Tides in sea

Main Question or Discussion Point

Tides in sea are caused not by moon but by sun. Though the force of attraction caused by sun is much more than that by moon.

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Doc Al
Mentor
Tides in sea are caused not by moon but by sun.
Why do you think that?

olivermsun
Was the OP a question?

Was the OP a question?
More of an assertion really...

you made a small mistake. you mistaken sun with moon.
Sun is at huge distance and big. so fairly it attracts a large part of crust with fairly equal force.
but moon is at small distance and is small. so its force is more concentrated on a part of earths crust causing tides.

you made a small mistake. you mistaken sun with moon.
Sun is at huge distance and big. so fairly it attracts a large part of crust with fairly equal force.
but moon is at small distance and is small. so its force is more concentrated on a part of earths crust causing tides.
Nope, try again.

AlephZero
Homework Helper
Nope, try again.
Actually, I think Phyreak has the kernel of the right answer there, though it's not very well explained in English-as-a-foreign-language.

Actually, I think Phyreak has the kernel of the right answer there, though it's not very well explained in English-as-a-foreign-language.
Right track, yes. Right answer, no.

I can sort of see what he's going for, but the wording is way out. It is the choice of words that make me feel he's further out than it appears.

Drakkith
Staff Emeritus
you made a small mistake. you mistaken sun with moon.
Sun is at huge distance and big. so fairly it attracts a large part of crust with fairly equal force.
but moon is at small distance and is small. so its force is more concentrated on a part of earths crust causing tides.
Also, the force of gravity from the moon on the earth is MORE than the sun is because of the extreme distance of the sun compared to the moon. (I think at least) The force due to gravity falls off greatly with distance. If you double the distance you have 1/4 the force. I forget the technical term for describing this, something like falling off at the square of the distance or something.

Also, the force of gravity from the moon on the earth is MORE than the sun is because of the extreme distance of the sun compared to the moon. (I think at least) The force due to gravity falls off greatly with distance. If you double the distance you have 1/4 the force. I forget the technical term for describing this, something like falling off at the square of the distance or something.
The earth revolves around the sun, so the sun's gravity is greater on the earth than the moon's gravity on the earth. However, the earth's gravity is greater on the moon than the sun's gravity is on the moon, so the moon rotates around the earth.

Drakkith
Staff Emeritus
The earth revolves around the sun, so the sun's gravity is greater on the earth than the moon's gravity on the earth. However, the earth's gravity is greater on the moon than the sun's gravity is on the moon, so the moon rotates around the earth.
I understand what you are saying, and I hope I wasn't interpreted as saying that the earth should orbit the moon. I haven't done any math, but does the force of gravity at this distance from the moon have more of an influence than the sun does at its distance? Obviously the earth would never orbit the moon simply due to the mass difference between the two. What's the value of the force of gravity from the sun on the earth, and of the moon on the earth?

Ugh, now I've gone and confused myself...

olivermsun
It's the gravitational gradient that matters to a tide, so the sun can do a perfectly good job of "capturing" the earth while the moon stirs up a hefty tide. I think that's what some of the posters have been describing above.

Rather than forces, I think it's easier to consider fields, and think of the fields as acting on test masses. So the distance from the moon to the earth is .002573 astronomical units. The distance from sun to earth is 1 astronomical unit. So based on distance alone, the ratio of the field created by the moon at earth divided by the field created by the sun at earth should be:

$$\frac{M_{moon}/(.002573)^2}{M_{sun}/1^2}=1.5105*10^5 \left(\frac{M_{moon}}{M_{sun}}\right)$$
So the sun must around 100,000 times the size of the moon to be as influential.

Drakkith
Staff Emeritus
It's the gravitational gradient that matters to a tide, so the sun can do a perfectly good job of "capturing" the earth while the moon stirs up a hefty tide. I think that's what some of the posters have been describing above.
Ah ok. I see what you mean now.

So the sun must around 100,000 times the size of the moon to be as influential.
Sun's mass is about 27000000 times greater than that of the Moon! Even the larger distance cannot compensate for it. If you apply Newton's law to point masses of Earth, Moon and Sun you will find that Sun attracts Earth much more than Moon attracts Earth.
As was already pointed out, the tidal forces do not follow simple inverse square law, but rather inverse cube law.

