Why are there two tides a day when there is only one moon?

[mack_10]

Something that has bothered me for a long time, if tides are due to the gravitational force of attraction of the moon, and clearly the moon transverses the sky once a day then there should only be one high tide a day not two.

If the second high tide is due to the attraction of the sun then I would expect not only a different height but that the interval between high tides would vary with the lunar cycle?

Is there a simple explanation for this or is it one of those things that everybody "knows" so no one questions it?

 

[JaredJames]

Extremely simply:

The moons gravity pulls the ocean towards it, giving you one tide.

But at the same time the moons gravity is weakest on the opposite side of the planet, which allows the centrifugal force from the Earth's rotation to create the second tide.

Nothing to do with the Sun (although it can have an effect).

A better explanation:

http://www.aztecsailing.co.uk/theory/ch2%20sect%202.html

The pictures should give you a very good idea of how it works.

I don't think many people realize the mechanics of the situation, but just go with "it's the moon". Certainly not a silly question, but perhaps a quick Google could have given you the info first.

 

[D H]

Extremely simply:

The moons gravity pulls the ocean towards it, giving you one tide.

But at the same time the moons gravity is weakest on the opposite side of the planet, which allows the centrifugal force from the Earth's rotation to create the second tide.

Nothing to do with the Sun (although it can have an effect).

A better explanation:

http://www.aztecsailing.co.uk/theory/ch2%20sect%202.html

Yech. There is no need to invoke centrifugal force to explain the tides. Using centrifugal force as an explanation in which the Moon and Earth are depicted as moving is particularly bad.

The clearest explanation to me is that the gravitational effect of the Moon on a drop of water on the surface of the Earth is the Newtonian acceleration of that drop toward the Moon less the acceleration of the Earth as a whole toward the Moon. This naturally leads to a bulge on the side of the Earth facing toward the Moon and another bulge on the side of the Earth facing away from the Moon. No fictitious forces are needed.

 

[Vanadium 50]

The moon pulls the oceans nearest it up a lot, and the bulk of the Earth farther away up less, and the oceans farthest away even less. Voila - two bulges.

 

[JaredJames]

Yech. There is no need to invoke centrifugal force to explain the tides. Using centrifugal force as an explanation in which the Moon and Earth are depicted as moving is particularly bad.

The clearest explanation to me is that the gravitational effect of the Moon on a drop of water on the surface of the Earth is the Newtonian acceleration of that drop toward the Moon less the acceleration of the Earth as a whole toward the Moon. This naturally leads to a bulge on the side of the Earth facing toward the Moon and another bulge on the side of the Earth facing away from the Moon. No fictitious forces are needed.

The CoG for the Earth-Moon system gives the tides. When the centrifugal force created by rotation around this point is greater than the moons gravity (when the moon is on the opposite side of the earth) you get the bulge.

Perhaps I tried to over simplify it previously.

 

[D H]

Rotation is not needed to explain the tides. Suppose an object of the Moon's mass was falling straight toward the Earth. That falling object would exert exactly the same tidal forces on the Earth as does our real Moon at the instant that this object was at a distance of 384,400 km from Earth's center of mass.

 

[russ_watters]

The moon pulls the oceans nearest it up a lot, and the bulk of the Earth farther away up less, and the oceans farthest away even less. Voila - two bulges.

Similar, but I like to say that the moon pulls the ocean on the near-side away from the Earth and pulls the Earth away from the ocean on the far-side.

 

[mack_10]

so your saying it is a coincedence that both tides are the same height, and the peaks of high tide not corresponding to the position of the moon in the sky is unimportant?

do you have any math to support this or is just something you know?

Does NASA launch its rockets at the peak of the tide to take advantage of the decrease in local gravity?

 

[JaredJames]

so your saying it is a coincedence that both tides are the same height, and the peaks of high tide not corresponding to the position of the moon in the sky is unimportant?

If the Earth wasn't spinning, the high tides would coincide directly with the moons location, but because of the rotation of the earth, which is quicker than the moons orbit, it means the high tides are slightly ahead of the moon.

do you have any math to support this or is just something you know?

This link gives everything you need on the subject:

http://www.pol.ac.uk/home/insight/tidefaq.html

 

[Vanadium 50]

so your saying it is a coincedence that both tides are the same height, and the peaks of high tide not corresponding to the position of the moon in the sky is unimportant?

Nobody said that.

do you have any math to support this or is just something you know?

