# What is the affect of moon on tides?

1. Jul 11, 2011

### khurram usman

i mean when the moon is full like during 10-20 of lunar month the tides are higher but y?even when the moon isnt full its still there whole in sky? what is the reason?

2. Jul 11, 2011

### Staff: Mentor

Welcome to PF!

The height of the tides is not affected by the phases of the moon. They are, however, affected by how far the moon is from the earth.

3. Jul 11, 2011

### nasu

The relative height of the tides depends on the relative positions of Sun, Moon and Earth.
The phases of the moon are given by the same thing.
Maximum height of the tides happens at new moon too (see neap and spring tides).

4. Jul 11, 2011

### jjustinn

5. Jul 11, 2011

### Staff: Mentor

Oops....I was thinking about the effects of the moon only. I'm sure you're right about what the OP was after.

6. Jul 11, 2011

### morrobay

The moon orbits the earth / 28 days. On the axis from center of earth to center of moon
there is a bulge of ocean water that moves with the moon.
So actually a high tide is experienced by a land mass as it rotates through this bulge
of water, and a low tide after it passes through it.
I can understand the bulge of water directly under the moon from gravitation.
What I dont understand is the lesser bulge of water on the opposite side of earth,
That accounts for the second, lesser high tide/24 hrs.

7. Jul 11, 2011

### Pythagorean

The sun has a higher force on Earth than the moon.

However, the gradient of the force from the moon across the Earth has a wider range than the sun. That means the moon pulls hardest on water near it than it does on the Earth, and harder on the Earth than it does the opposite bulge.

The lesser bulge is a result of the water wanting to travel straight while the earth is pulled slighty towards the moon.

8. Jul 12, 2011

### sophiecentaur

Actually. the forces 'away' and 'towards' the moon (ref the Earth's frame of reference) are more or less equal on either side. The Earth/Moon system is rotating around their common CM (ignore the Sun, for now) which is inside the Earth but offset from its centre (the Earth has a 28 day 'wobble'). If you calculate the sum of gravitational and rotational forces on either side of the Earth's surface, you get pretty well the same answer.

Adding the effect of the Sun's field produces Spring and Neap tides, depending on the alignment. Something that still amazes me, however, is that the small difference in direction, depending on how near the Moon and Sun are relative to the ecliptic, makes such a massive difference to the tidal ranges at the solstices compared with at the Equinoxes.

9. Jul 12, 2011

### nasu

However there is a difference between the height of the two tides produced in 24 hours. The difference seems to be of the order of 2-3 ft for both east and west coasts of N.America (even though the actual heights are quite different for the two coasts).
I also thought that the overall forces are quite the same on both sides and I looked for actual info to support this. Here is a nice site that shows both measured and predicted heights.

http://tidesonline.nos.noaa.gov/monitor.html

The predicted heights show the same difference so it must be some known factor in the model that makes the difference.
Unfortunately the data is shown for only a few days. It will be interesting to know if this difference changes during the month.

10. Jul 12, 2011

### sophiecentaur

There are many factors involved in the actual tidal heights in different parts of the world. The force on a mass of water, due to gravitational and rotational effects at a certain point on the globe is only the start of things.
@nasu: Interesting that your sample of tidal heights gives such a big difference because, in the English Channel, the tidal heights do not differ much for successive 'highs', most of the time.
I have, from time to time, looked at the tidal predictions around the World and there are huge differences from place to place, with some ranges being much greater than others. The tide, itself, consists of a wave which sweeps around the World about once a day with two main components with phases which depend on the relative Sun - Moon positions. On a water-covered sphere, the tides would be almost calculable, I guess, but when you stick land masses and channels in the way, plus 'resonant' large bodies of water, it's a nightmare to calculate. You need to rely on the analysis of records for any one place and come up with about a dozen 'harmonic constants'. These can be used to predict, 'accurately enough' the expected tidal heights all over the World. There are other factors, like atmospheric pressure and wind, that can make a huge difference to the heights and times of high tides and which limit the accuracy needed of safe navigation - you have to leave a good margin for safety.
All the published tide information is littered with disclaimers (same as with charts) - with good reason.

11. Jul 12, 2011

### nasu

These values are measured and they are for the same place. I mentioned the calculated ones as a secondary bit of info. I don't know the basis for their predictions.
The data seem to show a periodic, reproducible difference between the two tides from a 24 hour period. It appears to be a global effect and not a local one. The actual pattern and height varies wildly, as you mentioned. However the difference is there for most locations. Maybe the local conditions may reduce the difference. On some islands or bays in the Caribbean seems to be only one high tide every 24 hours (Lime Tree Bay is an example) which may be an overlap of two 12 hour peaks.

It seems that the two bulges may have different heights unless there is another effect.
Do you have any reference regarding the forces on the two sides being equal (or almost)?

