# Tides and centrifugal force

July 11th 2005

At many websites, but also in some textbooks, one finds plainly wrong explanations of the lunar and solar tides. In particular much confusion arises when authors try to explain the existence of the second tidal bulge, the one opposite to the Moon. Often they invoke the centrifugal force as an “explanation”. But centrifugal force is a fictitious force, and we can’t justify a real effect with a fictitious force, can we?

The necessary and sufficient reason for the tides is the fact that over the volume of the earth, the gravitational force (from Moon or Sun) has different size and direction over the earth’s volume as a result of the varying distance from the external body. This is the “differential” force, sometimes called the “tidal force” and is the gradient of the force due to the external body. The centrifugal force is required only when the gravitational problem is done in a rotating reference frame. Often it is incorrectly used, even when its use is appropriate to such a rotating frame. It is a quantity necessary for accounting purposes, without any real effect on tides.

We will demonstrate this by considering different reference frames, rotating and non-rotating. We will calculate the direction and the intensity of the total acceleration for just two trial points, the well known A and B in figure 2. They are, respectively, the sublunar point and the anti-lunar point. If the results are identical to the those emerging from the case in wich there is only the gravitational force, we may rule out the centrifugal force as necessary to explain tides.

Analysing the commonest incorrect explanation, we find that that confusion arises from the incorrect choice of the reference frame (RF). In physics, the first step to successfully deal with a problem, is the choice of the RF. If mistakes are made in this starting phase, one will surely arrive at plainly wrong results. In physics, as in many other systems, the “GIGO” rule applies: Garbage In, Garbage Out.

Another misleading thing is that to get the tidal field, we must subtract from the moon’s gravity a uniform force field. But we learned in college that when we deal with rotation, the fictitious forces are proportional to the distance from the rotation’s axis. So where does this uniform field arise? All this will be revealed in the examples which follow.

### Reference systems

Physics teaches us that every real situation can be described in different ways, depending upon the observer’s point of view. Each observer defines its own reference frame (RF). For instance, if I am at rest at the side of a road, I will define the road as my RF, but if I am inside a car, I will more naturally define the car itself as my RF.

If the car turns left, an observer standing at the roadside will say:

According to the principle of inertia, the passenger wants to go straight, however, the tires are steering the car to the left. Thus, the car pushes on the passenger forcing him to accelerate with the car. This is why the passenger feels a pressure on his right side.

The passenger, who chooses the car as a RF, will instead say:

The car is at rest with respect to me, but we both are immersed in a field of force pushing both the car and myself to the right, in a direction away from the center of the car’s curved path. It is a centrifugal force, that arises when the car moves along a curved path. This force pushes me against the right side of the car.

Both arguments are valid descriptions of what happens, however we mustn’t forget that the centrifugal force is a fictitious one. The only real force in this example (besides gravity) is the centripetal force exerted between the tyres and the asphalt. To be specific, it is one of the four fundamental forces, the electrical one.

It is clear that all the RFs are equivalent, in that they must give the same numeric predictions concerning forces and accelerations. To say this concisely, “for every reference system, I must be able to apply Newton’s second law:

$$a_{ine} = \frac{\vec{F}}{m}\tag{1}$$

The subscript indicates the reference system in use for the physical quantity it refers to. In this case we are equating the acceleration of a body, measured in an inertial reference system, with the ratio between the force acting upon it and its mass.

There are special RFs: those for accelerated motion of the frame. If we require that the Newton’s second law still be applicable to an accelerated frame, we must introduce fictitious forces. In the case of a rotating RF, they are called centrifugal force and Coriolis’ force, the latter named for Gaspard Coriolis, who announced his theorem in 1835.

Those forces are called fictitious because they have no physical reality, they emerge only in accelerating reference frames. The only forces worthy of being called “real” are the four fundamental forces: the gravitational, electrical, strong nuclear and weak nuclear. No physical effect can derive from a fictitious force: we can always choose a reference system in wich this force will disappear.

Another distinguishing difference between real forces and fictitious ones arises from consideration of Newton’s third law. Real forces always have identifiable reaction forces. Newton’s third law states: “If body A exerts a force on body B, then B exerts an equal size and oppositely directed force on A”. But fictitious forces are not forces exerted by a material body. They have no “source”. They are merely a convenient fiction for doing problems in accelerating reference frames. Therefore no reaction force can be found for fictitious forces, and Newton’s third law can’t be applied to fictitious forces. There is no “body A” that is exerting the fictitious force.

