Talk:Tidal force/Archive 1

Additional effect of rotation?
The article says "For two bodies rotating about their barycenter, the variation in centripetal force required for the rotation adds to the tidal force." I do not believe this. There is no "additional" effect of rotation. The high tides on both sides of the Earth are both perfectly well explained by the tidal effects of the Moon's gravity, and there is no additional effect caused by the fact the the Earth and the Moon are in orbit about their barycenter. Occultations 21:02, 20 December 2005 (UTC)


 * It is perhaps misphrased. What is means is, if you're trying to work out whether the body is going to fall apart or not, the rotation comes into it. William M. Connolley 14:30, 21 December 2005 (UTC).

Here's a thought experiment. Consider the positions of the Earth and the Moon at a particular moment in time. If the Earth and the Moon had somehow arrived at those positions and were stationary at the time (but of course free-falling towards each other), then the ocean tides on Earth would be exactly the same as normal. Someone else has made this point above under "Centrifugal force not necessary".

Next, suppose that the Earth and the Moon had somehow arrived at those positions but were passing each other at a high speed, way in excess of their actual orbital speeds about their barycenter. Again, the ocean tides on Earth would be exactly the same as normal.

The fact that in reality, the Earth and the Moon are, in a sense, passing each other at just the right speeds to keep them in roughly circular orbits about their barycenter makes no difference to the ocean tides on Earth.

Where do you get the formula



F_t = \omega^2mr + \frac{GMmr} {R^3} $$

from? It's not consistent with the earlier one:



F_t = \frac{2GMmr} {R^3} $$

for $$\omega = 0$$.

Also, you say that r is "the distance from the reference body's center along the axis". But surely, in the $$\omega^2mr$$ term, r has to be the distance from the center of rotation, ie. the barycenter. And where does the $$\omega^2mr$$ term come from? You can't say it comes from the fact that the object is moving in a circle. It's the force that makes the object move in a circle, but its origin must be something else.

So I don't see where you got this formula from, and I think this entire section on "Additional effect of rotation" is unnecessary and, moreover, incorrect.

I'm sorry if all this seems to break the rule on civility, but I think this section is wrong - I don't know how else to say it. I've presented my reasoning, let's discuss it.

Occultations 21:02, 20 December 2005 (UTC)


 * Gosh no that was very civil! But I think you've misunderstood - see top comment. William M. Connolley 14:30, 21 December 2005 (UTC).


 * I'm glad you weren't too offended. But a few questions ...


 * What have I misunderstood, and in what way?


 * Do you believe that there is an additional effect on tides that "adds to the tidal force" when two objects are rotating around their barycenter? (I don't.)


 * How do you arrive at the formula with the $$\omega^2mr$$ term?


 * How come it gives a different answer from the earlier formula when $$\omega = 0$$? Which one is correct?  (I think the earlier one is always correct.)


 * Occultations 13:06, 22 December 2005 (UTC)


 * OK, I think I know what the problem is but I may have misinterpreted you. So: this is nothing at all to do with tides, really. What it is, is about the forces that thend to break up the smaller body (for simplicity). Regardless of the relative motions of the bodies (in Newtonian physics...) there is the std tidal forces. *In addition* if the smaller body is rotating, there are additional forces (centrifugal). As it happens, the most common situation is for the two bodies to be tide-locked which affects omega.


 * having said that, I haven't checked the formula, and I don't see why it should be labelled as F_t. William M. Connolley 16:35, 22 December 2005 (UTC).


 * Ok, I think I understand what you're saying. The Earth and the Moon are rotating about their barycenter, and so there's a centrifugal effect - water anywhere on the Earth seems to weigh less than normal because of "centrifugal force", or, to avoid mentioning fictitious forces, because some of the net force on it is "used up" as the centripetal force making it move round in a circle once every sidereal month.


 * This is presumably where your $$\omega^2r$$ term comes from. And in that term, $$r$$ can only be the distance from the centre of rotation, ie. the barycenter of the Earth/Moon system.


 * So how big is this effect? One rotation per sidereal month is about 2.7 radians/s.  So $$\omega^2r$$ at the surface of the Earth facing away from the Moon (where $$r$$ is about 1.75 Earth radii) is about 8 m/s2, and at the sub-lunar point on the surface of the Earth (where $$r$$ is about 0.25 Earth radii) it's about 1 m/s2.   In comparison, the tidal effect of the Moon at these two points is $$\frac{2GMr} {R^3}$$, which is about 1 m/s2.  The centrifugal effect is much bigger than the tidal effect.  So if there really is an "additional effect of rotation", then the high tide at the surface of the Earth facing away from the Moon would be about 8 times bigger than the high tide at the sub-lunar point.  This is not what is observed.


 * For a theoretical approach, see Tides, which says "the centrifugal force is uniform and does not contribute to the tides".


 * Where you talk about the forces that tend to break up a body, here the rotation is obviously important, any rocky body will fly apart if it's rotating fast enough. But what rotation are we talking about?  In the case of the Earth, it's its daily rotation relative to the fixed stars.  Its monthly rotation around the barycenter of the Earth/Moon system does not contribute (look again at the picture at Tides - "the Earth has no "rotation" around this point. It just "displaces" around this point in a circular way").  This effect, where "centrifugal force" tries to pull a body apart, has nothing to do with tidal forces.


 * So I've presented theoretical and experimental arguments that there is no "additional effect of rotation" on tides or tidal forces. Do you agree?


 * --Occultations 21:24, 23 December 2005 (UTC)

Umm, I think we're getting closer, but not there yet. To repeat: the centrifugal effect has nothing whatever to do with tides (err). Tides and tidal forces are differenct. The only effect is on the break-up point of the smaller body, assuming that the smaller body has no internal cohesive forces other than gravity. Tidal forces tend to break the body up. So do centrifugal forces. William M. Connolley 21:55, 23 December 2005 (UTC).


 * Yes, I think we're getting there. You're saying that rotation produces a  centrifugal effect, which has nothing to do with tides or tidal forces.  But the centrifugal force does add to the tidal force to produce a net force that's larger than either of them individually, and this net force is what tries to break to body up.


 * But there's still one problem. When you add the $$\omega^2mr$$ term to $$\frac{2GMmr} {R^3}$$, how do you manage to end up with $$\omega^2mr + \frac{GMmr} {R^3}$$?  A coefficient of 2 has disappeared.  Is this a typo?


 * --Occultations 14:31, 24 December 2005 (UTC)

- - - - -

Let's define what the tidal force (say from the Sun) is. It is minus the force that the ground must exert in order to hold some small body on its surface, after we subtract the force that it would exert anyway due to effects unrelated to the Sun (Earth's gravity etc).

Since you're talking about the centrifugal force, you're probably working in a reference frame rotating around the Sun together with the Earth. OK, let's do so.

Let's define:

$$r$$ is the radius of the Earth

$$R$$ is the Sun-Earth distance

$$F_t$$ is the tidal force which was calculated in the article without rotation

In the rotating reference frame, the body is moving in a circle around the center of the Earth with velocity $$v = \omega r$$ (similarly to how the North pole points towards the Sun in the summer, and away from the Sun in the winter). Let's say the body is on the far side of the Earth. Its (centripetal) acceleration is $$a = -\omega^2 r$$, where we defined the direction away from the Sun as positive.

The forces acting on the body are:

(1) Gravity from the Sun = [Sun's gravity at the center of the Earth]$$ + F_t = -m\omega^2 R + F_t$$

(2) Centrifugal force = $$m\omega^2(R+r)$$

(3) Coriolis force = $$-2m\omega^2 r$$

(4) Force from the ground = [usual force] $$-$$ [tidal force]

(5) Forces unrelated to the Sun

Note that (5) and the first term of (4) cancel each other, by the definition of the tidal force.

