Understanding Tides

There has been a long animated discussion about tides at Wattsupwiththat which highlighted a confusion about both the causes of tides and their strength. Several people are adamant that tides are a universal phenomena experienced by any object near a large gravitational source. They argue that tides are caused by the gradient of the 1/r^2 field.

While an extended object falling into a star  or black hole will experience a transient  tidal force, it is only bodies in orbit around each other that experience long term tidal forces acting on the surface. The earth is in orbit about the earth-moon barycenter. It is not in an inertial frame. It is in an accelerating frame of reference, similar to the way that the rotation of the earth about its axis causes a reduction in g at the equator.

So what is the real cause of tides on earth and why is the lunar tide larger than the solar tide ? Here is my derivation for  the formulae for tides.

For 2 bodies in orbit the “centrifugal force” must balance the gravitational force. The centrifugal force is constant within a solid body in the same way that it is for a plate spinning on a stick.

Fig 1

Fig 1


\frac{GMm}{r^2} = M\omega^2d

Therefore the centrifugal acceleration on body M is \omega^2d = \frac{Gm}{r^2}

Now consider the net force per unit mass acting on point a). Assuming that \phi = 0 we get

\frac{Gm}{(r-R \cos \theta)^2}  - \frac{Gm}{r^2}

= \frac{Gm}{r^2}( (1 - \frac{R}{r \cos \theta})^{-2} - 1)

assuming that r>>R we can do a binomial expansion to get

= \frac{Gm}{r^2}(1 +\frac{2R}{r}\cos \theta - 1)

= \frac{2GmR \cos \theta}{r^3}

This derives the approximate formula for the tidal force. There are two bulges centered on \theta = 0 (and) \theta =\pi

Now we can do the full calculation where the angle \phi is no longer zero and thereby identify how the tidal force acquires a vertical component.

Calculation of the angle phi. Distance a-m is the hypotomuse

Calculation of the angle phi. Distance a-m is the hypotomuse

The distance a-m is by Pythagorus

\sqrt{R^2 \sin^2 \theta + (r - R \cos \theta)^2}

= \sqrt{R^2 +r^2 -2rR \cos \theta}

Gravity acting on point a) is therefore =  \frac{Gm}{R^2 + r^2 -2rR \cos \theta}

Net tidal force now has 2 components

Fx = Gm( \frac{ \cos \phi}{R^2 + r^2 - 2rR \cos \theta} - \frac{1}{r^2})

Fy = \frac{ -Gm \sin \phi}{R^2 + r^2 - 2rR \cos \theta}


\cos \phi = \frac{r-R \cos \theta}{\sqrt{R^2 + r^2 - 2rR \cos \theta}}


\sin \phi = \frac{R \sin \theta}{\sqrt{R^2 + r^2 - 2rR \cos \theta}}

Now let’s compare the two solutions by calculating the effective tidal acceleration. Figure 3 shows the approximate formula in red and the two components of the exact result in blue.

Fig 3: Comparison of the exact solution for tides with the approximation. There is now a significant vertical (y) component to the tidal force.

Fig 3: Comparison of the exact solution for tides with the approximation. There is now a significant vertical (y) component to the tidal force.

On earth the gravitational acceleration ‘g’ = 9.8 m/s^2  This can be compared with the above “tidal” accelerations of ~ 10^-6 m/s^2. So the moon’s tidal force is 10 million times less than the earth’s gravitational force at the surface.  This is not going to do any direct heavy lifting of the oceans! Instead it is the tractional component of the tidal force parallel to the surface which moves vast quantities of ocean.  Now with the addition of the vertical component Fy the vector diagrams of tidal forces looks more like this:

Fig 4: Vector diagram showing resultant tidal force (fx,Fy)

Fig 4: Vector diagram showing resultant tidal force (fx,Fy)

Note that the largest tractional forces are at larger theta angles. This drag of water currents throughout the depth of the ocean results in both tidal bulges. Figure 5 shows the tractional north-south force parallel to the earth’s surface which is unaffected by the earth’s gravity and therefore moves water from outside the bulge towards the centre.

Fig 5: The tractional force moving vast quantities of water in the oceans to cause the tides.

Fig 5: The tractional force moving vast quantities of water in the oceans to cause the tides.

The position and strength of these tractional forces is constantly changing during the lunar month, during the 18.6 year precession cycle and during longer time scale astronomical cycles. So the way I like to understand tides is that both the moon and the sun exert a horizontal drag force on the oceans and atmosphere. Twice a month we get spring tides when the sun and moon line up at  new moon and the full moon, whose strength depends on the coincidence with both orbital perigees.  Long term astronomical variations must have an effect on  climate as orbital parameters slowly change increasing or decreasing ocean mixing and atmospheric dynamics.


There is a new argument about whether centrifugal forces play any role at all in tides. In some sense both sides in this argument are correct. You don’t need to use the centrifugal force to derive the formula for tides. This is because there is a perfect balance between the centrifugal force and the gravitational force at the center of the earth when in orbit around the earth-moon barycenter. This balance also determines the strength of tides on earth. When in doubt see what Feynman says.