In non-inertial reference systems, Newton’s second law must also be slightly modified. For a reference frame that has motion of translation as well as motion with constant angular velocity, we can write the acceleration of a point seen from an inertial RF as follows:

$$a_{ine}=a_{sr}+a_{rot}+\vec{\omega}\times(\vec{\omega}\times\vec{r})+2\,\vec{\omega}\times\vec v_{rot}\tag{2}$$

The cross indicates the vector product.

The acceleration seen from an inertial RF is the sum of the acceleration of the RF itself, (sr), also called dragging acceleration, plus the acceleration shown in the rotating reference system (rot). We still have to add two more forces called, respectively, centripetal acceleration and Coriolis’ acceleration.

Thus, if we want Newton’s second law to hold also in a non-inertial RS, recalling the equations (1) and (2) we write:

$$a_{rot}=\frac{\vec{F}}{m}-a_{sr}-\vec{\omega}\times(\vec{\omega}\times\vec{r})+2\,\vec{\omega}\times\vec v_{rot}\tag{3}$$

In this article we will see five cases, that we deal with using the same approach:

1. we will choose a RF, noting its accelerations with respect to an inertial RF (fixed stars)
2. we will study the kinematics in the new RF, noting the accelerations of our two test points A and B
3. we will write the fundamental equation (3) equating the acceleration observed in the previous step to the sum of different components:
1. the real acceleration: the gravitational one $$\frac{F_g}{m}$$
2. in non-inertial RF in translatory motion a fictitious acceleration appears, opposite to the SR’s acceleration ($$-a_{sr}$$).
3. in non-inertial RF in rotary motion the centrifugal acceleration $$\,-\vec{\omega}\times(\vec{\omega}\times\vec{r})$$ will arise, always opposite to the center of rotation, and the Coriolis’ acceleration $$-2\,\vec{\omega}\times\vec v_{rot}$$, whose direction depends upon the velocity of the point with respect to the RF.
4. in the formula (3) the equality will not be respected because we are thinking about A and B as points belonging to the Earth’s shape. The earth’s shape is not spherical, since the tidal acceleration will deform the Earth, turning a sphere into an ellipsoid. Thus, we will add a residual acceleration Δ, that will represent the direction and intensity of the tidal acceleration. To summarize, the fundamental equation will be:

Real forces – fictitious forces = m * (acceleration + $$\Delta$$)

Last but not least, we must deal with the classic pitfall of reference systems. We invite the reader to answer to the following question: does the reference system shown in figure 1, rotate or translate? The answer will be given elsewhere in this article. Stay tuned!

Fig. 1 – does the reference system rotate or translate?

### Our physical model

To be honest, we have to point out that a complete explanation of tides, explaining their periodicity and intensity, is very difficult. We here limit the treatise to consider a simplified model with a spherical, smooth and homogeneous Earth, covered with a uniform film of water. Around that Earth, revolves a point-like Moon. Well, actually both revolve around the common center of mass (CM). We intentionally leave the Sun out of the picture, because it doesn’t add anything to the model, from a qualitative point of view.

Since the Earth’s rotation don’t have any influence on the existence of the two tidal bulges, we will suppose that the Earth is not rotating. Our planet will maintain its orientation with respect to the fixed stars. If the reader is still doubtful, we invite him to consider that the Earth’s rotation will deform the Earth only in the sense of producing the equatorial bulge. That deformation is symmetrical with respect to the Earth’s axis: it does not affects the tidal bulges at all.

Some physical data: in our model we have a planet, the Earth, whose mass is $$M_T=5.9742\cdot10^{24}\,\mathrm{kg}$$ and its natural satellite, the Moon, whose mass is $$M_L=7.36\cdot10^{22}\,\mathrm{kg}$$.