So the net force is $$\Sigma F = F_t - m\omega^2 r - $$ [tidal force]

Using $$\Sigma F = ma$$ we get [tidal force] = $$F_t$$

What do you think?

Happy New Year!

Yevgeny Kats 09:41, 1 January 2006 (UTC)

Roche limit?
Should the Roche limit be introduced on this page? It mentions break-up of celestial objects due to tidal forces in several places, but Roche limit is relegated to a "see also" at the bottom of the page. It ought to be fairly easy to give it a 1 sentence introduction somewhere in the text...? JulesH 11:40, 26 September 2006 (UTC)

Effects of tidal forces
I would like to suggest that this section be expanded to include the effect of tidal forces on climate.

Michael H 34 02:48, 14 June 2007 (UTC) Michael H 34

Notes:

Combined effect of tidal forces and the topography of the ocean floor:

[]

Critical of a link between the variation in tides and the variation in climate:

[]

Michael H 34 04:26, 14 June 2007 (UTC) Michael H 34

More analysis:

In conclusion, we favor processes with millennial time constants (not yet identified) inherent in ocean–atmosphere dynamics as the source of the millennial climate variability; millennial variability in solar radiation (not yet discovered) is a possibility. But low beat frequencies between tidal harmonics (rather than repeat coincidence, the traditional view) cannot be ruled out by any evidence known to us; if indeed these are a factor, the low amplitude notwithstanding, then the {-1, 6, 6} combination proposed by KW is the most likely candidate.

[]

Michael H 34 04:42, 14 June 2007 (UTC) Michael H 34

This article is not about tides, as in waves crashing on the seashore tides, but the tidal force of gravity. The tidal force is what gives rise to tides, but isn't the same thing as tides. For instance, the tidal force is given as the reason for several other phenomena on this page. I think it is something like mistaking gravity and falling. Gravity causes falling, and therefore causes the effects of falling, but I don't think you would have a detailed discussion of the effects of falling on a page about gravity, would you?

At any rate, Wikipedia has a page about tides where your ideas might fit better. —Preceding unsigned comment added by 150.243.67.7 (talk) 22:34, 6 November 2007 (UTC)

tidal force/tidal height
if the tidal force is doubled then what is the effect on the height of the tides? would the tides be twice as heigh? —Preceding unsigned comment added by Em3ryguy (talk • contribs) 13:09, 14 May 2009 (UTC)

Removed Misleading diagram
I have removed the misleading figure 5, showing a force field on the entire space. Note that the tidal forces act on points of a material body, and are calculated by subtracting the forces (vectors) working on two diametrically opposed points. There are therefore no tidal forces where there is no planet. The tidal force field is defined on the planet only. DVdm (talk) 13:24, 1 January 2009 (UTC)


 * I don't agree - well I agree that technically there are no forces, but there is still the differential gravity field everywhere. The existing figure was accurate (though the caption shouldn't have said force). What is misleading is the existing figure, which implies a force only at the surface, which is definitely wrong William M. Connolley (talk) 22:05, 4 September 2009 (UTC)

Dubious measure
A recently-added fact has been marked dubious: it adds numbers to be 'more precise', but in the absence of still further definition he numbers can't be right. The alleged fact is : More precisely, the lunar tidal force along the Moon-Earth axis gives an acceleration of 1.121 × 10-7 g, while the solar tidal force along the Sun-Earth axis gives an acceleration of 0.515 × 10-7 g, where g is the gravitational acceleration at the Earth's surface.

To be precise, the wrong thing about this alleged fact arises because, when a first body is perturbed (in its motion relative to a second body) by the gravitational action of a third body, then the perturbing or tidal force, on the perturbed first body, is the (vector) difference between the third body's gravitational attractions on the first and second bodies. That vector difference has two relevant dependences: (a) on the angular relation between the three bodies, and (b) it is also directly proportional to the linear distance of separation between the centers of the first and second bodies.

The statement of the alleged fact caters for the angular-dependence point, by specifying that the perturbing or tidal force is taken at some position along a defined axis, but it says nothing about the linear distance between the relevant first and second bodies at which the quoted numbers arise.

The specific numbers can only relate to a specific distance of separation, and for correctness, the fact-statement needs to say correctly what that is for the particular numbers cited.

Terry0051 (talk) 15:19, 6 October 2009 (UTC)


 * The section to which the remark was added gives the simple mathematical description of purely static tidal forces. So motion, angular momentum and perturbation theory are not considered. Isn't it perfectly obvious that "distance of separation" is 2 Delta(r), i.e. 2*6400 km, and that M and R represent the mass and typical distance to Moon (resp. Sun), resulting in the first (resp. second) value? DVdm (talk) 16:09, 6 October 2009 (UTC)
 * I suppose one could add words along the form ... the earth's surface is offset by x Km from the Earth-Moon centre of gravity, so that the magnitudes of the tidal accelerations are this and that which compared to g are ... And I suppose a further note that x depends on the location of the surface position with regard to the Earth-Moon orientation. Presumably the supplier of the numbers has calculated correctly. There are also nice images of the tidal acceleration vectors on a grid on the surface of a sphere, but this depiction does not report the magnitudes, only the directions, nicely projected onto a flat screen.NickyMcLean (talk) 20:40, 6 October 2009 (UTC)

[From Terry0051] I'm sorry if I offended you, that wasn't intentional, but no, it's not quite obvious. It would just be helpful to state at what distance, from what body, each of these specimens of tidal (perturbing) forces applies, because they only apply at the respective distance for which they have been calculated, and they are not valid at any other.

And it really doesn't have anything to do with motions and angular momentum, because as I'm sure you appreciate, gravitational forces (aside from relativistic corrections not considered here) arise only from masses and distances, not velocities. The motions that the forces produce are another matter. Nor does it have anything to do with the Earth-Moon center of gravity, which is not a centre of attraction. On the other hand, it is all about perturbation theory, because the tidal force is only another name for a resultant perturbing force. Terry0051 (talk) 22:47, 6 October 2009 (UTC)


 * Perhaps a rewording like this?
 * For example, writing in the above equation &Delta;r as Re for Earth radius, Me for Earth mass, and M resp. R for the masses resp. average distances to Moon and Sun, the magnitude of Earth's surface of the tidal acceleration can be expressed as
 * $$ a_t = \frac{F_t}{m}\approx G ~ \frac{2 M}{R^2} ~ \frac{\Delta r}{R} = ~ 2 ~ G ~ \frac{M}{R^3} ~ R_e = 2 ~ \frac{R_e^3 ~ M}{M_e ~ R^3} ~ g$$
 * where g is the gravitational acceleration at the Earth's surface.
 * In this case the lunar tidal force along the Moon-Earth axis gives an acceleration of 1.121 × 10-7 g, while the solar tidal force along the Sun-Earth axis gives an acceleration of 0.515 × 10-7 g.
 * Free free to hone. But... isn't this beginning to smell a bit like OR? Perhaps we should either find a proper source, or throw the example away... :-)
 * DVdm (talk) 08:23, 7 October 2009 (UTC)
 * Free free to hone. But... isn't this beginning to smell a bit like OR? Perhaps we should either find a proper source, or throw the example away... :-)
 * DVdm (talk) 08:23, 7 October 2009 (UTC)

[From Terry0051] Thank you for the formulae, and for the specification 'at the Earth's surface' or '1 earth-radius distant form the center of the Earth'. I just calculated the numbers another way, and agree with yours, as long as they are only quoted to 1 or 2 sigificant places, because of the numerical approximations involved, also because of the real variability, as compared with use of mean values. Perhaps they could be given in something like this way: 'about 1.1' and 'about 0.5', both of them being × 10-7 g. From the point of view of a reader of the article, I'd suggest not including the formulae without explanation, as their rationale does not seem self-explanatory. I don't know what to suggest about the possible OR point, maybe there is a source? With good wishes, Terry0051 (talk) 08:57, 7 October 2009 (UTC)

[From Terry0051] Here's an afterthought about sources, to deal with the possible 'OR' point: How about this for a source :) for the tidal force due to the Sun at the Earth's surface along the Sun-Earth axis?

