View Full Version : Earth causing lunar 'tides'

wd40

2012-Sep-08, 05:38 PM

The Moon orbiting the Earth causes tides on Earth, sometimes moving water 100s of yards.

In theory, if the Moon was a smaller identical Earth, with the same atmosphere, water oceans and continents, then can one say that in accordance with Mach's Principle & Relativity, that whether it's the Earth going round the Moon or vice versa, that identical tides will be caused on the Moon, albeit with smaller movements and taking in to account the reduced gravity?

Jeff Root

2012-Sep-08, 06:11 PM

I'm pretty sure the tidal forces of Earth on the Moon are

much greater than the tidal forces of the Moon on Earth.

The effect is greater, too: It locked the Moon's rotation

to the Earth.

-- Jeff, in Minneapolis

Jason Thompson

2012-Sep-08, 09:56 PM

Yes, Earth does cause tides on the Moon. In fact that is why the Moon is now locked into showing us one face all the time. The tidel bulge of the Moon 'hangs' down towards Earth.

wd40

2012-Sep-08, 10:39 PM

If there were was an atmosphere. oceans and land masses on the Moon, what would a lunar tide look like?

Would it vary "daily"?

Or would the water be perpetually heaped to one area?

grapes

2012-Sep-08, 11:46 PM

If there were was an atmosphere. oceans and land masses on the Moon, what would a lunar tide look like?

Would it vary "daily"?

What would you mean by "daily"? Would you mean, over the time period that the earth would appear to go around the moon, from the point of view of someone on the moon? :)

wd40

2012-Sep-09, 12:00 AM

From the point of someone on the Moon.

Van Rijn

2012-Sep-09, 12:47 AM

From the point of someone on the Moon.

From the point of view of someone on the Moon, what do they see the Earth do through a lunar day? Contrast that with what someone on the Earth sees the Moon doing through an Earth day.

wd40

2012-Sep-09, 01:13 AM

17507

http://en.wikipedia.org/wiki/Tide

My question is: would the daily tidal effect on the Earth's waters due to the Moon be mirrored vice versa if the Moon had geographically identical oceans and continents, albeit reduced in size?

Jeff Root

2012-Sep-09, 01:25 AM

The Sun would raise tides twice every lunar day (month)

very much like it does on Earth. I'm sure the longer period

would make some kind of difference, but that might be

obscured by other differences.

The tidal bulges raised in water by the Earth would be

much higher, but they wouldn't move around very much.

The water would be heaped up on both the near and far

sides. Tides on Earth are ahead of the Moon by quite

a large distance, not directly under the Moon, but tides

on the Moon would be very close to directly under the

Earth (and directly opposite the Earth).

Tides raised by the Moon on Earth's oceans are on the

order of a foot high. I'll guess that tides on the Moon

raised by the Earth would be on the order of a hundred

feet high. But as I said, they wouldn't move much.

-- Jeff, in Minneapolis

Jeff Root

2012-Sep-09, 01:33 AM

My question is: would the daily tidal effect on the

Earth's waters due to the Moon be mirrored vice

versa if the Moon had geographically identical

oceans and continents, albeit reduced in size?

Why should it be?

Your attachment didn't work. That may be a result

of problems with the recent forum software change.

-- Jeff, in Minneapolis

Jeff Root

2012-Sep-09, 01:39 AM

The Sun would raise tides twice every lunar day (month)

very much like it does on Earth. I'm sure the longer period

would make some kind of difference, but that might be

obscured by other differences.

Come to think of it, while the pile of water directly

caused by the Sun's tide would be similar (though

probably higher because of the Moon's lower gravity),

the much slower rotation means it wouldn't pile up

against shorelines anywhere near as much.

-- Jeff, in Minneapolis

pzkpfw

2012-Sep-09, 01:49 AM

I can imagine that - if it had this water and such - that due to the effects of the tides (the effects of the water itself "sloshing around") the Moon might not yet be tidally locked.

That is, there'd perhaps not be the same uneven mass distribution that helps get it tidally locked (so "soon") in the first place.

grapes

2012-Sep-09, 03:47 AM

From the point of someone on the Moon.

A "day" like that would last more than a million years. :)

The Sun would raise tides twice every lunar day (month)

very much like it does on Earth. I'm sure the longer period

would make some kind of difference, but that might be

obscured by other differences.

The sun tidal force on earth is half the lunar tidal force, on the moon it would be an eighth of that.

Tides raised by the Moon on Earth's oceans are on the

order of a foot high.

Depends on what you mean by order--they're around a meter high.

I'll guess that tides on the Moon

raised by the Earth would be on the order of a hundred

feet high.Because the Earth is around 80 times more massive than the moon? But the moon has one fourth the radius.

chornedsnorkack

2012-Sep-09, 09:24 AM

Letīs normalize the upwards acceleration of mean lunar high tide as 1,00.

Then the downwards acceleration of mean lunar low tide is -0,50 (The high tide bulges are on sublunar and antilunar spots, the low tide circles the globe along meridian - it is always low tide on both poles because Moon is always on horizon).

Upwards acceleration of mean solar high tide is then +0,45 - and it also is always low tide on both poles.

