that number or something close to it, but I forget where. --142.112.221.43 (talk) 04:37, 8 July 2023 (UTC)[reply]

Models predict that liquid water will disappear on Earth in approximately one billion years.^[1] It seems that the estimate of 50 billion years for the Earth to tidally lock to the Moon^[2] is based on the assumption the oceans remain more or less as they are now.  --Lambiam 09:21, 6 July 2023 (UTC)[reply]

I'd guess that the presence of oceans makes little difference, but it is only a guess. The solid ground is affected by tides too. --142.112.221.43 (talk) 04:37, 8 July 2023 (UTC)[reply]

At least it should not be hard to compute the asymptote, given conservation of angular momentum: solve an expression in which the different ωs are equal. (By Kepler's third law, orbital distance can be expressed as a function of ω.) —Tamfang (talk) 19:09, 6 July 2023 (UTC)[reply]

@142.112: Info here [3]. 2A00:23C7:A103:CF01:F40E:89FF:FD2F:57D (talk) 12:34, 8 July 2023 (UTC)[reply]

I remember reading an article on this many years ago, and I do recall that after giving the 47 day figure it went on to say that the moon would move back towards the earth and eventually smash into it. 2A00:23C7:A103:CF01:F40E:89FF:FD2F:57D (talk) 12:41, 8 July 2023 (UTC)[reply]

Here goes. The angular momenta of Earth and Moon are the rotational momentum ${\displaystyle I_{\oplus }\omega _{\oplus }}$  of Earth and the orbital momentum ${\displaystyle I_{\text{☾}}\omega _{\text{☾}}}$  of the Moon, in which the ${\displaystyle I}$ -factors stand for moments of inertia. (We can ignore the Moon's own rotational momentum, since it is minute compared to the other two values.) The components are time-dependent but their present values are known and their sum ${\displaystyle L_{\text{tot}}=I_{\oplus }\omega _{\oplus }+I_{\text{☾}}\omega _{\text{☾}}}$  is preserved. We wish to find the value of ${\displaystyle \omega }$  for which ${\displaystyle I_{\oplus }\omega +I_{\text{☾}}\omega =L_{\text{tot}}.}$  While ${\displaystyle I_{\oplus }}$  is constant, this is not the case for ${\displaystyle I_{\text{☾}}=m_{\text{☾}}r_{\text{☾}}^{2},}$  in which ${\displaystyle m_{\text{☾}}}$  is the mass of the Moon and ${\displaystyle r_{\text{☾}}}$  is the radius of its orbit. Although ${\displaystyle m_{\text{☾}}}$  is constant, ${\displaystyle r_{\text{☾}}}$  is not. We can eliminate ${\displaystyle r_{\text{☾}}}$  by equating the Moon's orbital centrifugal acceleration with the gravitational centripetal acceleration:

${\displaystyle \omega _{\text{☾}}^{2}r_{\text{☾}}=Gm_{\oplus }r_{\text{☾}}^{{-}2},}$ 

in which ${\displaystyle G=6.674{\times }10^{{-}11}\,{\text{m}}^{3}{\text{kg}}^{{-}1}{\text{s}}^{{-}2}}$  is the gravitational constant of Newton's law of gravitation and ${\displaystyle m_{\oplus }}$  is Earth's mass. (This implies Kepler's third law. I find a discrepancy of 0.95% when I use the current values, so one or more of my sources are slightly off.) This results in the equation

${\displaystyle I_{\oplus }\omega +m_{\text{☾}}(G^{2}m_{\oplus }^{2}\omega ^{{-}1})^{1/3}=L_{\text{tot}}.}$ 

This can be rewritten as a quartic equation in ${\displaystyle \omega ^{1/3},}$  solvable algebraically, but only with an unwieldy formula. The best approach is to resort to a numerical method. We need numerical values for the various constants. The most problematic is the value of ${\displaystyle I_{\oplus }.}$  In general, the rotational moment of inertia of a sphere of radius ${\displaystyle R}$  and mass ${\displaystyle m}$  equals ${\displaystyle I=\alpha mR^{2},}$  in which the moment-of-inertia factor ${\displaystyle \alpha ,}$  a dimensionless quantity, depends on the density distribution of the mass. For a sphere of uniform density, ${\displaystyle \alpha ={\tfrac {2}{5}},}$  but the Earth's core is denser than the mantle. A value for the Earth found in the literature is ${\displaystyle \alpha _{\oplus }=0.331\,.}$ ^[4][5][6] Further using ${\displaystyle m_{\oplus }=5.972168{\times }10^{24}\,{\text{kg}},}$  ${\displaystyle R_{\oplus }=6371{\times }10^{3}\,{\text{m}},}$  current value of ${\displaystyle \omega _{\oplus }=2\pi (86400\,{\text{s}})^{{-}1},}$  ${\displaystyle m_{\text{☾}}=7.342{\times }10^{22}\,{\text{kg}},}$  current value of ${\displaystyle r_{\text{☾}}=3.844{\times }10^{8}\,{\text{m}},}$  and current value of ${\displaystyle \omega _{\text{☾}}=2\pi (2.3606{\times }10^{6}\,{\text{s}})^{{-}1},}$  we find

${\displaystyle L_{\text{tot}}=5.835{\times }10^{33}\,{\text{kg}}\,{\text{m}}\,{\text{s}}^{{-}1}+2.888{\times }10^{34}\,{\text{kg}}\,{\text{m}}\,{\text{s}}^{{-}1}=3.471{\times }10^{34}\,{\text{kg}}\,{\text{m}}\,{\text{s}}^{{-}1}.}$ 

Now we can solve the equation numerically. It has in fact several solutions: ${\displaystyle \omega =1.5194{\times }10^{{-}6}\,{\text{s}}^{{-}1},}$  corresponding to a new day and month of ${\displaystyle 47.86}$  old 24-hour days, and ${\displaystyle \omega =3.6313{\times }10^{{-}4}\,{\text{s}}^{{-}1},}$  corresponding to a new day and month of ${\displaystyle 4.8}$  hours. For the first, the new orbital radius for the Moon is ${\displaystyle 5.568{\times }10^{8}\,{\text{m}},}$  while the second gives ${\displaystyle 1.446{\times }10^{7}\,{\text{m}},}$  about ${\displaystyle 1.53}$  times the Roche limit. In the absence of forces speeding up the angular velocities, the second solution is spurious.  --Lambiam 07:08, 10 July 2023 (UTC)[reply]