Newton writes first that the force due to the Sun to depress the sea at places 90 degrees from the Sun is 1 to 38604600 in terms of 'the force of gravity with us', and then that the corresponding tide-raising force in places directly under the Sun (or their antipodes) is twice as great as that. The ratio 2 to 38604600, neglecting completely unrealistic tail-figures, comes to 0.518×10-7 g, to compare with our calculations of 0.515×10-7 g. (Pretty good for a gravitational estimate from 1687! The corresponding 1687 Latin first-edition passage is here).

Newton's estimate for the lunar contibution was a long way out, though. That was partly because of the very poor observational basis back in the 1680s for determining the Earth-Moon mass ratio, leading to an estimate of about double what it should be for the relative strengths of Moon and Sun. Here, perhaps, a usable reliable source would be one that gives the modern figure for the relation between the tide-raising efects of the Sun and the Moon as about 45%, e.g. here. The two source-derived figures in combination then give something quite close enough to your figure for the lunar effect. Terry0051 (talk) 11:31, 7 October 2009 (UTC)


 * (Not having taken into account your afterthought), ok, good idea. Specifying (with wikilink) the origin of the gravitational acceleration and using 2 significant digits, but still expressing it "in units of g", perhaps like this:
 * For example, writing in the above equation &Delta;r as Re for Earth radius, Me for Earth mass, and M resp. R for the masses resp. average distances to Moon and Sun, and using the gravitational acceleration g at the Earth's surface given by
 * $$ g = G ~ \frac{M_e}{R_e^2},$$
 * the magnitude of the tidal acceleration at the Earth's surface can be expressed as
 * $$ a_t = \frac{F_t}{m}\approx G ~ \frac{2 M}{R^2} ~ \frac{\Delta r}{R} = ~ 2 ~ G ~ \frac{M}{R^3} ~ R_e = 2 ~ \frac{R_e^3 ~ M}{M_e ~ R^3} ~ g$$
 * In this case the lunar tidal force along the Moon-Earth axis gives an acceleration of 1.1 × 10-7g, while the solar tidal force along the Sun-Earth axis gives an acceleration of 0.52 × 10-7g.
 * As for the sources you gave, perhaps you can find a way to incorporate it in the article...
 * DVdm (talk) 11:48, 7 October 2009 (UTC)
 * As for the sources you gave, perhaps you can find a way to incorporate it in the article...
 * DVdm (talk) 11:48, 7 October 2009 (UTC)

[From Terry0051] Ok, I've offered (actually made) an amendment (and also taken off the tag), please feel free to hone, as you say.

At the same time, I notice that there is something seriously mixed-up about the formulae given a little earlier in the article, above where you placed your numerical examples. That is because they are basically couched in terms of the force on the body of mass m, instead of what would be more relevant here, in terms of the acceleration experienced by m. While that does come to an equivalent thing in respect of m itself, because one divides the force of course by the mass of m to get the accleration, it is no longer equivalent when talking about the acceleration on what might be a small particle at some distance delta-r from the center of the body of mass m, and here the mass m should not enter the equations at all, they should be expressed in terms of accelerations. But I don't have the time to fix that right now. Terry0051 (talk) 13:35, 7 October 2009 (UTC)


 * Your amendment looks good to me, and it is footnote/sourced. Unless we find a source for the derivation I proposed, we don't really need it here.
 * By the way, It wasn't me, but user Ferred who placed these numerical examples in Tidal force and Tide. :-)
 * Do you take care of amending and untagging the Tide article as well?
 * Thanks and cheers, DVdm (talk) 14:41, 7 October 2009 (UTC)

Tidal forces causing the bulge on the other side
I'm surprised no one before Feynman seems to have suggested the obvious centrifugal force from the earth's orbit (one rotation per month) around the center of mass of the moon and the earth. Common geography texts and even encyclopedia britannica are missing this.

This centrifugal force is obviously most significant as it is what balances the gravitation pull from the moon. Without it the earth and moon would fall in on each other. I would think Newton took this into account in his 1686 paper?

It is the DIFFERENCE between this centrifugal force and the the moon's gravitational pull on any material object on the earth's surface that make up the tidal forces "ellipsoid" (similarly for the earth-sun "ellipsoid". From the average values that balance each other, the centrifugal force increases linearly with the distance from, and in a direction away from the earth-moon center of mass. The gravitational pull decreases as one over the square of the distance from the moon. —Preceding unsigned comment added by 129.37.172.100 (talk • contribs) 01:23, 4 August 2004


 * if the moon were not rotating at all then the cetrifugal force at any point on its equator, due to its motion around the earth, would be a vector equal to  the vector of earths gravity at the center of the moon. but the earths gravity upon the near and far sides of the moon differs from this. this difference is the tidal force. the moons rotation does produce a centrifugal force but it is canceled out by a uniform bulge at the moons equator and is therefore irrelevant. just-emery (talk) 05:59, 13 May 2009 (UTC)

A simple back of the envelop calculation shows that indeed the centrifugal force due to the motion of the earth around the center of mass of the earth-moon system (which 1700km below the earth surface facing the moon) will give a resultant force much larger (by up to 30 times larger) than the simple change in the moon gravity from one side of the earth facing the moon to the side opposite to it. THe fact that popular books only site gravity is because it is easier to explain, but the contribution of the centrigual force is the main component of the force responsible for tides on earth. —Preceding unsigned comment added by Just an astrophysicist (talk • contribs) 15:50, 31 March 2010 (UTC)

OK, I am not sure how all this work to make a correction to the VERY WRONG article on WIKI sbout tidal forces that says "mathematical treatement"... I am trying to correct it but I am refrain from doing so and I am asked to first discuss it ? and give references... it is simple astro I with simple calculus:

ERRATA

The (main) component of the tidal force on earth (therefore viewed in a rotating frame of reference, where the rotation considered here is the one about the motion of the center of mass of the earth-moon system) is from the centrifugal acceleration due to the motion of the earth about the center of mass of the Earth-Moon system. A detailed treatment taking into account both the gravitational attraction of the moon and the centrifugal acceleration is being worked on currently below (work in progress).

The angular velocity of the Earth-Moon system is given by $$ \omega $$

$$ \omega^2 = G \frac{ M_m + M_e }{d^3}, $$ where $$ M_m $$ is the mass of the Moon, $$ M_e $$ is the mass of the Earth, and $$ d $$ is the distance from the center of mass of the Earth to the center of mass of the Moon.

Let the Earth have a radius :$$ R_e $$, and the center of mass of the Earth-Moon system (point $$ O $$) is located on the line joining the center of mass of the Moon to the center of mass of the Earth (point $$ A $$) a distance $$ d_1 $$ from the center of mass of the earth. When the moon is at the zenith the point $$ O $$ is located under your feet about 1,700km under the crust of the earth. In other words we have

$$ d_1 < R_e ; ~ R_e -d_1 \approx 1,700km $$

Let us now denote by $$ B $$ the point on the earth surface exactly facing away from the Moon, and :$$ C $$ the point on the earth surface where the Moon is at the zenith. Thus the points $$ B, A, O, C $$ are on a straight line, the line joining the Earth to the Moon.