However Sun DOES have influence on Earth's tides. It is smaller than Moon's influence for sure, but it is there.

I think for understanding tides, it's very helpful to understand that a tide is kind of less a 'lifting' of the water due to the gravity of the moon, and more of a wave. It's a wave on the surface of the ocean where the 'crest' of the wave is, more or less, at the point on the surface of the ocean which is closest to the moon.

This wave is hundreds or thousands of miles across (from east to west; that is, the wavelength of the wave is like a thousand miles), but the height/amplitude of the wave is only a few feet high. The "rising tide" effect happens as the crest of the wave gets nearer and nearer to the shore. Just like when 'regular' small waves (which might be 10 feet across) come ashore, they seemingly 'rise up' - because they are in motion, the water is getting shallower, and so the shallow coast line 'pushes' the wave upwards as it moves in).

The gravitational attraction from the moon, while fairly weak, is *strong enough* to cause the water to flow and create the "tidal wave", and the tidal wave then causes the tide to rise and fall. At least, that's how I remember it being explained to me, and makes sense to me.

I think for understanding tides, it's very helpful to understand that a tide is kind of less a 'lifting' of the water due to the gravity of the moon, and more of a wave. It's a wave on the surface of the ocean where the 'crest' of the wave is, more or less, at the point on the surface of the ocean which is closest to the moon.

This wave is hundreds or thousands of miles across (from east to west; that is, the wavelength of the wave is like a thousand miles), but the height/amplitude of the wave is only a few feet high. The "rising tide" effect happens as the crest of the wave gets nearer and nearer to the shore. Just like when 'regular' small waves (which might be 10 feet across) come ashore, they seemingly 'rise up' - because they are in motion, the water is getting shallower, and so the shallow coast line 'pushes' the wave upwards as it moves in).

The gravitational attraction from the moon, while fairly weak, is *strong enough* to cause the water to flow and create the "tidal wave", and the tidal wave then causes the tide to rise and fall. At least, that's how I remember it being explained to me, and makes sense to me.
So would this imply that the angular speed at which the moon rotates around the earth (about 24 hours for one revolution) is equal to the angular speed of the tidal wave, since the crest is always more or less on the surface of the ocean closest to the moon? If you dip a stick into the water, isn't the speed at which ripples propagate (the speed of the wave) determined by the medium itself, rather than on how you dip the stick in the water? So how can the moon affect the speed of the ripples it causes?

Drakkith
Staff Emeritus
So would this imply that the angular speed at which the moon rotates around the earth (about 24 hours for one revolution) is equal to the angular speed of the tidal wave, since the crest is always more or less on the surface of the ocean closest to the moon? If you dip a stick into the water, isn't the speed at which ripples propagate (the speed of the wave) determined by the medium itself, rather than on how you dip the stick in the water? So how can the moon affect the speed of the ripples it causes?
The thing is that gravity is pulling on ALL of the water, not just some of it. I don't agree that it's like a "wave". For me, it is much easier to understand it as gravity pulling the water.

olivermsun
The moon's (or sun's) gravity is pulling more on the side of the earth closer to it than the other side. This causes an elongation of the "earth sandwich" in that direction. Hence two bulges in the ocean, one on each side. The earth turns "under" the moon (sun), but the bulges try to stay stationary relative to the moon (sun), so they move around the surface of the ocean with some lag. It gets more complicated, since the oceans have irregular shapes which do all sorts of funny things to the moving bulge of water (and in fact some of the sea basins are have modes which are resonant with one or more tidal constituents).

So yes the surface tide is a forced wave which is affected by the gravitational gradient of the moon (and the sun), the rotation of the earth, and the shapes of the ocean basins.

The moon's (or sun's) gravity is pulling more on the side of the earth closer to it than the other side. This causes an elongation of the "earth sandwich" in that direction. Hence two bulges in the ocean, one on each side. The earth turns "under" the moon (sun), but the bulges try to stay stationary relative to the moon (sun), so they move around the surface of the ocean with some lag.
So do the tides flow from West to East? Also, how much is the lag? If the moon is directly over you, or on the other side from you, how long will it take for the tide to reach its highest point?

olivermsun