Fig. 2

The mass ratio is 81,17 their distance R is assumed equal to the average distance: 384.400 km.
The center of mass of the Earth-Moon system CM is 4678 km offset from the Earth’s center

$$c=R\cdot\frac{M_L}{M_T+M_L}=4678\,\text{km}\tag{4}$$

Since the Earth’s equatorial radius r is 6378 km, the center of mass is within the Earth, $$r-c=1700\,\mathrm{km}$$ deep.
The Earth’s center will then describe a circle of radius c around the CM, the Moon, too, will describe a circle, whose radius will be $$R-c$$.
The angular velocity of these motions can be calculated from Kepler’s third law:

$$\omega^2=\frac{G(M_T+M_L)}{R^3}=7.101877\cdot10^{-12}\,\text{rad/s}\tag{5}$$

### Case 1: the free fall

We face now the problem of explaining tides in the simplest way: as the Earth was in free fall towards the Moon. Let’s imagine stopping the motion at the instant where the distance between Earth and Moon centres is R. To study this situation we choose the RF at the Earth’s center. Since it’s a non-inertial RF, because it is accelerated by the Moon’s gravity, we shall consider a fictitious acceleration equal and opposite:

$$-\frac{GM_L}{R^2}=-3.322\cdot10^{-5}\,\mathrm{m/s^2}\tag{6}$$

pointing in the direction opposite to the Moon.

Kinematics

From a kinematic point of view, in this reference system, stopped at the moment in which the distance between the Earth and the Moon is R, all the points of the Earth’s surface are stationary, the Moon runs [falls] towards the Earth.

Dynamics

We calculate the total acceleration of A and B, recalling the equation (3) as the sum of:
– Moon’s gravitational acceleration
– fictitious acceleration
The result will be equated to the kinematic acceleration of that point, considered belonging to the Earth’s surface, plus a component, called $$\Delta$$, that actually represents the tidal force, the force that tends to deform the Earth’s surface.

Paying attention to vectors’ direction, for A we can write:

$$\frac{GM_L}{(R-r_A)^2}-\frac{GM_L}{R^2}=0+\Delta\tag{7}$$
$$\Delta=1.13\cdot10^{-6}\,\mathrm{m/s^2}$$

And for B we can write:

$$\frac{GM_L}{(R+r_B)^2}-\frac{GM_L}{R^2}=0+\Delta\tag{8}$$
$$\Delta=-1.07\cdot10^{-6}\,\mathrm{m/s^2}$$

The order of magnitude is $$10^{-7}\,\mathrm{g}$$: one ten-millionth of the gravity acceleration.

Fig. 3: The gravitational acceleration due to the Moon for three points of the Earth.

Fig. 4: After having removed the fictitious acceleration of that particular RF, we see emerging forces that want to deformate the planet along the Earth-Moon line.

Fig. 5: By doing the calculation for all the points of the Earth’s surface, emerges the well-known tidal force field.

We have then demonstrated the existence of two tidal bulges, in opposite directions and slightly different intensity, being stronger the one towards the Moon. It’s all we need to explain tides: the tidal effect is only due to the gravity gradient.

We emphasize that tidal bulges are not due to the shrinking of water masses. As we can see in figure 5, in some points of the Earth’s surface, the tidal forces are tangent to a sphere; that implies a force displacing the water towards the sublunar or anti-lunar points, where it collects.

The fact that Earth and Moon are orbiting one around the other implies a centripetal (or centrifugal, it depends upon the RF) acceleration. It is interesting to show that this force plays no role in the formation of tides, as we will show in the following cases.

### Case 2: the inertial reference system

In this second example and in the following ones, we will suppose that Earth and Moon are in circular orbits around each other.

We choose the RF at the center of mass of the Earth-Moon system. This point is at rest with respect to the Universe. This RS does not rotate, therefore, in our analysis, we will not have fictitious forces.

Fig. 6

Fig. 7 – Left: the situation seen from an inertial RF. The RF chosen is marked with orange arrows.
Right, the visual demonstration that every point of the Earth moves in the same way, by describing circular paths all of the same radius as the motion of the Earth’s center around the CM.

Kinematics

The Earth’s center describes a circular orbit of radius c around the CM with angular velocity ω.

Thinking about the points of the Earth’s surface, we have to avoid a common misunderstanding.

Saying that the Earth’s center rotates around the CM, does not imply that the Earth as a whole is rotating around that point! We cannot forget that, according to our model, the Earth’s orientation is fixed with respect to fixed stars. It is not rotating, it is translating. All its points describe circular paths of equal radius c, but having different centers. Now you have the answer to the question of figure 1.