Thus we have :

$$ B A = R_e , $$

$$ A O = d_1 , $$

$$ O C = R_e - d_1. $$

Let us now consider the effect of both the centrifugal force (due to the rotation of the earth about the point $$ O $$ at an angular speed $$ \omega $$) and the gravitational attraction of moon on points $$ A, B, C $$.

On point $$ A $$

The gravitational attraction due to the Moon only on a mass $$ M $$ located in A is

$$ F_g^A = \frac{G M M_m}{d^2} , $$ directed towards the Moon.

The centrifugal force in A due to the motion of the earth about the point O is

$$ F_{cent}^A = M \omega^2 d_1, $$ directed away from the Moon.

If we denote by $$ d_2 $$ the distance from O to the center of mass of the Moon, then we have

$$ d_1 + d_2 = d , $$

and

$$ d_1 M_e = d_2 M_m $$

which gives

$$ d = d_1 ( 1 + \frac{M_e}{M_m} ) $$

Replacing $$ d_1 $$ and $$ \omega $$ into the expressions for the centrifugal and gravitatiaonal forces we find simply (taking care of the signs)

$$ F_g^A + F_{cent}^A = 0. $$

Which is not a surprise since this is how the circular orbits of the Moon and Earth about the common center of mass is obtained, namely by assuming that the centrifugal acceleration and gravitational accelleration ballanced each other for the point masses Earth and Moon.

Now for point B and C, it is a little more algebra.

In point B, the gravitational force on a mass M due to the Moon is given by

$$ F^B_g = \frac{G M M_m}{d+R_e}^2. $$

And the centrifugal force there is

$$ F^B_{cent} = M \omega^2 ( R_e + d_1 ) $$

We now use

$$ \omega^2 = \frac{G (M_m + M_e)}{d^3} , $$

and get $$ F^B_{cent} = \frac{MG(M_m+M_e)d_1}{d^3} + \frac{MG(M_m+M_e)R_e}{d^3}. $$

We also replace $$ d_1 $$ by its expression $$ d_1 = \frac{M_m}{M_m+M_e} d $$

and obtain for the centrifugal force on M at B

$$ F^B_{cent} = \frac{GMM_m}{d^2} + \frac{GMM_m}{d^2}\frac{R_e}{d_1}. $$

Since $$ R_e - d_1 =1700km $$ we write

$$ \delta = R_e - d_1 $$

or

$$ R_e = d_1 + \delta $$

and obtain

$$ F^B_{cent} = \frac{GMM_m}{d^2} + \frac{GMM_m}{d^2}\frac{d_1 + \delta }{d_1}. $$

which is then written: $$ F^B_{cent} = \frac{GMM_m}{d^2} + \frac{GMM_m}{d^2} + \frac{GMM_m}{d^2} \frac{\delta }{d_1}. $$

We now write $$ \eta = \frac{\delta}{d_1} $$ and finally get $$ F^B_{cent} = \frac{GMM_m}{d^2} (2 + \eta ). $$

This force is directed away from the Moon, while the gravitational force is towards the Moon. Substracting the centrifugal force to the gravitational force gives:

$$ F^B_g - F^B_{cent} = \frac{GMM_m}{(d+R_e)^2} - ( 2 + \eta ) \frac{GMM_m}{d^2} $$

We expand the first term using the rule $$ \frac{1}{(1+\alpha)^a} \approx 1 - a \alpha + ... $$

where $$ \alpha << 1 $$.

or

$$ \frac{1}{(d+R_e)^2} = \frac{1}{d^2(d + (R_e/d))^2} \approx \frac{1}{d^2} ( 1 - 2R_e/d ). $$

and we obtain

$$ F^B_g - F^B_{cent} \approx -\frac{GMM_m}{d^2} ( 2 + \eta - 1 + 2 \frac{R_e}{d} ) , $$ where the first 2 terms in the parenthesis are due to the centrifugal force while the last two terms are due to the gravitational pull of the Moon. We can write this as

$$ F^B_g - F^B_{cent} \approx -\frac{GMM_m}{d^2} ( 1 + \frac{R_e-d_1}{d_1} + 2 \frac{R_e}{d} ) , $$

and note that now the first two terms are from the centrifugal acceleration while the last term in the parenthesis is from the gravitational pull of the Moon. Putting in numbers, we get for the acceleration (dividing by the mass)

$$ a_{tide}^B = \frac{GM_m}{d^2} (1 + 0.3634 + 0.03122). $$

For the point at C we obtain (to the same order of approximation) exactly the same expression. We see that the the centrifugal acceleration in points B and C contribute 1.3634 while the gravitational pull of the Moon contributes only 0.03122.

Every 1st year student who took some calculus class and can expand using some Taylor series will be able to follow. So I am starting here and will finish a little later as I need some time to type all this down. —Preceding unsigned comment added by Just an astrophysicist (talk • contribs) 17:42, 31 March 2010 (UTC)


 * Hi. As is explained on your talk page, you need to provide a reliable source for this, otherwise it will be considerd as original research and removed. That is how things work here. Cheers, DVdm (talk) 17:50, 31 March 2010 (UTC)

OK, Thanks for pointing this out. I'll what I can do. In the meanwhile I am putting here the rest.


 * I think, DVdm, you have been just a little severe with JaA. A bit of basic calculus is hardly "original research", and JaA is attempting to fill in a fundamental gap (failure!) in the article. We should be assisting him, not making it difficult for him. I've been aware of this issue for a while, but with "so much to do and so little time", I've done nothing. Mea culpa. So thank you JaA. If you have relevant references at your fingertips, then so much the better, and do please hang in here and continue contributing. Also, please sign your comments by typing four tildes ~ . --Epipelagic (talk) 03:39, 1 April 2010 (UTC)


 * We can't invent physical justifications that do not exist in sources and then claim to be doing only basic calculus. Not only is it original research, but in this case it is also clearly incorrect. For example, see this page, as well as the section entitled "Tide Generating forces and Tidal Constituents" on this page. ▻Tim Shuba (talk) 17:59, 1 April 2010 (UTC)


 * See also this and this and this. DVdm (talk) 08:01, 2 April 2010 (UTC)

Centrifugal force not necessary
While convienient to ease explanation, centrifugal force (or rotation) is not necessary for tidal forces to occur. If the moon was falling "directly towards" earth, the tidal forces would also be felt. (Of course, as the two bodies come closer, tidal forces varies). —Preceding unsigned comment added by 80.201.174.61 (talk • contribs) 11:53, 12 August 2004


 * possible, but would the tidal effect on earth as is, be as strong without the centrifugical force? or am i missing the point here —Preceding unsigned comment added by Paulroyaux (talk • contribs) 04:16, 14 August 2004

This just shows that even Feynman was not always right. The centrifugal effect varies linearly with position, so it can be modelled as a constant plus a radial force that increases with the distance from the centre of the Earth. That means it plays no direct part in the formation of any of the bulges.


 * Actually, Feynman was completely right, and the centrifugal force is quite important. If the moon would turn around the earth in another orbit, the tides would be different. In fact, the earth is rotating around itself, the moon, and the sun, and we don't live in an inertial system. It is completely appropriate to describe our real life situation with the pseudo forces of a rotating system. NicholB (talk) —Preceding undated comment added 11:27, 25 November 2010 (UTC).

Thus, the cause of the difference between the bulges towards and away-from the moon has to be due to a non-linear variation in the effective field - such as is produced by the inverse square law of the moon's gravitation. If we expand the 1/r2 law locally around the centre of the Earth, we have:

a constant term and a linear term that play no part in the tidal generation;

a second order term that is responsible for the existence of the bulges and dips, but would give equal bulges towards and away from the moon; and

a third order term that is responsible for the bulge facing the moon being larger than the one in the opposite direction (plus higher order terms, of course).

