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Would this be because tidal deceleration causing the Earth to spin faster or are there other actions in play that I haven't considered? Would the Earth even spin faster because of the tidal deceleration?

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    $\begingroup$ doesn't sound right to me, if the moon had a retro grade orbit its gravitational pull would still be the same on the face of the earth wouldn't it? $\endgroup$ – Daz Hawley Dec 23 '14 at 18:09
  • $\begingroup$ What's your source? $\endgroup$ – Daz Hawley Dec 23 '14 at 18:12
  • $\begingroup$ A kid in my class asked what would happen if the moon orbited the other way. My teacher said that nothing would really change except that the tides would seem to be 7% more often. I have no idea what source. $\endgroup$ – Marcello Nicoletti Dec 23 '14 at 18:15
  • $\begingroup$ @DazHawley how is that even related? $\endgroup$ – hobbs Dec 24 '14 at 5:32
  • $\begingroup$ @hobbs usc.edu/org/seagrant/Education/IELessons/Docs/MoonAndTides.pdf $\endgroup$ – Daz Hawley Dec 24 '14 at 8:03
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Suppose the Moon didn't orbit the Earth at all, so it just stayed at some fixed point while the Earth rotated underneath it:

Stationary Moon

In this case every point on the equator would pass directly under the Moon every 24 hours, and we'd get a high tide every 24 hours. (There's another high tide when we're exactly on the opposite side of the earth to the moon, but let's ignore that for now.)

But the Moon does orbit the Earth in a prograde orbit, and that means after 24 hours the Earth has revolved once but the Moon has also moved on a bit:

Prograde orbit

So to get directly under the Moon again takes a bit longer than 24 hours. From memory it's about 45 minutes longer, so the interval between every two high tides is about 24 hours and 45 minutes.

Now suppose the Moon moved in a retrograde orbit:

Retrograde orbit

This time as you're revolving around with the Earth the Moon moves towards you, so it takes about 45 minutes less to get directly under the Moon again. If the Moon were in a retrograde orbit every two high tides would be separated by about 23 hours and 15 minutes.

That's why tides would be more frequent if the Moon were in a retrograde orbit. The diffrence between the prograde and retrograde orbit tide timings is about 90 minutes per 24 hours, which is about 7%. That's where your teacher gets the 7% figure from.

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  • $\begingroup$ This is basically the difference between a sidereal and synodic day (wrt moon) $\endgroup$ – k_g Dec 23 '14 at 21:49
  • $\begingroup$ Do your diagrams portray the rotations as viewed from above the South Pole? $\endgroup$ – DJohnM Dec 23 '14 at 23:39
  • $\begingroup$ @User58220 Yes. Most major bodies in the Solar System rotate and revolve clockwise as viewed from Earth's South. $\endgroup$ – user10851 Dec 24 '14 at 3:35
  • $\begingroup$ Very good explanation. The only quibble I have is that there are not one, but two 'tide bulges': One on Earth under the moon, and one on Earth opposite the moon. So high tide happens twice as often: about once every 12-and-a-bit hours, instead of once every 24-and-a-bit. Picture of the double bulge is here: en.wikipedia.org/wiki/Tide $\endgroup$ – Esteis Dec 24 '14 at 11:52
  • $\begingroup$ @Esteis: I did mention just after the first diagram that there are two high tides in 24 hours, but I played this down because I wanted to focus on my main message. $\endgroup$ – John Rennie Dec 24 '14 at 11:55
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I think you're overthinking what is meant by the claim.

Consider the Earth spinning for one sidereal day, about 23.9 hours. In the rotating frame of the Earth, the Moon has made a complete orbit overhead, causing the tidal forcing function to have two maxima and two minima.

But did the Moon make exactly one complete cycle from this point of view? No.

The Moon goes around the Earth once every sidereal month, about 27.3 days. So after one sidereal day, the Earth has spun around once, but the Moon, being prograde, has advanced a little. Thus the period of the Moon passing overhead is a bit longer than 23.9 hours. In a retrograde orbit this effect would be reversed, and the Moon would pass overhead more frequently than once a sidereal day.

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The tides would have a faster rhythm, but barely. Most of the 'tidal rhythm' comes from the rotation of the earth (24 hours) not the orbit of the moon (700 hours).

Indeed in that case tidal deceleration would cause the earth to spin faster. That energy would come from the orbit of the moon, which would be slowed by the tidal bulge.

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  • $\begingroup$ Ok, that's what I was expecting. My teacher said only 7% faster as a rough estimate. $\endgroup$ – Marcello Nicoletti Dec 23 '14 at 18:32
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    $\begingroup$ How would tidal deceleration cause the earth to spin faster with a retrograde moon? $\endgroup$ – Peter Shor Dec 23 '14 at 23:37
  • $\begingroup$ @CupricWolf it's quite a good estimate in fact; (24 hours + 50 minutes) / (24 hours - 50 minutes) is 1.0719, so the lunar component of tides would be about 7.19% faster. $\endgroup$ – hobbs Dec 25 '14 at 4:41

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