Fig. 8

The Earth moves around the CM holding its orientation relative to the fixed stars. Thus, there is no single fictitious force (centrifugal) associated to the motion of the Earth around the CM.
So, the following calculation is plain wrong. Some authors want to calculate the effect of the centrifugal force operating as follows: if it’s true that the Earth rotates around the CM, then the centrifugal acceleration for the point A is $$\omega^2(r-c)=1.207\cdot10^{-5}\,\mathrm{m/s^2}$$ and for the point B is $$\omega^2(r+c)=7.852\cdot10^{-5}\,\mathrm{m/s^2}$$.

Fig. 9

We would get the nonsensical consequence that the centrifugal acceleration in B is about 6,5 times the one for A. Probably this is the fact that deceives those who claim that the tidal bulge opposite to the Moon is created by the centrifugal force!

Besides, this calculation gives a tidal acceleration whose magnitude is about 70 times the real one, so we are really on a slippery slope.
However, it is correct to say that all the points of the Earth describe a circle whose radius is c and thus undergo an acceleration $$\omega^2c=3.322\cdot10^{-5}\,\mathrm{m/s^2}$$ pointing towards the circle’s center, i.e. towards the Moon.

Dynamics

Since the RF chosen is inertial, we will not take in account fictitious forces. The only force is real, the gravitational one.

For A we can write:

$$\frac{GM_L}{(R-r_A)^2}=\omega^2c+\Delta\tag{9}$$
$$\Delta=1.13\cdot10^{-6}\,\mathrm{m/s^2}$$

The tidal force is then directed towards the Moon, as in the previous case.

For B we can write:

$$\frac{GM_L}{(R+r_B)^2}=\omega^2c+\Delta\tag{10}$$
$$\Delta=-1.07\cdot10^{-6}\,\mathrm{m/s^2}$$

In this case the tidal force on the Earth is opposite to the Moon.

Let’s get back to the quantity $$\omega^2c$$. We can rewrite it, taking in account the definitions of c and $$\omega^2$$ we have seen in equations (4) and (5).

$$\omega^2c=\frac{G(M_T+M_L)}{R^3}\cdot R\cdot\frac{M_L}{M_T+M_L}=\frac{GM_L}{R^2}\tag{11}$$

The interesting result is that the uniform field of centripetal accelerations has the same magnitude of the gravitational attraction of the Earth’s center towards the Moon!
Naturally this is not a coincidence, it is the mathematical reformulation of the fact that the Earth’s center is in orbit, attracted by the Moon.

We then got the same results even if the Earth and the Moon are orbiting each other.

### Case 3: tricky but educational

We again choose the RF at the center of mass, but this time it will be rotating with the same angular velocity of Earth-Moon revolution. This time both the Earth and moon centres will appear stationary.
As a consequence, in our analysis we will have to consider, for every point of the Earth, two fictitious accelerations: the centrifugal one and the Coriolis’ one. The first has the origin in CM and intensity $$\omega^2P$$ where P is the distance between the point being analysed and CM.
The second has intensity $$2\,\omega v=2\,\omega^2\,r$$ and its direction is given by the vector product between ω and v, inverted.

Fig. 10

Fig. 11 – Left: the situation seen from an inertial RF. This time the RF rotates, holding both the Earth and the Moon stationary in the frame.
Right: it is clear that in this RF the Earth shows a rotation on its axis. A field of centrifugal forces arises, originating from CM. In this representation the field of Coriolis’ forces is not shown.

We can express this field of centrifugal accelerations in a simpler way: as the sum of a radial field, with origin in the center of the Earth, and an uniform field of accelerations.

Fig. 12

Fig. 13

In fact the vector P can be decomposed in two vectors: $$\vec{P}=\vec{c}+\vec{r}$$.

And thus also $$\omega^2\vec{P}=\omega^2\vec{c}+\omega^2\vec{r}$$.

We note that $$\omega^2\vec{c}$$ is a constant quantity: it does not depend upon the point P being considered. Instead, the quantity $$\omega^2\vec{r}$$ is a radial isotropical acceleration.

Some authors in Usenet newsgroups have been deceived by this latter field: they give a physical meaning to the radial field, as a force contributing to expand the Earth’s equatorial bulge! Of course, it’s not the case.