This was already addressed in a slightly different way below PhysicistQuery 20:15, 21 July 2007 (UTC)

Formula
Having pondered this a bit, I don't think I believe the formula. Certainly the one on the page has a typo: its literally meaningless to have the formula immeadiately followed by "<< r". Also, dr is undefined. Check back to before it was TeX'd: a comma is now missing.

But also, I'm not sure the intended formula is accurate either. I think that if the outside grav field is:


 * a = GM/R^2

(a = acceleration; M mass of central body; R distance from) then the "tidal force" between the center and edge of a body radius r at diatance R is:


 * delta-a = GM ( 1/R^2 - 1/(R-r)^2)

which is then *approximately* equal to:


 * delta-a = GM ( 2/R^3 ) . r

if r << R (modulo a sign convention or two...). (William M. Connolley 09:35, 27 Aug 2004 (UTC))


 * A clear derivation can be found at fr:Discuter:Sphère de Hill.


 * Urhixidur 14:47, 2004 Aug 27 (UTC)

A look at the history shows an interesting migration in the formula over time, but it was never actually right; an extra factor of 2 got inserted early on. I rewrote most of the article, and I hope it's right now. Fpahl 21:24, 6 Oct 2004 (UTC)

I think the article needs to explain better why there is a force in two directions instead of simply adding together in one direction. --ShaunMacPherson 03:19, 17 Oct 2004 (UTC)

THis is elementary, and

Saturn's rings
The caption under the Saturn photo has been changed. Saturn has moons both large and small within the rings; they are responsible for the gaps in the rings (see moons of Saturn). Tidal disruption caused by the gravity of Saturn itself, and not the gravity of its moons, is responsible for the existence of the rings. This is discussed in the rings of Saturn article. Piperh (talk) 11:17, 14 July 2009 (UTC)


 * In support of the above, this modification to the caption is still not quite accurate; there are moons inside the rings, e.g. Pan (moon). I have tweaked the caption to make it more factual. kostmo (talk) 07:11, 13 March 2011 (UTC)

Centrifugal force
The formula :$$\vec a_t$$(axial) $$ ~ \approx ~ \pm ~ \hat r ~ 2 \Delta r ~ G ~ \frac{M}{R^3} $$ must be multiplied with 1.5 respecting the occuring centrifugalacceleration. See the Swedish page of Tidal Force! The right formula is : :$$\vec a_t$$(axial) $$ ~ \approx ~ \pm ~ \hat r ~ 3 \Delta r ~ G ~ \frac{M}{R^3} $$ —Preceding unsigned comment added by 217.208.122.59 (talk) 17:12, 31 March 2011 (UTC)


 * The Swedish article on tidal force, Tidvattenkrafter, is wrong (Google translate yields a good translation). It adds a correction for a variation in centrifugal force across a cross section of a perturbed body such as Earth to the existing variation in gravitational force. But no such variation in centrifugal force exists according to George Howard Darwin, a world renowned authority on tides, who states that centrifugal force is constant inThe tides and kindred phenomena in the solar system (1898, pages 91–106, non-mathematical). A similar non-mathematical treatment by NOAA makes the same statement relative to the centrifugal force due to the Earth and Moon revolving around their common center of mass. Because Earth is a coherent body, all points within or on its surface acquire this centrifugal force . A mathematical treatment in the Canadian Tidal Manual says the same thing, except that it uses centripetal force directed toward the center of a circular orbit around the Sun, the opposite of centrifugal force. Furthermore, the Swedish article has no references at all, so its hypothesis (added by Bekos who promoted it on its talk page) is original research. — Joe Kress (talk) 04:48, 2 April 2011 (UTC)

If your sources not have mentioned a variation in centrifugal force, they must have missed a central physical fact. Everyone who has been sitting on a chair in a merrygoround are sure there is a centrifugal force (in those cases not neutralized by the gravity) and they are even sure that the centralfugal force increase the farther they sit from the rotationcentre and decrease the nearer they sit from the rotationcentre. The sun-earth-system is a merrygoround. Why is this system not influenced of physical facts? The earth-moon-system is a rotatingsystem round their barycenter, their center of gravity, like a merrygoround. Why is this system not influenced of physical facts? I have changed the arguments in Tidvattenkrafter to make the understanding easier. I beg your pardon for my bad english, but I hope you understand my meanings. Still I insist that the right formula is :$$\vec a_t$$(axial) $$ ~ \approx ~ \pm ~ \hat r ~ 3 \Delta r ~ G ~ \frac{M}{R^3} $$ — Preceding unsigned comment added by Bekos (talk • contribs) 16:29, 5 April 2011 (UTC)


 * You are ignoring a central principle of Wikipedia. You must cite your source. I have cited several reliable sources for the classic position that for tidal force, centrifugal force is constant across Earth, but you have not cited any source for yours. Any argument that you make that is not already in some reliable source (trovärdiga källor) is your own opinion, that is, it is original research (originalforskning), which is not allowed in Wikipedia. — Joe Kress (talk) 18:52, 5 April 2011 (UTC)

if the earths rotation with respect to the stars is eliminated then the centrifugal force due to its motion around the earth-moon barycenter is the same everywhere on earth.

Just granpa (talk) 22:39, 5 April 2011 (UTC)


 * In the beginning of the article "Matematisk diskussion" i declare that it follows the gravitymodel of Newton, the the formula ag = -GM/R^2 for gravityacceleration is wellknown and the formula for the reactiv centrifugalacceleration ac = +v^2/R is wellknown as well. If you handle a body there you are thinking all masses in singularity ther is no problems. ag + ac = 0. The accelerations are neutralizing each other and the body is in a free fall. But ag is not the same all over the body if you add +/-&Delta;r to R. G and M are constants and not chances but R^2 chances and ag whith it. ac are not the same all over the body if you add +/-&Delta;r to R. Both v and R changes and with them ac. This is wellkown matematic facts and no original research. But if you say ac not chance if you chance v and R it is original research, if you don´t have a good explanation for that non-physical fenomen. — Preceding unsigned comment added by Bekos (talk • contribs) 08:04, 7 April 2011 (UTC)


 * ac is a function of r and v but if the orbit is perfectly circular and the body isnt rotating with respect to the stars then r and |v| are the same at every point on the earth and therefore ac is the same at every point on the earth (only the direction changes with time).
 * Just granpa (talk) 00:54, 8 April 2011 (UTC)

I added the section Relation with non-uniform centripetal force.--Patrick (talk) 11:18, 8 April 2011 (UTC)

To granpa: I can´t see what the stars have to do with it. The earth is rotating with respect to the sun in the sun-earth system and the earth and the moon are rotating with respect to their epicenter. A merrygoround on earth is rotating with respect to its rotatingcenter. Either of them are rotating with respect to the stars. What´s the difference? Is´nt there the same physical effect at every one of them? Am I missing something?