The upwards acceleration of lunar high tide varies between +0,85 and +1,15 due to eccentricity of Moon.

NOW, goto Moon.

If Moon were freely rotating, the tidal acceleration gradient on Moon should be 81 times the gradient on Earth, because Earth is 81 times the mass.

Since Moon itself is 0,27 times the diametre of Earth, the upward acceleration due to tides would be 22 times bigger on subterran and antiterran points than it is on sublunar and antilunar points. Since the general acceleration due to gravity is 6 times smaller on Moon, the height of equilibrium tidal wave would be 130 times higher.

If Moon were tidally locked on a perfectly circular orbit then the only tides would be solar. Same acceleration gradient as on Earth, 0,27 times diametre, 1/6 gravity means 1,6 times the height of solar tides on Earth, which is about 0,7 times the height of lunar tides on Earth.

But Moon is tidally locked on an elliptic orbit.

The upwards acceleration on the subterran point is on average +22 times the mean lunar tide on Earth - but it varies by +- 15%, so +- 3,3 - whereas the range on earth equator is from +1,0 to -0,5. Thus 4,4 times the range of lunar tides on Earth by acceleration - and 26 times the range by height. Note that since the maximum range of tides on Earth is 1,6 times the mean lunar - 0,15 times for perigean tides, 0,45 times for spring solar tides, whereas solar tides on Moon are negligible - the equilibrium tidal height is about 16 times that of perigean spring tides on Earth.

But the range in acceleration is even bigger because, remember, the subterran point moves by libration. If the tidal acceleration at subterran point is +25,3, then at apogee it would be +18,7 - at a different subterran point. (Or would these actually be the same? Can anyone figure out the effects?)

Also note that while on Earth, there are 2 high tides a day (sublunar and antilunar), on Moon there is 1 high tide per month (perigee at subterran and antiterran points, apogee at the edge...). On Earth, tidal wave in oceans is not fast enough to catch the acceleration bulge, but is left behind. On Moon, even in quite shallow oceans the waters can flow to equilibrium with acceleration.

wd40

2012-Sep-18, 12:55 AM

Is there any way that the Moon can cease, catastrophically or gradually, being tidally locked to the Earth i.e. that the far side of the Moon become visible from Earth?

Was the far side of the Moon ever visible from Earth in the past?

pzkpfw

2012-Sep-18, 01:03 AM

... Was the far side of the Moon ever visible from Earth in the past?

It wasn't always tidally locked, so, yes.

(Visible by whom?)

Is there any way that the Moon can cease, catastrophically or gradually, being tidally locked to the Earth i.e. that the far side of the Moon become visible from Earth?...

Impact by asteroid??? It'd be something "impressive".

Jens

2012-Sep-18, 03:55 AM

Is there any way that the Moon can cease, catastrophically or gradually, being tidally locked to the Earth i.e. that the far side of the Moon become visible from Earth?

Just land a bunch of astronauts, and ask them to all start running in one direction at the same time. Then, when the moon hits the high point of its libration, tell them to start running in the other direction, and repeat as long as necessary. In about a billion years it would probably work.

grapes

2012-Sep-20, 03:25 AM

NOW, goto Moon.

If Moon were freely rotating, the tidal acceleration gradient on Moon should be 81 times the gradient on Earth, because Earth is 81 times the mass.

Since Moon itself is 0,27 times the diametre of Earth, the upward acceleration due to tides would be 22 times bigger on subterran and antiterran points than it is on sublunar and antilunar points. Since the general acceleration due to gravity is 6 times smaller on Moon, the height of equilibrium tidal wave would be 130 times higher.

The variation in height of the equipotential surface (Equation 3.101, Physics of the Earth, 3rd ed., Stacey) is given by:

\delta a = \frac{W_2}{g} = \frac{m}{M} \Bigl( \frac{a}{R} \Bigr)^3 a \Bigl( \frac{3}{2} cos^2 \psi - \frac{1}{2} \Bigr) *

Using that formula, the height of the tide on the moon generated by the earth is 36.6 times the height of the tide on the earth generated by the moon.

W is the disturbance to the static potential, g is acceleration of local gravity, m is the mass of the disturbing body, M is the mass of the body, a is the radius of the body, and R is the distance between the two bodies.

grapes

2012-Sep-23, 08:51 AM

Depends on what you mean by order--they're around a meter high.

The variation in height of the equipotential surface (Equation 3.101, Physics of the Earth, 3rd ed., Stacey) is given by:

\delta a = \frac{W_2}{g} = \frac{m}{M} \Bigl( \frac{a}{R} \Bigr)^3 a \Bigl( \frac{3}{2} cos^2 \psi - \frac{1}{2} \Bigr) *

Using that formula, I get .534 meters (21 inches), closer to Jeff's value than mine.

In terms of the ocean tide, though, that's further reduced by the amount the solid body responds to the change in potential, measures by the Love number. Earth Love number h2 is about .62, so the ocean tide is about .534 times .38, which is .2 meters (8 inches).

Powered by vBulletin® Version 4.2.3 Copyright © 2019 vBulletin Solutions, Inc. All rights reserved.