What do we mean by “balanced”? What balances? If the moon pulls the whole earth toward it, why doesn’t the earth fall right “up” to the moon? Because the earth does the same trick as the moon, it goes in a circle around a point which is inside the earth but not at its center. The moon does not just go around the earth, the earth and the moon both go around a central position, each falling toward this common position. This motion around the common center is what balances the fall of each. So the earth is not going in a straight line either; it travels in a circle. The water on the far side is “unbalanced” because the moon’s attraction there is weaker than it is at the center of the earth, where it just balances the “centrifugal force.” The result of this imbalance is that the water rises up, away from the center of the earth. On the near side, the attraction from the moon is stronger, and the imbalance is in the opposite direction in space, but again away from the center of the earth. The net result is that we get two tidal bulges.

A body in free fall into the sun experiences ever increasing tidal forces until it is torn apart. A body in orbit eperiences varying tidal forces depending on the eccentricity of the orbit. So on a purely logical basis Willis and Greg are correct because centrifugal forces don’t enter into the formula for tides. However in order to calculate the variations of tides on earth you need to include orbital dynamics because they change the earth-moon distance.

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15 Responses to Understanding Tides

  1. Greg Goodman says:

    Nice thorough explanation. It’s good to have solid equations for the off axis variation too. That is usually skipped totally.

    However, one thing I don’t think is technically correct.

    “For 2 bodies in orbit the “centrifugal force” must balance the gravitational force. The centrifugal force is constant within a solid body in the same way that it is for a plate spinning on a stick.”

    “Therefore the centrifugal acceleration on body M is…”

    For two bodies in orbit, gravity is providing the centripetal force causing the centripetal acceleration that characterises circular movement. There is no centrifugal fictitious force in the way you have presented this. (Which is just as well because you’d need Coriolis too in that case)

    The centripetal distinction can appear pedantic but it’s important to be clear whether you are presenting an inertial frame (centripetal) or rotating frame (centrifugal+Coriolis).

    Generally very useful. Probably the most thorough generalised explanation I’ve seen. I’m sure I’ll come back to this as a reference.

  2. Clive Best says:

    Thanks for the comments Greg,

    I think it is best to annalyse the earth-moon system from an external inertial frame of reference. Then Newton’s second and third laws explain the origin of the centrifugal force. F=ma and to every action there is an equal and opposite reaction. The centripetal force is simply gravity and the reaction to it is the centrifugal force. Einstein based general relativity on the equivalence principal – an observer experiencing an acceleration is the same as a gravitational field. A lift in free fall does not experience gravity at least until it hits the ground. It will get stretched so that a unit of length increases but time will also go slower !

    Where I don’t agree with Willis is his insistence that tidal forces are universal and that every body exerts a tidal force on every other body. One single electron does not exert a tidal force on a quark. Gravity is the fundamental force.

  3. Greg says:

    “I think it is best to annalyse the earth-moon system from an external inertial frame of reference.”

    “… an observer experiencing an acceleration is the same as a gravitational field. ”
    So the centripetal acceleration IS gravity by equivalence. There’s no need for an equal and opposite.

    I would have said the equal and opposite bit was each end of the gravitational attraction. A gravitational force on the earth and the one on the moon.

    Don’t forget centrifugal force is a fictitious force . It’s purely a convenience to make newtonian mechanics work in a rotational frame. Since you chose an inertial frame there are NO centrifugal or Coriolis forces.

    • Clive Best says:

      Centrifuges separate U256 from U238. Nobody told the machines what they were doing was fictitious or impossible. The engineers observing them working from the lab – an inertial external frame don’t stand around scratching their heads wondering how the hell it all works.

      This is getting silly.

  4. Clive Best says:


    In principal you are right because the centrifugal force is not needed to derive the formula for tidal forces – See the postscript I added above in the post. I think it is more of a philosophical argument than a real argument.

  5. Greg Goodman says:

    Thanks Clive. I didn’t want to get into a long messy discussion on your blog, so I didn’t insist but I’m glad we are agree about it.

    The centripetal/centrifugal thing can seem pedantic but I’ve always found paying knit-picking attention to things like that helped me gain a deeper understanding. Especially in maths and physics.

    Maybe others who are smarter than me see the core issues straight away. On the other hand I’ve met lots of people who _are_ smarter than me that lack a deeper understanding of what they’re doing.

  6. John F. Hultquist says:

    I’m still working through the following site. You may be interested.

    Also uses the “centrifugal” view and has a slight terminology difference; versus the “tractional component” you have used, this says:
    This second component, know as the tractive (“drawing”) component of force is the actual mechanism for producing the tides.
    There are numerous spelling problems (note the ‘know’ for ‘known’ in the quote).

