Tides due to the moon vs. the sun?

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  • #26
Nereid
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To answer the earlier question - coastlines do matter, esp north-south ones, and bays. I don't recall how much they matter, in terms of deviations from a homogeneous Earth, or one with only oceans.

You can get an idea of how big the effect should be - OOM - by working out a) the average water tide, b) the average land tide, c) the difference in gravitational field (at a distance of the average orbit of the Moon) between an ocean covered Earth and one with no ocean, then d) put in various lags.

To see how big the lag is, consult your favourite tide table, and compare the times of local high and low tide with the times of meridian passage of the Moon. Of course, there are several adjustments that you'd need to make ...
 
  • #27
selfAdjoint
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In some places coastline effects are large enough to reduce the normal two high tides a day to one.
 
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Yes, semi-diurnal tides can be reduced to diurnal tides. For example in the South China Sea (see picture attach).

This is a very complex phenomenon, in which many factors like coastal configuration and ocean depth play a role. But what is the basic principle? Is it that the tidal bulge moves over the earth (for example the Pasific Ocean), then meets a coast line (like the Philipines), and has to push itself through the straits as through a funnel, leading to reduced tides??

If that is the case, then why does not the whole of the South China Sea have diurnal tide? Why are there these little red spots where tides are semi-diurnal (see attach)? And why on that locations?

And, one last question about the forces: If centrifugal force is equal on all points on earth and the gravitational force is different on the near and far side of the earth, why is then the resultant tidal force still equal on both sides, leading to tidal bulges with the same height (is that so? are they of equal height?)? Or is that the story of the boy and the girl on the skateboard? Are actually the gravitational forces equal on both sides?
 

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  • #29
selfAdjoint
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And, one last question about the forces: If centrifugal force is equal on all points on earth

It's not. Centrifugal force is proportional to the distance from the axis of rotation. It's a maximum at the equator, zero at the poles, and has a unique value for each circle of latitude.

and the gravitational force is different on the near and far side of the earth, why is then the resultant tidal force still equal on both sides, leading to tidal bulges with the same height (is that so? are they of equal height?)?[/b]

Yes, the bulges are the same height. On the near side to the moon the water is closer to the moon than the earth and is more strongly pulled, hence it is pulled up into a bulge. On the far side the earth is closer to the moon than the water, so it is pulled away from the water, making a water bulge. Tidal forces are all about differential effects of gravitation at different distances.
 
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Huston we have a problem. We have landed on a very big planet with very strong winds, They are called solarwinds. Have we forgotten something? I get a strong feeling we have. Perhaps the sun is a bit warm.

-Keep searching for intelligent life. Aaouch, sun of a star.
 
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Yes, the bulges are the same height. On the near side to the moon the water is closer to the moon than the earth and is more strongly pulled, hence it is pulled up into a bulge. On the far side the earth is closer to the moon than the water, so it is pulled away from the water, making a water bulge. Tidal forces are all about differential effects of gravitation at different distances.


Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the earth and moon revolving around their centre of mass?
 
  • #32
sylas
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Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the earth and moon revolving around their centre of mass?

It's not centrifugal force; that has nothing to do with it.

The answer you have quoted is correct and it explains the bulges on both sides and why they are about equal. Read it again. It says: Tidal forces are all about differential effects of gravitation at different distances. It stretches out a fluid sphere into an ellipsoid shape. Relevant to the center of the sphere, you get bulges on each side. There's a difference between the Moon's gravity at the center of the Earth and the nearside bulge, and there's about the same difference between the Moon's gravity at the center of the Earth and the far side bulge.
 
  • #33
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I found this to be a very insightful and interesting undergraduate thesis on tides by a world-class student (currently I think a PhD student at Caltech).

http://www.gps.caltech.edu/~carltape/research/pubs/thesis/tides_CHT_thesis.pdf" [Broken] (6.7 MB)
 
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  • #34
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Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the earth and moon revolving around their centre of mass?
That is a common but completely erroneous explanation. First off, what centrifugal force?

