How does the moon effect tides?

How does the moon effect/control tides?
does it effect the likes of the great lakes also?
(I think you guys have doubled my IQ since I started posting!)

The moon and the earth have a mutual gravitational attraction, right? That is, the moon pulls on the earth and vice-versa. Gravity is proportional to the inverse square of the distance between the bodies. That means that if you double the distance between the bodies, the gravitational force between them drops to one-fourth of what it was.

Now, the moon is closer to one side of the earth than the other. Granted, the difference is just the diameter of the earth, but it’s enough to make the gravitational pull on the “near” side of the earth stronger than it is on the “far” side. Since the earth isn’t perfectly rigid, it deforms a bit due to this differential force. Since the oceans are especially non-rigid, they deform even more, so the moon in effect pulls the oceans towards themselves. This will affect any body of water connected to the oceans, so I think that it would affect the Great Lakes to a small extent, but not as much as it would affect the oceans.

The gravity from the moon is a little stronger on the side of the Earth nearest to it, and a little weaker on the far side of the Earth.

If it were just the Earth, gravity tends to pull it into a sphere shape. The moon’s gravity would tend to make the Earth slightly elliptical, longer on the axis that’s lined up with the moon. But the spinning of the Earth makes the long axis of that ideal shape change as it spins, so that six hours later, the wide axis of the ellipse is 90 degrees of from where it was (OK, a little more, because the moon moves some in six hours). The land is too solid to move in six hours, but the water in the oceans can slosh around.

The great lakes are too small to have a noticeable effect.

By the way, the energy of moving that water around (and very very slightly moving the land) is a frictional force slowing the Earth down. Eventually, the Earth’s spinning will slow down to be in synch with the moon. The moon’s spinning has already slowed down to be in synch with the Earth due to these tidal forces, that’s why we see the same face of the moon all the time. And that’s why the moon is an elliptical shape, with its long axis towards the Earth.

Another by the way, the force on the Earth due to the Sun’s gravity is stronger than that from the moon, but the Sun has a very small tidal effect in comparison. Can you figure out why? Read the first sentence of this post again.

*CurtC: By the way, the energy of moving that water around (and very very slightly moving the land) is a frictional force slowing the Earth down. *

So if we started using the wave power machines often proposed by environmentalists, we actually might be helping to slow the Earth down? Cool! :D:D

If the oceans are being ‘pulled’ by the moon, why doesn’t every liquid regardless of size move in the direction of the pull of the moon?
I realise that the force of the ‘pull’ will depend on the size of the body of water but I don’t notice my glass of water moving at all on a flat surface, is this because the earths ‘pull’ is stronger upon the smaller body of water(my glass of water) as opposed to the larger(the oceans)?

Your glass of water does move like the tides, but the angle is so small you won’t notice it.

Let’s say you’re on the equator. The water level in your glass should be perfectly flat when you are nearest the moon, and farthest away, and I hope it’s obvious why. But as the moon is either rising or setting from your point of view, the water level in the glass will be slightly angled, with the deeper end towards the moon.

The angle that it makes with “level” is the same independent of the size of the body of water. With the oceans which are thousands of km across, you’ll see a change in the elevation at the edges of maybe 2 m. Scale that down to the size of a glass, which is around 80 mm across, and you’ll see maybe 30 nanometers of elevation change at the edge of the glass, if my back-of-a-bar-napkin calculations are right. 30 nm is pretty hard to detect, being a small fraction of the wavelength of light.

By the way, for the Great Lakes, this same calculation predicts around a 5 mm elevation change. Also hard to detect (in a lake with waves).

Your glass of water does move like the tides, but the angle is so small you won’t notice it.

Let’s say you’re on the equator. The water level in your glass should be perfectly flat when you are nearest the moon, and farthest away, and I hope it’s obvious why. But as the moon is either rising or setting from your point of view, the water level in the glass will be slightly angled, with the deeper end towards the moon.

The angle that it makes with “level” is the same independent of the size of the body of water. With the oceans which are thousands of km across, you’ll see a change in the elevation at the edges of maybe 2 m. Scale that down to the size of a glass, which is around 80 mm across, and you’ll see maybe 30 nanometers of elevation change at the edge of the glass, if my back-of-a-bar-napkin calculations are right. 30 nm is pretty hard to detect, being a small fraction of the wavelength of light.

By the way, for the Great Lakes, this same calculation predicts around a 5 mm elevation change. Also hard to detect (in a lake with waves).

To add a little fuel to the fire…all the above posts are correct. However this is an unproven theory (I’ve been saying that a lot today). Could be some other reason altogether the tides are the way they are, although this theory seems pretty good and will probably hold out…HOWEVER, I don’t know of any experiments that actually back this up by testing for microscopic “tides” in standing bodies of water in a laboratory. If anyone knows of any such experiments, I would be curious to hear of it

I seem to recall reading that the Great Lakes have measurable tides on the order of about 2 inches. I can’t find a reference but I’ll keep looking.