    • Clive Best says:

      I still think it is perfectly reasonable to use the centrifugal force to describe the earth-moon orbit and the resultant tides. The centripetal force acting on the center of the earth is the gravitational attraction GmM/r^2 . The reaction to this is the centrifugal force which is equal and opposite -GmM/r^2. The differential gravitational force felt at the earth’s surface just depends on the distance from the ocean element to the centre of the moon.

      You get exactly the same “instantaneous” answer as a body in free fall but now at least we include the fact that the earth’s orbit determines what r is.

      So it is more of a philosophical question. Zealots like Willis insist that only he is correct and everyone else is wrong. Perhaps he should first study general relativity before making any absolute assertions about such things.

      • Greg says:

        I don’t think it’s zealotry. It reminds me of school teachers. They know enough to give lessons to those that know nothing but get very defensive when someone asks a difficult question about an inconsistency .

        To maintain the illusion of the teacher who has all the answers, they will rarely correct their errors but bluster through and try to make a fool of the smart pupil who points out an error or asks a probing question.

        Willis’ free-fall tidal forces piece is just an attempt to justify his errors in his previous article’s plotting of tidal forces, that you quickly pointed out and for which I provided a two-line correction to his code.

        He seems more motivated by saving face than getting the right answer, which is unfortunate. He’s inquisitive, and smart enough to make useful insights at times.

  7. Greg says:

    Yes, Clive I’ve seen the NOAA account. I think it is flawed in several ways.

    First is mixing frames of reference. Either chose a rotating EM system (inertial frame), which their fig 1 clearly does, OR use centrifugal and Coriolis forces and a non-inertial rotating frame. It’s the mathematical equivalent of mixed metaphors. It’s not just naming convention or pedantic, if its centripetal , it points the other way. This is simply a confused account.

    Then we look a fig 2 and they have a static moon and earth , ie. a rotational frame of reference, equating Fg to Fc , which is correct in itself but why change reference frame half way through?

    Probably because the author did not even realise this is what he was doing !i

    We then get to the ‘tractive forces”, still in the rotational frame. They start to invoke the off-axis components that give rise to horizontal movement. Now you simply can’t do that in a rotational frame of reference without invoking Corolis forces. They are not an optional extra. They are part and parcel of trying to use newtonian mechanics in a rotating frame.

    We clearly see the effects of C. forces in the horizontal components of both atmospheric and oceanic circulation, they are not insignificant.

    Now there is no reason why all this can not be worked out equally correctly using a rotating frame if coreolis is part of the calculation. That seems mathematically masochistic and unnecessarily confusing as a means of explaining the effects.

    The NOAA account manages to fudge through a hand-waving account of what happens and get the right general description because it does not try to lay down any maths and they know what the answer should be.

    I think your account is much more satisfactory. And my ‘hand-waving’ account much simpler and clearer.

  8. Philip Kuntz says:

    There is a clear and concise algebraic derivation of tidal forces but many people may not be aware of it.
    In his book “New Foundations for Classical Mechanics”, second edition, David Hestenes points out in the chapter on celestial mechanics that tidal forces in a two-body system can be derived from essentially one equation invoking nothing else than F=ma and a few lines of vector algebra. He considers the acceleration (force per unit mass) experienced by a unit mass somewhere on the surface of a spherical Earth. On the left side of the equation is the acceleration of Earth in the barycentric system; on the right side is the sum of three forces acting on the unit mass: the attraction to Earth, the attraction to Moon and a constraint force,f, that ensures that the unit mass is at rest with respect to Earth. In the notation of your figure 1, R is a vector pointing from the centre of Earth to the point on the surface and r is a vector pointing from the centre of Earth to the centre of Moon. The equation is

    Gmr/|r|^3 = -GMR/|R|^3 -Gm(r-R)/|r-R|^3 + f

    (Note that the term on the lhs equals the second time derivative of the vector from the centre-of-mass of the Earth-Moon system to the centre of Earth.)

    The constraing force is therefore

    f = GMR/|R|^3 + Gmr/|r|^3 – Gm(r-R)/|r-R|^3

    The first term on the right simply prevents the unit mass from falling to the centre of Earth. The next two terms oppose the tidal force (by definition). So, the tidal force, g(R), is the negative of those terms:

    g(R) = -Gmr/|r|^3 + Gm(r-R)/|r-R|^3

    That completes the derivation. To go further, the second term on the right is expanded using the binomial theorem after neglecting quantities involving (|R|/|r|)^2.

    |r-R|^(-3) = [(r-R)^2]^(-3/2)
    = [|r|^2 -2r.R +|R|^2]^(-3/2)
    –> |r|^(-3)(1 – 2r.R/|r|^2)^(-3/2)
    = |r|^(-3)(1 + 3r.R/|r|^2 +…)

    To within this approximation, the tidal force is (r^ means a unit vector in the r-direction)

    g(R) = Gm|R|(3R^.r^ r^ – R^)/|r|^3,

    which describes the distribution of forces shown in your figure 4, thereby explaining the two bulges in the tides.

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