I don't understand the hostility here. But in any event, this certainly can be done quantitatively. You will end up with something like

$$F_x = 2mg\frac{r\cos\theta}{R} , F_y = 2mg\frac{r\sin\theta}{R}$$

with x directed away from the moon, r being the radius of the earth, and R being the distance from the Earth to the moon. Note that for $\theta = 0$, F_x is positive, so there is a bulge away from the moon, and if it's 180, it becomes negative so there is a bulge towards the moon.

Does NASA launch its rockets at the peak of the tide to take advantage of the decrease in local gravity?

No - to see why, calculate the strength of the moon's pull on an object on the Earth's surface.

 

[D H]

so your saying it is a coincedence that both tides are the same height, and the peaks of high tide not corresponding to the position of the moon in the sky is unimportant?

What makes you think that? Nobody has said anything of the sort.

do you have any math to support this or is just something you know?

See [thread=278034]this thread[/thread].

Does NASA launch its rockets at the peak of the tide to take advantage of the decrease in local gravity?

No. Those rockets have to reach a specific orbit. It's hard enough doing that without worrying about the Moon. For example, the launch windows for the Shuttle going to the Space Station are only a few minutes wide. Adding in the Moon would make the windows even smaller and very, very far apart. There is no reason to do that. The tidal forces are too small to take advantage of and too small to worry about.

 

[mack_10]

Why refer me to that thread, they seem just as confused about the explanation there as here?

As for the idea that tidal forces are too small to worry about - consider the mass of water being lifted by those tidal forces.

http://www.pol.ac.uk/home/insight/tidefaq.html was interesting but still seems to be making some assumptions.

Why isn't every second high tide occurring directly under the moon? If two different effects are causing each of the tides then why are they the same height?

 

[JaredJames]

Why refer me to that thread, they seem just as confused about the explanation there as here?

Cut the hostility. There's no need to be nasty, people are only trying to help you.

As for the idea that tidal forces are too small to worry about - consider the mass of water being lifted by those tidal forces.

Precisely. The mass of water. The gravitational pull between the oceans and the moon is significantly stronger than the pull between the moon and a teeny tiny rocket.

http://www.pol.ac.uk/home/insight/tidefaq.html was interesting but still seems to be making some assumptions.

Such as?

Why isn't every second high tide occurring directly under the moon?

For the reason given in that link. The Earth rotates faster than the moons orbit so the tides are ahead of the moon.

If two different effects are causing each of the tides then why are they the same height?

Two different effects? It's the moons gravity doing both. Ignore my first posts, they were based on incorrect information and I can't edit them now.

 

[Vanadium 50]

I think you are misattributing who is confused.

Again, why the hostility?

 

[mack_10]

I think you are misattributing who is confused.

why the hostility?

 

[DaveC426913]

I think you are misattributing who is confused.

why the hostility?

Mack, this line:

Why refer me to that thread, they seem just as confused about the explanation there as here?

reads as if we are responsible for ensuring you grasp the concepts, and that, since you don't, we have failed you.

Is that the message you meant to send?


Tides are a concept that don't have a good human-scale equivalent. No one on PF is confused, it's just difficult to describe if you don't have the foundations to build on. Kind of like trying to describe Special Relativity, there's some background that needs to go into it.

There are more lengthy descriptions out there if you're interested, that have multiple steps and accompanying diagrams.

 

[berkeman]

I think you are misattributing who is confused.

why the hostility?

mack, when quoting another user, please use the Quote button, so that attribution is given to the original user. It is confusing to have you quote V50's post with just clear text like that. Thank you.

 

[DaveC426913]

mack, when quoting another user, please use the Quote button, so that attribution is given to the original user. It is confusing to have you quote V50's post with just clear text like that. Thank you.

(I may be wrong, but I interpreted Mack's post - not as quoting V50 - but as parroting him. As in: tossing the same question(s) back in his face.)

Anyway, who knows. It's degenerated now. Perhaps Mack will do some reading and come back with some specific questions about things he still doesn't understand.

 

[JaredJames]

(I may be wrong, but I interpreted Mack's post - not as quoting V50 - but as parroting him. As in: tossing the same question(s) back in his face.)

That's exactly how I viewed it, particularly given the removal of "again, " which is a rather specific item. Plus the fact there were no other remarks / comments made that would imply it is a response to Vanadium's post.

 

[berkeman]

That's a possiblity. I'm inclined to let mack reply, before deciding what to do.