The tides are caused by the combination of lunar gravitational attraction, solar gravitational attraction, and centrifugal “force”. The moon’s tide-causing force is more than twice the sun’s. There are generally two tidal bulges, on opposite sides of the earth. Tides are greatest (spring tides) twice a month when the sun, moon, and earth are in a line (full or new moon). In these configuaration, the effects of sun and moon tend to add. They are least when the sun moon and earth form a right triangle (first quarter and third quarter). In these configurations, the effects of sun and moon tend to cancel.

In mid-ocean tides are typically only a foot or two in height. The height is magnified by local conditions near shore (slope of continental shelf, a narrowing channel, etc.)

The water at the tidal bulge has to come from somewhere. Each of a multitude of water molecules moves a short distance toward the bulge. The net effect is for a large volume of water to move from low-tide areas toward high-tide areas. The tide in your glass of water doesn’t rise noticibly because there is nowhere for it to come from. After all, the average slope on the ocean caused by the tidal bulge would be only about be two feet over six thousand miles.

To elaborate on CurtC’s post, the size of the body of water affects the amount of the tide. Inland seas typically don’t have noticeable tides; the Great Lakes are the same way. Being connected to an ocean by a small strait doesn’t change this too much, The Baltic Sea, separated from ocean by peninsulas and islands, doesn’t have noticeable tides since the amount of water being pulled on is simply less, and all the fractions commensurately smaller. I would imagine the Black Sea has even less tidal action; something like the East China Sea probably has tides comparable to open ocean.

I don’t know of any measurements on bodies of water, but you can measure the differences in gravitational force that appears to cause tides in your basement (assuming you have a basement and are willing to set up a fairly finicky experiment). Bending Spacetime in the Basement.

CurtC is not correct. The deeper side is not necessarly the side closer to the moon. High tides exist on the side nearest and farthest from the moon. The earth and moon orbit about their common center of mass. The orbital speed is such that the average centifugal force exactly equals the gravitational pull. On the side of the earth closer to the moon gravity is stronger than centrifugal force giving us high tide on the side of the earth toward the moon. On the side of the earth away from the moon, centrifugal force is greater than the pull of gravity giving us high tide on the other side of the earth.

He is correct in saying that the angle of deflection in every body of water is the same from a glass of water to the oceans.

Or in other words, Why are there two high tides a day when the moon is only overhead once a day?

avalongod says…

…and you’ve been wrong a lot today :slight_smile:

Seriously, do you think that gravity is an unproven theory? As everyone has pointed out, tides are a result of gravitational forces.

My memory bank recalls that the tide in Lake Superior is about two inches and the tide in the Medeterranian Sea is about six inches.

All you guys are missing something. The law of gravity does apply here but mass does not move instantly. This causes delays, amplifications, resonant phenomena, etc. It is VERY complex. Some places have two tides while other have just one. Conceivably you could design bays with more tides but the friction dampens them out.

As an example: The tides in the Chesapeake Bay are NOT directly caused by any force of gravity. The Bay runs roughly from N to S so you would expect a miniscule tide from E to W but this is NOT what happens. What happens is that the tidal differences at the mouth of the bay create tidal waves that travel up the Bay so that you have high tide and low tide at the same time on different points of the bay even though they are at the same geographical longitude.

So, though gravity from the moon and sun are the ultimate cause for tides, there is much more at work and you cannot just say the water rises where it is nearer to the moon.

I forgot to mention: The tides in the open ocean are NOT even. There are pleces in the middle of the ocean where the tidal waves cancel each other out and there is NO tide. These points have a name which I cannot recall now.

I would point out that today’s tide is not the effect of today’s moon but mostly of oscillations that have built up over time. If the moon were suddenly to vanish the tides would continue for quite a while and if after they stop you put the moon back in place, it would take a while for the tides to build up and settle into a stable pattern.

Is “nodes” the word you’re looking for?

“The tide is noticeable, in at least Lake Superior…” is what I started to write. There is what is called a seiche, which is oscillation of the water level in lakes and harbors. When I was on Isle Royale, I learned this term because the river which flowed into Washington Harbor, near Windigo, would periodicly flow backwards due to the seiche. I recall that this was attributed to tides (or at least tides were used as an analogue).

As I looked into it on the web, when I found sites which talked about seiches on the Great Lakes, they mentioned winds and differences in atmospheric pressure as causes of seiches, but were silent about tides. My guess is that this means the tides, while they have to exist at some level (avalongod notwithstanding), are so small as to be negligable, or not measuireable.

Forgot to mention, here is a good site with some graphs of Great Lake level variations on several time scales.