To Patrick: Where can I see the section Relation with non-uniform centripetal force? — Preceding unsigned comment added by Bekos (talk • contribs) 10:41, 10 April 2011 (UTC)


 * . Did you refresh the page?--Patrick (talk) 12:30, 10 April 2011 (UTC)


 * Rotation is not just relative like velocity. Absolute rotation ("with respect to the stars") involves "centrifugal force".--Patrick (talk) 12:39, 10 April 2011 (UTC)

Now I can see your section Relation with non-uniform centripetal force. If I understand you right, you mean that you adding 50% only in extreme cases. I mean that you should do it in all cases when a body rotate round another body. I have an example. Suppose that you have a body consisting of three bodies with masses of 1 kg attached to each other with weihgtless stiff bars with length of 3.5797E7 meters between them. The middle body is rotating round the Earth on the geostationary height, 4.2175E7 meters, over the Equator. The gravityforce, Fg, is 2.2427E-1 N on the middle body and on the body near the ground, but not touching it, it is 9.8060 N. It is the same gravityforce as on a body of the same mass at the same distance from the Earth center on the North Pole. The centrifugalforce, Fc, on the middle body is -2.2427E-1 and it is weihgtless because Fg + Fc = 0. My formula Fc(1-&Delta;r/R)gives -3.3916E-2 N for the nearground body and Fg + Fc gives 9.7720 N, the same as for a stone of same mass lying on the ground on the Equtorline. If Fc is the same all over the the threepart body is Fg + Fc for the nearground body 9.5817 N. That gives a indication that I am right when I say that Fc changes and not is homogen all over the body.

introducttion messed up
the introduction of this article is messed up .. rewrite please !! 82.66.112.67 (talk) 03:50, 8 February 2012 (UTC)

Centrifugal
User removed. Note that the edit summary is however not related to the content of the section. The edit summary talks about "...lunar centrifugal forces are uniform everywhere on earth..." whereas the section is about two forces acting upon the secondary body (i.e. the moon): the tidal forces caused by the earth on the one hand, and the centrifugal bulging force caused by the secondary bod body's own rotation about its own axis on the other hand. So I have restored the section—in a somewhat clumsy way, for which my apologies. It would be a good idea to have the current section sourced however. - DVdm (talk) 13:00, 23 February 2012 (UTC)

Note. I have changed the and added a CN-tag. - DVdm (talk) 13:16, 23 February 2012 (UTC)


 * Thank you for your comment. I'm sorry but the commentation field was short and did not allow a more thorough explanation. In fact it is my understanding that centrifugal forces play no role in tides and that this has been known since Newton. For a clear demonstration see Griffin's paper, which ends as follows (p. 131): "So Newton was entirely correct. The tides are entirely due to his inverse square law of gravity, which predicts that the gravitational field of the Moon or Sun will be greater on one side of the Earth than the field on the other. It has nothing to do with the rotation of the system or with centripetal or centrifugal forces." Although Griffin himself only treats the moon's effect on the earth's tide, his reasoning and method when applied to the sun's effect leads the same conclusion: differential gravity and not centrifugal forces are responsible for tides. This is something which Griffin points out is often misunderstood: "A number of websites give completely incorrect explanations for the tides" (p. 129). Take a look and see what you think. Thanks!
 * Griffin, A., 2008. Tides, as explained by Newton. Physics Education 43, 129–131. — Preceding unsigned comment added by Tomukas (talk • contribs) 17:59, 23 February 2012 (UTC)
 * I agree with all that, and I did some searchin' and readin' too (e.g. here on page 156), but the little section does not say in any way that centrifugl forces are resposible for or play a role in tidal forces. It just says that a body can be torn apart by a combination of tidal forces and centrifugal forces, and it gives a specific example, which is i.m.o. sufficiently interesting to be mentioned here. But a source would be very welcome :-) - DVdm (talk) 19:11, 23 February 2012 (UTC)


 * The Roche limit for a satellite is the orbital diameter at which the surface gravity of the satellite is offset by the tidal force - that is, the centre of the satellite is orbiting at the proper distance, its surface is closer to the central body and so is attracted more, the tidal force, and for a large enough diameter and a close enough orbit this might be enough that a pebble at the surface would float away. If the satellite were rotating then the centrifugal force would add to this effect, with appropriate considerations of alignment of rotation and such details. On the earth, the acceleration of gravity is greater at the poles than at the equator due to the poles being closer to the centre of mass, but also, there is a centrifugal effect due to the earth's rotation. This has no connection to the tides except in that the level of the equipotential surface relates to altitude at a rate due to the nett force at a location. The centrifugal force due to the earth as a whole orbiting around the earth-moon centre of mass would be the same in magnitude and direction for the earth as a whole (but, as the surface's orientation varies, its direction wrt. the surface varies) except that the earth's shape will be deformed slightly due to it, just as its shape is deformedf roma sphere due to the daily revolution. To proceed further, careful calculation of actual forces is required, with a deision as to where the point of view is... NickyMcLean (talk) 20:04, 23 February 2012 (UTC)

Hi again and my apologize for the delay in continuing the discussion. My objection was to the line

this "centrifugal force" adds 50% to the tidal force. This I believe could easily be wrongly interpreted as meaning that the centrifugal force adds, that is contributes to tidal force, which is not true. (Centrifugal forces/effects do not contribute to tidal forces/effects). However it is true in this case it should be added to the tidal force to produce the total force capable of tearing a rotating body and indeed that for synchronous rotation the term's magnitude is 50 % of the tidal term. I encountered this entry while studying tidal forces and it puzzled me. However, the section I believe is valid, if the line above is slightly altered.

" ....this "centrifugal force" contributes an amount equal to 50% of the tidal force contribution." I attach a version with this alteration, and which also includes a mathematical treatment which supports the section.

Relation with centrifugal force

If a secondary body orbits a primary body, the forces that could tear the second body apart if its strength and internal gravity are not enough, are the tidal force and the "centrifugal force" associated with any rotation of the secondary body about its axis. In the case of synchronous rotation with the secondary body much smaller than the primary, on the line through the centers of the two bodies this "centrifugal force" contributes an amount equal to 50% of the tidal force contribution.

Mathematical treatment

Any rotation of the secondary body (in an inertial reference frame) will lead to centrifugal forces which can contribute along with the tidal forces to the total forces which might tear a body apart. A special case arises when a secondary body achieves synchronous motion about a larger primary body. The problem can be treated by changing from an inertial reference frame to a rotating reference frame. In synchronous orbit, the secondary body is now fixed (that is, not rotating) in the rotating reference frame. Under these circumstances the centrifugal force associated with the secondary body can be treated easily beginning with the equation of motion in the rotating co-ordinates system: [1]

$$\mu \ddot{r} = gravitational force + centrifugal force$$

or

$$\mu \ddot{r} = \dfrac{GM_1M_2}{r^2},+ \mu\omega^2 r $$

where r is the magnitude of the vector from the secondary to the primary, that is the distance between (the centers of gravity of) the two bodies, $$\omega$$ is the angular velocity and $$\mu$$ is the reduced mass given by

$$\dfrac{1}{\mu}=\dfrac{1}{M_1}+ \dfrac{1}{M_2}$$

which is approximately the mass of the smaller body when $$M_1 \gg M_2$$. For the simplest case of a circular orbit, a balance applies at the center of gravity of the secondary body and the acceleration is zero. Thus we can write

$$\dfrac{GM_1M_2}{R^2} = \mu\omega^2\mathit{R}$$

where R is the radius of the orbit.

Considering the line between the two bodies we observe that the balance of forces present at the center of gravity would not exist elsewhere on the line. The gravity differential amounts to the tidal force and the centrifugal differential can be easily calculated in the rotating reference frame since the angular velocity is constant and thus the differential centrifugal force simply varies with distance. Thus, in particular, at the points where this line goes through surface of the secondary body we can calculate a force

$$F=\dfrac{GM_1M_2}{(R\pm \mathit{\Delta} r)^2}+\mu\omega^2(R\pm \mathit{\Delta} r)$$

where $$\mathit{\Delta} r$$is essentially the equatorial radius of the secondary body. This expression can be simplified to

$$F \approx \dfrac{\pm GM_1M_2}{(R^3}(2\Delta r+ \mathit{\Delta} r)$$

where first term is the tidal contribution and the second the centrifugal term. Thus on both sides of the secondary body the centrifugal term is seen to add to the total force by an amount which is half that of the tidal contribution.