12. Jul 12, 2011

### olivermsun

13. Jul 12, 2011

### sophiecentaur

Measured or predicted would show much the same - because the predictions are so reliable, once there had been enough data collected. We virtually stake our lives on the predictions.

I believe there are some places which are dominated by an approximate 24hr resonance and some places where there is a resonance nearer 25 hours. This will produce exaggerated effects from either Sun or Moon, depending on the resonant frequency. See the tides in the Bay of Fundy - 13m range!!.

Do the figures you have seen show a higher or lower high tide on the side towards or away from the Moon or the are they day /night related?

As for the forces being the same, I did the sums myself, once, for Earth / Moon and used standard figures for masses, distances and orbit times. I did it numerically (I chickened out of the analysis but it's not really too grisly). It involved finding a small difference between two numbers, as I remember and the values on each side came out more or less equal. But I think it is only to be expected because the 'offset' of the CM of the two bodies from Earth's centre is directly related to their relative masses and separation and the rotational force difference depends on exactly that as well. There may need to be some assumption about average density but I'm not too sure / can't be naffed to follow it up that closely.
I'd bet the analytical way would show they're identical. I was very surprised by the result at first but it now seems quite reasonable. It's easy to be wise after the event!

I think the same sort of argument applies for the 'microgravity' inside orbiting space ships. There is a net 'upward' force for the parts of the ship away from Earth and a net 'downward' force for the parts on the Earth side. The CM is the only bit that is in precisely the right position for a sustained orbit at a given angular velocity.

EDIT @ Oliversum - cheers for that reference. I won't have to do the calculation for myself now!

14. Jul 12, 2011

### D H

Staff Emeritus
That is a simplified (over simplified) view of the tides. There is no such wave. Think about it in terms of North America+South America and Africa+Eurasia. These constitute two huge north-south barriers that utterly preclude the existence of such a wave.

Actually, over three dozen of them. This is a much better model of how the tides work.

Here are the 37 coefficients that NOAA uses to model the tides:
1. M2 - Principal lunar semidiurnal constituent
2. S2 - Principal solar semidiurnal constituent
3. N2 - Larger lunar elliptic semidiurnal constituent
4. K1 - Lunar diurnal constituent
5. M4 - Shallow water overtides of principal lunar constituent
6. O1 - Lunar diurnal constituent
7. M6 - Shallow water overtides of principal lunar constituent
8. MK3 - Shallow water terdiurnal
9. S4 - Shallow water overtides of principal solar constituent
10. MN4 - Shallow water quarter diurnal constituent
11. NU2 - Larger lunar evectional constituent
12. S6 - Shallow water overtides of principal solar constituent
13. MU2 - Variational constituent
14. 2N2 - Lunar elliptical semidiurnal second-order constituent
15. OO1 - Lunar diurnal
16. LAM2 - Smaller lunar evectional constituent
17. S1 - Solar diurnal constituent
18. M1 - Smaller lunar elliptic diurnal constituent
19. J1 - Smaller lunar elliptic diurnal constituent
20. MM - Lunar monthly constituent
21. SSA - Solar semiannual constituent
22. SA - Solar annual constituent
23. MSF - Lunisolar synodic fortnightly constituent
24. MF - Lunisolar fortnightly constituent
25. RHO - Larger lunar evectional diurnal constituent
26. Q1 - Larger lunar elliptic diurnal constituent
27. T2 - Larger solar elliptic constituent
28. R2 - Smaller solar elliptic constituent
29. 2Q1 - Larger elliptic diurnal
30. P1 - Solar diurnal constituent
31. 2SM2 - Shallow water semidiurnal constituent
32. M3 - Lunar terdiurnal constituent
33. L2 - Smaller lunar elliptic semidiurnal constituent
34. 2MK3 - Shallow water terdiurnal constituent
35. K2 - Lunisolar semidiurnal constituent
36. M8 - Shallow water eighth diurnal constituent
37. MS4 - Shallow water quarter diurnal constituent

For those that have a number after them, the number roughly indicates the frequency in terms of cycles per day (the exact frequency is given by tidal theory). Some don't have a number. Some of those have a very low frequency, some up to a year.

For a given location, NOAA develops a pair of values for each of those 37 coefficients: A phase and a magnitude. The predicted tidal height at that location at a given time is simply

$$h(t) = \sum_i h_i \cos(\omega_i (t-t_{0,i}) + \phi_i)$$

where ωi is the frequency for the ith coefficient and t0,i is the theoretical time of a peak of that coefficient, both of which are given by tidal theory. Note that this is a mechanistic prediction based on theory plus a locale-specific fit of observations to the theory. It ignores things like weather, which at times can have a significant impact on the tides. (Think hurricane.)

