Understanding High Tides: The Role of the Moon and Earth's Rigid Crust

[shiraztk]

I was just wondering how high tides occur and happen to read the wiki article at http://en.wikipedia.org/wiki/Tide but I am bit confused like if the moon caused the tide then why there should be a high tide on the opposite side of the Earth also. The figure in the article is just confusing and I believe it must be like my attached figure if moon is the one who causes the High Tide.

 

[russ_watters]

Welcome to PF!

No, your diagram is wrong. The moon is also pulling the Earth away from the water on the far side of earth. That's why there are tides on both sides.

 

[DrGreg]

shiraztk, if your diagram was correct, we'd get high tides every 25 hours, but actually we get them every 12½ hours.

 

[HallsofIvy]

The earth, as a whole, is rigid. The ocean is liquid. The reason that's important is that the moon's pull on the entire Earth can be calculated as the pull at the center of the Earth while the moon's pull on every molecule of water is separate. Also the pull is proportional to $1/r^2$. That means that water on the surface of the Earth closer to the moon is pulled a little more than the Earth itself, even that part of the Earth at the surface. But the Earth is then pulled a little more than the water on the opposite side of the earth.

 

[shiraztk]

Ok, I agree that practically its not happening as I have depicted.
If tides happen because of gravitational pull of Moon,
1. Will my weight be different if I measure it beneath the surface of water on high Tides and low tides? I agree that if it was on the surface of the Earth My mass is too small to be considered. But since water is affected by gravitational pull of Moon and I am now inside water me too should be affected?

 

[Bandersnatch]

Yes, you will weigh less during high tide than during low tide.

When thinking about tides, it's important to realize that the Earth and the Moon are both orbiting their mutual centre of mass (a.k.a. the barycentre: http://en.wikipedia.org/wiki/Barycentre#Barycenter_in_astronomy), which is somewhat displaced from the centre of the Eearth(but still deep under the surface).
In a non-inertial reference frame of the Earth's surface, this gives rise to centrifugal forces proportional to the distance from the barycentre towards and away from the general direction of the Moon.

The bit facing the Moon is closer to the barycentre than the one facing away, so it experiences lower centrifugal force. However, it is also attracted by the Moon's gravity more strongly(it's closer after all), so in the end the net outward forces on both sides of the Earth are more or less the same.

These forces act on every particle/object, including you. It doesn't matter whether you are beneath surface of water or standing firmly on the ground. Yet, they are minuscle enough to not be noticeable without doing proper measurements, or looking at the oceans.

 

[shiraztk]

OK, thanks for the answer.
Does it mean that my weight is resultant of Centrifugal force, Gravitational force of Earth & Gravitational force of moon?

 

[Bandersnatch]

And buoyancy. You weigh less when submerged in water, or in air, than you would in a vacuum.

Also, there's the problem of the Earth not being an ideal sphere, and not having perfectly uniform density, causing the gravitational force to vary due to more/less of more/less dense material under your feet pulling you down.

Additionally, remember there's the centrifugal force due to the Earth's diurnal rotation, AND due to the orbit around the barycentre of the Moon-Earth system.

Then there's similar, but smaller effect from the Sun's influence(i.e., the Sun causes tides as well). If you want to be very anal about it, you could add the influence of other planets in the solar system, no matter how minuscle these might be.

Finally, remember that the centrifugal forces are there only as long as you take the surface of the Earth as your reference frame, which is non-inertial.

This link has a nice explanation of the tides, with pictures. Everybody loves pictures.
http://www.aztecsailing.co.uk/theory/ch2%20sect%202.html

 

[256bits]

Ok, I agree that practically its not happening as I have depicted.
If tides happen because of gravitational pull of Moon,
1. Will my weight be different if I measure it beneath the surface of water on high Tides and low tides? I agree that if it was on the surface of the Earth My mass is too small to be considered. But since water is affected by gravitational pull of Moon and I am now inside water me too should be affected?

You are able to see the affect of the tides more easily with oceaens and large lakes since the water is a liquid and will flow easily as the Earth rotates, and you can notice the change in height of water in relation to the land level. The more solid Earth is also affected by the lunar gravitation but the affect is a vertical dispalcement and not a horizontal flow. You are affected by the lunar gravitation whether you are on land or on or under water.

Will you weigh less at high tide versus low tide? Speaking of water only,
To answer this question consider an Earth covered with a deep sea of water, and with no moon. Also add in the fact that this water- Earth is non-rotating. The surface of the water would be spherical ( if the Earth did not rotate ), and all locations of this sphere would be at the same gravitaional potential so the water would not have any tendency or reason to flow from one location to the other.

If you add a gravitating object such as the moon, the equipotential surface of the water is now changed, so water will flow towards the side facing the moon and flow towards the side opposite the moon. While the gravitational force of the moon varies as 1/r^2, it is the difference in gravitational potential that causes the tides. This force varies as 1/r^3. This is the difference of the gravitational force of the moon with that of an object located at the center of mass of the earth, with that of an object on its surface. This has the affect of weakening the Earth's gravitation on the surface object, on the side facing the moon and on the opposite the moon. At right angles to the moon-earth the tidal force reinforces the Earth's gravitation.

The water covered Earth will take on the appearance or shape of as ellipsoid, or more basically a spheroid. On this surface you will find all locations having a equipotential. so no matter where you are on the surface you are attracted to the center of mass of the water- Earth with the same gravitational potential. At a high tide location your weight would be the same as at a low tide location. If your weight would be different you can surmise, then the water would also experience this difference in gravitation and have a potential to flow from one location to the other.,

For the real earth, with its land mass, rotation, the ocean level still tends to follow an equipotential. But the rotation flattens the poles and the water abuts against the land mass, and frequency affects of the tides may or may not make you weigh less or more at high tide from that at low tide, if it is the height of the tide near a land mass to which you are referring.

 

[Bandersnatch]

But since unlike water, the Earth is somewhat rigid, the crust doesn't fully follow the equipotential like water does, right? So a person standing on land would be able to measure the difference in force between tides.

 

[sophiecentaur]

But since unlike water, the Earth is somewhat rigid, the crust doesn't fully follow the equipotential like water does, right? So a person standing on land would be able to measure the difference in force between tides.

Some moons of the larger planets have their 'solid' crust deformed so much by tidal action that there is volcanic activity due to the heat generated and the expected craters are not there because they have 'melted' flat over the years and the surface is very smooth.
Sorry - can't remember the names; I'm sure someone can do that off the top of their heads!