[1] Mechanics, vol 1, pp275-281, Charles Kittel, Walter D Wright, Malvin A Ruderman, McGraw-Hill 1965. --Tomukas (talk) 11:08, 4 March 2012 (UTC)


 * A few remarks:
 * It's not Walter D Wright, but Walter D. Knight (see http://books.google.be/books?id=bDkZYAAACAAJ)
 * I don't find the above derivation (nor the final equations, nor any mention of tidal forces tearing bodies apart) in my copy of the book. Perhaps you have another edition. Can you point to the exact page and to the chapter number and section title where we find the derivation and, specifically, the final equation?
 * Taking into account the equation
 * $$\dfrac{GM_1M_2}{R^2} = \mu\omega^2\mathit{R},$$
 * I checked the first order Taylor approximation of your equation
 * $$F=\dfrac{GM_1M_2}{(R\pm \mathit{\Delta} r)^2}+\mu\omega^2(R\pm \mathit{\Delta} r).$$
 * I get a rather different result, for &Delta;r << R:
 * $$F \approx \dfrac{GM_1M_2}{R^3}(2R \mp \mathit{\Delta} r),$$
 * so the second term can only be 50% of the first term when &Delta;r ~ R, in which case the approximation is not valid, and furthermore meaning that the secondary body's equatorial radius &Delta;r equals its orbit radius R around the primary body, which is of course nonsense :-)
 * Unless the above derivation and equations can be found in the book, it looks like this is wp:original research, which, per policy, can of course not be included here. Furthermore, if the derivation is correct, it clearly invalidates the second sentence of the text paragraph. Perhaps we should remove it, pending a proper source. DVdm (talk) 13:45, 4 March 2012 (UTC)


 * Note - I and removed the dubious unsourced statement until we have a good source for it. I left the cn-tag on the first statement, although that one is pretty obvious. Let's find the source(s). - DVdm (talk) 12:53, 5 March 2012 (UTC)

Figure 4 Caption Confusing
I am not an expert, but surely in the caption the phrase "...once the field of the sphere is subtracted..." is completely misleading. There seem to be present here "a body to the right" and also the sphere actually depicted in this figure. The "field of the sphere", then, surely would refer to the gravitational field generated by the presence of the depicted sphere itself. Yes, this field is neglected (I presume on the grounds that the sphere is less massive than the "body to the right", and that the "body to the right" is not too distant). It is neglected from both parts of the figure, but this neglected field is not that which we subtract in order to go from the force (or acceleration) vectors shown in the top part of the figure to those vectors shown in the bottom part. What is subtracted is something different: the single gravitational force vector at the centre of the sphere due to the "body to the right" (whose shape has never been discussed, although, incidentally, I suppose it is usually assumed for simplicity to be a concentrated point mass or else a uniform 3D sphere - or else at least a 3D sphere composed of nested uniform 2D spherical shells) is subtracted at all points of the field depicted in the top part, to produce the field depicted in the lower part. — Preceding unsigned comment added by 83.217.170.175 (talk) 18:19, 8 August 2012 (UTC)


 * I agree that the caption is very poorly formulated. Wikipedia is all yours, so feel free to fix it. - DVdm (talk) 18:55, 8 August 2012 (UTC)

I've had a go. The text associated with the image file gives a better description. NickyMcLean (talk) 21:49, 8 August 2012 (UTC)


 * Good catch and ditto job. - DVdm (talk) 07:16, 9 August 2012 (UTC)

Subtraction of vectors
Again on figure 4 caption. WHY is the central arrow/force subtracted ? Force vectors never subtract. Only can add an opposite vector. So which is the left-pointing vector, constant on all sides of the central body, that is added to the right-pointing vectors to produce the difference, if it is not the centrifugal force, balanced with the gravity attraction on the center of the body? 83.208.147.243 (talk) 22:20, 17 January 2013 (UTC)
 * It is not true at all that forces cannot be subtracted. Subtracting a vector is exactly equivalent to subtracting an opposite vector, just as subtracting a number is equivalent to adding its negative, but that doesn't mean that numbers can't be subtracted. JamesBWatson (talk) 22:37, 17 January 2013 (UTC)

Consider re-write of entire article
Please consider rewriting the entire article. It is a shame that one of the most fundamental and also important phenomenology in physics is presented in such a poor manner! As a physicist and teacher I think I know what I am talking about here. The article are missing a great deal of very important detail including a several typos and inconsistent statements. A particularly irritating feature, are the missing and confusing pictures. If you ever hope that this article will be at least minimally useful, please add material and fix all the problems. There should be a tag on this article warning for incomplete material. Jahibadkaret (talk) 19:08, 2 September 2010 (UTC)


 * You are invited to rewrite the article yourself, but beware that copyrighted images are forbidden on Wikipedia—you must either draw them yourself or copy them from a pre-1923 source.— Joe Kress (talk) 02:59, 3 September 2010 (UTC)


 * Point taken. The article have just grown a bit too fast by too many excited contributors. It has a lot of meat, but no bone structure. But I really don't think the only images available are from pre-1923! Jahibadkaret (talk) 23:44, 4 September 2010 (UTC)


 * Yes, any image in the public domain is allowed. A prime source is anything written by an agency of the United States government, which are by law in the public domain. However, this is relatively unique to the the US government. Many (most?) other governments retain a copyright in their works, including governments of states of the United States, the British government, and other commonwealth countries. Unfortunately, I've found some works that are co-written by an ageny of the US government and an agency of the British government. Because the latter retains a Crown copyright, that work would not be in the public domain. Another problem is that many images and other works (Wikisource) placed on Wikipedia, although released into the public domain by the uploading editor, are actually copyrighted images or works which should never have been uplaoded. Such images are very difficult to identify. It is relativey easy to identify copyrighted texts, which are often uploaded from an online source which can be Googled. — Joe Kress (talk) 02:45, 5 September 2010 (UTC)

This article sounds like it is written by someone who is not native speaker or has trouble writing clearly.

Please do not revert every attempt to improve the language. If there is a factual problem (with your understanding) discuss that rather than reverting wholesale any improvements to article. — Preceding unsigned comment added by 82.253.182.227 (talk) 21:22, 28 April 2013 (UTC)

@ Flyer22 (talk | contribs)‎ "Revert. That's not the correct order, per WP:TALK." Not according to this. Rosemary Cheese (talk) 12:07, 19 May 2013 (UTC)
 * I reverted you because you moved this section out of its talk page order. Look above: This section was started in 2010; it's in the correct order with regard to the sections that were started before and after it. A new reply in an old section does not mean that we place the old section at the bottom. Per Talk page guidelines, the newest section goes at the bottom, not the newest reply. That's why the talk layout guideline states "Make a new heading for a new topic." This section is not a new topic. Flyer22 (talk) 16:49, 19 May 2013 (UTC)

Trolls?
It seems that the article has somehow been railroaded by people with an anticentreptal penchant. But their diagram showing strange gravitational forces -- with some vectors actually pointing away from BOTH of the two bodies involved -- is obviously wrong. If you can't explain the effects coherently yourselves, stop railroading the discussion. You aren't doing the science of tides any favors. Or is that the whole aim -- we have some science trolls here? — Preceding unsigned comment added by AtomAnt (talk • contribs) 13:29, 22 September 2013 (UTC)


 * Nah, no trolls. Don't forget what the arrows represent. They are not in any way related to or depending on the attraction by the (on-picture) planet. They arise by the differences in attraction between the forces exerted by the (off-picture) satellite only. They are the vector differences of the attraction of the matter elements by the satellite, minus the attraction of the elements if they would be at the center of the planet.
 * For instance, L being an element on the left, C an element at the center, R an element on the right, the net attractive forces by the satellite only are like this:

L--->       C->      R--->                             (satellite)
 * Subtracting the arrows at C, this results in this:

<--L           C            R-->                                  (satellite)
 * You can easily see how this works for the elements at the top, at the bottom and at intermediate places, but they are too difficult to draw here. Hope this helps. - DVdm (talk) 15:13, 22 September 2013 (UTC)

Unhelpful diagram


This picture in the section "Explanation" shows only a single body (is it the Earth?) with arrows on it. Doesn't the direction of the force depend on the position of the moon? Axl ¤  [Talk]  11:28, 11 December 2008 (UTC)


 * Yes, the moon can be on either side (here right or left). The effect is the same. See also talk page section Talk:Spaghettification on article Spaghettification. DVdm (talk) 13:24, 1 January 2009 (UTC)

The problem with this diagram is with the accompanying text. This is not a diagram of "Moon's (or Sun's) gravity differential field at the surface of the earth". If it was ALL the arrows showing gravitational force from the satellite would point TOWARD the satellite. Yes, the ones on the far side would be smaller arrows than the ones on the side near the satellite (and yes this would produce distortion) -- but the gravitational field produced by a body ALWAYS point toward that body.

What this diagram actually shows is the balance of the satellite's gravitational field against the centrifugal force. The equal and opposite arrows show how the gravitational field and and centrifugal force cancel out -- thus the distance between the two bodies does not change. Though equal the centrifugal force is strongest on the side away from center of rotation (the left side in this case) and the gravitational pull of the satellite is strongest on the side closest to itself (the right)- which leads to tidal forces outward on those two sides of the earth.

I recognize that this diagram is meant to show the 'gravity differential field' by showing the gravitational force at a given point after you subtract out the 'average' field strength from the satellite found at the center of the planet. But that is not what the caption describes. It just says this is a gravity differential field, and anyone not already familiar with planetary field differential be left thinking that the satellite's gravitational field points both left and right somehow.

I came here to learn why ocean tides can be diurnal and can't make any sense of it here.

But its not just this diagram -- why is the centrifugal force getting short-shift throughout this whole article? Centrifugal force has equal affect on the planet as the moons gravitational force. Don't we all realize that the centrifugal force HAS to be equal and opposite to the moons gravitational pull -- otherwise the orbit would not balance and the earth and moon would fall into each other. — Preceding unsigned comment added by 98.165.151.66 (talk) 10:03, 15 April 2014 (UTC)


 * See below in Talk:Tidal_force. - DVdm (talk) 20:53, 19 December 2014 (UTC)

Readability edit reverted
I suggested this edit for increasing the readability of a sentence: https://en.wikipedia.org/w/index.php?title=Tidal_force&diff=693111612&oldid=693110042

It was reverted for "introducing factual errors". I am interested in getting my edit approved or having my error explained to me. — Preceding unsigned comment added by Wojje (talk • contribs) 30 November 2015‎ 15:30 (UTC)


 * Please sign all your talk page messages with four tildes ( ~ ). Thanks.
 * I explained on my talk page: . Please do read the sentence that you are editing . It is about "the perturbing force on the Moon" (ON the moon) and a colon follows it. - DVdm (talk) 15:34, 30 November 2015 (UTC)

Too complex
This is too complex and elitist. It seems to be written by someone who wants to show off rather than explain. — Preceding unsigned comment added by 62.107.57.180 (talk) 00:49, 3 May 2016 (UTC)


 * Please sign all your talk page messages with four tildes ( ~ ). Thanks.
 * As far as I can see the article explains well, and does not suffer much from complexity or elitism. As you can verify, it is written by over 250 different editors over a period of almost 15 years. Do you have anyone or anything in particular in mind?
 * By the way, as this wasn't mentioned yet in the text, I have added a pointer to Spacetime curvature in the See also section. Search for "tidal effects" in that article. - DVdm (talk) 09:55, 3 May 2016 (UTC)

No calculus needed.
Why in the world does the derivation use a Maclaurin series expansion? Simple FOIL algebra is all that is needed. Why make the article less accessible to those who haven't studied calculus? Throwing out the HOTs also obscures the fact that the force is not linearly symmetrical, which is interesting I think. One could just ignore the 1/R^2 in the denominator and get the same approximate result, but why? For Earth-Moon I calculated (average distances/ diameters) the force is 1.095e-7 g opposite the moon and 1.151e-7 g nearest to the moon. Jesse1919 (talk) 23:17, 13 April 2014 (UTC)


 * I guess you mean like this:
 * $$\vec a_g = - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{(1 \pm \Delta r / R)^2} = - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{(1 \pm \Delta r / R)(1 \pm \Delta r / R)} \stackrel{\text{(FOIL)}}{=} - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{1 \pm 2\Delta r / R + (\Delta r / R)^2} \stackrel{\text{(ignore denom } \tfrac{1}{R^2} \text{)}}{\approx} - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{1 \pm 2\Delta r / R}$$
 * followed by
 * $$- \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{1 \pm 2\Delta r / R} \approx - \hat r ~ G ~ \frac{M}{R^2} \pm \hat r ~ G ~ \frac{2 M }{R^2} ~ \frac{\Delta r}{R} - \cdots$$
 * Other than by using the Maclaurin series of 1/(1 + 2x) being 1 – 2x + 4x2 – ..., how would you justify the last (additional) approximation with simple FOIL algebra? - DVdm (talk) 08:41, 14 April 2014 (UTC)
 * Oh, I give in, what's FOIL then? William M. Connolley (talk) 21:30, 14 April 2014 (UTC)
 * I was baffled too. Googling FOIL algebra brought me to FOIL method. Obviously for pre-calculus students . - DVdm (talk) 21:43, 14 April 2014 (UTC)
 * Thanks. I agree with you, it doesn't help William M. Connolley (talk) 22:15, 15 April 2014 (UTC)
 * Heh... being just in the mood for juggling with a few equations here...  looking at this again after two years, there is indeed a perfect pre-calculus no-Maclaurin way for the last step:
 * $$\frac{1}{1+2x} = \frac{1-2x}{(1+2x)(1-2x)} = \frac{1-2x}{1 - 4x^2} \stackrel{\text{(ignore } 4x^2 \text{ in demom)}}{\approx} 1-2x$$
 * giving
 * $$- \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1}{1 \pm 2\Delta r / R} = - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1 \mp 2\Delta r / R}{(1 \pm 2\Delta r / R)(1 \mp 2\Delta r / R)} = - \hat r ~ G ~ \frac{M}{R^2} ~ \frac{1 \mp 2\Delta r / R}{1 - 4(\Delta r / R)^2} \stackrel{\text{(ignore denom } \tfrac{1}{R^2} \text{)}}{\approx} - \hat r ~ G ~ \frac{M}{R^2} ~ ( 1 \mp 2\Delta r / R ) = - \hat r ~ G ~ \frac{M}{R^2} \pm \hat r ~ G ~ \frac{2 M }{R^2} ~ \frac{\Delta r}{R}$$
 * Better late than never - DVdm (talk) 10:49, 5 May 2016 (UTC)

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Size and distance, Centrifugal force myth
I have undone user's 's well meant edits, as the two two essential cited sources and  don't qualify as wp:RS. Please bring solid sources, such as text book references for this. - DVdm (talk) 08:35, 19 September 2018 (UTC)


 * , thanks for your observation. I cited the peer reviewed version of this paper.  I liked this paper because it is so readable, and it explains some things better than the article.  Thanks again.    Comfr (talk) 19:37, 19 September 2018 (UTC)