Last edited: Jul 12, 2011
15. Jul 12, 2011

### sophiecentaur

Yes, I agree. But you've got to start somewhere and, after the simple discussion of the forces, the notion of a wave sloshing around the Earth is the next thing to think of. It's only in the Southern Ocean that any such wave could hope to propagate. I wonder . . .

16. Jul 12, 2011

### turbo

Fundy tides are outrageous! When my wife and I took her mother and aunt on a tour of the Maritimes, we managed to time our visit to Hopewell Rocks so that we arrived at low tide and got to walk around on the sea-floor with those "flowerpots" looming above us. It was pretty darned humbling!

http://www.thehopewellrocks.ca/

17. Jul 12, 2011

### Pythagorean

So I guess what you're saying is that the gradient comes more from the distribution of centripetal force across the Earth (i.e, as R away from earth-moon COM system)?

18. Jul 12, 2011

### nasu

19. Jul 12, 2011

### sophiecentaur

Sorry. I don't understand. Could I have it in words of one syllable or less?

20. Jul 12, 2011

### D H

Staff Emeritus
Center of earth, not Earth-Moon CoM.

To me, the easiest way to look at things is to remember that the Earth as a whole is accelerating moonward (and sunward). Because gravitation is a 1/r2, the gravitational acceleration at some point on the surface of the Earth is slightly different than the acceleration of the Earth as a whole.

Mathematically,

$$\vec a_{\mbox{rel}} = G M_{\mbox{moon}} \left(\frac{\vec R - \vec r}{||\vec R-\vec r||^3} - \frac{\vec R}{||\vec R||^3}\right)$$

where R is the vector from the center of the Earth to the center of the Moon and r is the vector from the center of the Earth to some point on the surface of the Earth.

This simplifies when these two vectors are parallel to

$$\vec a_{\mbox{rel}} = G M_{\mbox{moon}} \left(\frac{R - r}{(R-r)^3} - \frac{R}{R^3}\right) \approx G M_{\mbox{moon}}\frac {2r}{R^3}$$

When the vectors are antiparallel this simplifies to

$$\vec a_{\mbox{rel}} = -\,G M_{\mbox{moon}} \left(\frac{R + r}{(R+r)^3} - \frac{R}{R^3}\right) \approx \,G M_{\mbox{moon}}\frac {2r}{R^3}$$

In both cases a positive value means "in the direction of r" -- in other words, outward (away from the center of the Earth). The omitted terms are on the order of the included term times r/R times some near unity number, so very small.

21. Jul 12, 2011

### nasu

As they are shown only for about two days, they are both. Or it's hard to say. Taking Boston, on July 11-12 as an example, the smaller maxima are in the morning (around 9-10 AM, it's hard to read the graphs) and the larger ones in the evening (9-10 PM).
If I remember correctly, last night the moon was about 3/4 and increasing in size so it rises near evening time. So the moon will be up during the first part of the night and this is pretty much the situation for both days. The Moon will be high in the sky during the evening tide and on the other side of the Earth during the morning tide.
This is a direct link to the data for Boston:
http://tidesonline.nos.noaa.gov/plotcomp.shtml?station_info=8443970+Boston,+MA+&flag=0

Here I found a program to calculate the tides for many locations around the world
http://www.sailingusa.info/tides.htm [Broken]
I looked only at a few locations in the English channel (French side). It looks a little more complicated. Maybe due to local configuration.

Last edited by a moderator: May 5, 2017
22. Jul 12, 2011

### turbo

Local configuration can be critical, as in the Bay of Fundy situation. Width-restriction and slope of the sea-bottom can result in massive tidal variations. These large variations helped settlers on the west shore of Nova Scotia to flush salt out of salt-marshes very quickly with their clever dikes and flapper-valves, using local streams to rinse out the salt.

23. Jul 12, 2011

### D H

Staff Emeritus
That is just a function of the fact that tides in Boston are a function of the principal lunar semidiurnal (M2) plus, to a lesser extent, the larger lunar elliptic semidiurnal (N2), principal solar semidiurnal (S2), and the two lunar diurnal terms (K1 and O1). The relation between the smaller and larger maxima will vary over the month, year, and even longer. It is similar to the concept of beats in acoustics (wiki: http://en.wikipedia.org/wiki/Beat_(acoustics)).

This site specifies harmonic constituents of the tides for Boston: http://co-ops.nos.noaa.gov/data_menu.shtml?stn=8443970%20Boston,%20MA&type=Harmonic%20Constituents [Broken]

Last edited by a moderator: May 5, 2017
24. Jul 12, 2011

### nasu

This make sense. Using that program that I mentioned I saw that the relationship changes over longer periods of time.

Last edited by a moderator: May 5, 2017