Secondly, and more importantly, suppose an object of the Moon's mass is falling straight toward the Earth with zero tangential velocity. There is no centrifugal force in this case; the object is falling straight into the Earth. At the point in time that this object reaches the Moon's orbital distance the tidal forces will be exactly equal to those exerted by the Moon. The tidal forces exerted by the Moon has nothing per se to do with Moon's orbit. It is solely a function of where the Moon is.

So, how to explain that there are two bulges? The answer is simple. The Earth as a whole is accelerating gravitationally toward the Moon:

[tex]a_e = \frac {GM_m}{{R_m}^2}[/tex]

where ae is the acceleration of the Earth toward the Moon, Mm is the mass of the Moon, and Rm is the distance between the centers of the Earth and the Moon.

The distances between the center of the Moon and the points on the surface of the Earth directly opposite the Moon and directly between the Earth and Moon are Rm+re and Rm-re, respectively. Here re[/i] is the radius of the Earth

Because these points are a bit further from/closer to the Moon than the center of the Earth, the acceleration toward the Moon at these points will be slightly different than that of the Earth as a whole. In particular,

[tex]a_p = \frac {GM_m}{(R_m\pm r_e)^2}[/tex]

It is the difference in acceleration that is important here. Calculating this,

[tex]a_{p,\text{rel}} = \frac {GM_m}{(R_m\pm r_e)^2} - \frac {GM_m}{{R_m}^2}
\approx \mp\, 2\,\frac{GM_mr_e}{R_m^3}[/tex]

The point on the surface of the Earth directly between the centers of the Earth and the Moon experiences a bit more acceleration toward the Moon that does the Earth as a whole, so this differential acceleration is directed toward the Moon -- that is, away from the center of the Earth. The point on the surface of the Earth directly opposite the Moon experiences a bit less acceleration toward the Moon that does the Earth as a whole, so this differential acceleration is directed away from the Moon -- which is, once more, away from the center of the Earth. The Moon pulls the submoon point and its antipode apart, like a piece of taffy. For the points on the surface of the Earth where the Moon is at the horizon, the action is to squeeze those points inward a tiny bit.

For the Moon these tidal forces are very, very small. For a neutron star or black hole they are not. There is a cool technical term for how black holes pull the extremes of some object apart and squeeze in at the middle: http://en.wikipedia.org/wiki/Spaghettification" [Broken].
 
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  • #35
russ_watters
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Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the earth and moon revolving around their centre of mass?
The easiest way to deal with this is to get rid of the rotational motion altogether. Without the moon being in orbit and the earth rotating, the tides would still be there (for a little while, anyway - until the moon crashed into the earth!).
 
  • #36
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The easiest way to deal with this is to get rid of the rotational motion altogether. Without the moon being in orbit and the earth rotating, the tides would still be there (for a little while, anyway - until the moon crashed into the earth!).

Exactly! The only thing necessary condition for tides (or differential acceleration) is that the object be in free fall. The effect of rotation enables us to "see" the tides by standing in one place, and the effect of orbital free fall keeps the whole experience going on and on.
 
  • #37
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Exactly! The only thing necessary condition for tides (or differential acceleration) is that the object be in free fall.
The only necessary condition for tides to exist is that the object be where it is. It does not have to be in free fall. Outfit the Moon with some BIG honking rocket engines that are capable of making the Moon hover 384,399 km away from the center of the Earth without orbiting and the tides will still exist.
 
  • #38


Alright, so this topic is old. But there is one thing I've been puzzling over for the past half day.

Say we ignore the sun's effect on the tides and just went with the moon-earth system. I understand why the tides happen, but I don't understand why, in an idealized model of the earth (sphere covered completely with water , no variation in water depth), the tides on either side of the earth are equal.
 
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