This radial field will be canceled out by Coriolis’ forces, as we will show later.

We should pay attention to the fact that if our RF is rotating but the Earth is fixed with respect to stars, in the new reference system the Earth will rotate in an opposite direction (see figure 14 and the animation of figure 11).

Fig. 14

Kinematics

The Earth’s center is at rest; all the points of the surface (in particular, our A and B) are rotating. This rotation implies a centripetal acceleration $$\omega^2\vec{r}$$ towards the Earth’s center.

Dynamics

Let’s calculate the accelerations of the points A and B as the sum of:
– Moon’s gravitational acceleration
– the uniform component of the centrifugal acceleration
– the radial component of the centrifugal acceleration
– the Coriolis’ acceleration.

Paying attention to the direction of vectors, for A we can write:

$$\frac{GM_L}{(R-r_A)^2}-\omega^2c+\omega^2r_A-2\omega^2r_A=-\omega^2r_A+\Delta\tag{12}$$

but all the $$\omega^2r_A$$ cancels, so we get again

$$\Delta=1.13\cdot10^{-6}\,\mathrm{m/s^2}$$

For B we can write, by remembering that vectors $$\vec{r_A}$$ and $$\vec{r_B}$$ are opposite,

$$\frac{GM_L}{(R+r_B)^2}-\omega^2c-\omega^2r_B+2\omega^2r_B=\omega^2r_B+\Delta\tag{13}$$

this time too, all the $$\omega^2r_B$$ terms cancels, thus we get again:

$$\Delta=-1.07\cdot10^{-6}\,\mathrm{m/s^2}$$

The cancellation of $$\omega^2r$$-like terms shouldn’t surprise us: a little bit of thought shows us that the rotation of the RF must be balanced by the counter-rotation of the Earth in the new RF. It is a complication due to the choice of the RF, but if we perform the calculations properly, we get the correct result.

The radial centrifugal field is canceled by the Coriolis’ acceleration. The only field remaining is the uniform one, just as in the second case!

The Earth’s center feels the uniform field of acceleration $$\omega^2c$$ and the acceleration due to the Moon’s gravity. As we have seen in the equation(11), they are equal in size. Their sum is zero, so the Earth’s center is at rest. This is perfectly reasonable: the fictitious forces are useful right to correctly apply Newton’s second law $$F=ma$$. If the Earth’s center is at rest in the chosen RF, it is because the gravitational attraction due to the Moon and the fictitious force are identical.

Again, the numerical values are identical, despite the fact that the RF is pretty complicated.

Perhaps the reader is perplexed about the rotation of the Earth. We want to point out again that even if we introduced the daily rotation for the Earth, it would just enlarge the equatorial Earth’s diameter, turning it into a oblate ellipsoid. This deformation is symmetric about the earth’s rotation axis: it does not create tidal bulges.

### Case 4: at the Earth’s center

We now choose the RF in the center of the Earth having a fixed orientation with respect to fixed stars. It’s the most used RF to represent the vector field of tidal forces. The only mishap is that this vector field will be rotating: it will be sufficient for us to imagine it frozen in an instant.

We recall what we saw in the case 2: the Earth is not rotating, it is translating along a circular path! As a consequence, in our analysis we will have to add just an uniform fictitious acceleration, opposite to the Moon, whose intensity is $$\omega^2c$$. It is due to the motion of the RF around the CM.
The axes of the RS in effect do not vary their orientation with respect to the fixed stars: this reference system is translating. Thus, no centrifugal nor Coriolis’ force will be added.

Fig. 15

Fig. 16 – Left, the situation seen from an inertial RS. In this fourth case the RF chosen is in the center of the Earth. The situation is similar to the second case.
Right, in effect, it is shown the uniform centrifugal forces field.

Kinematics

The CM moves around the Earth’s center having an angular velocity ω. All the Earth’s points are still.

Dynamics

Let’s calculate the accelerations of A and B as the sum of:
– Moon’s gravitational acceleration
– the fictitious acceleration.

Paying attention to vectors’ direction, the acceleration for A will be:

$$\frac{GM_L}{(R-r_A)^2}-\omega^2c=0+\Delta\tag{14}$$
$$\Delta=1.13\cdot10^{-6}\,\mathrm{m/s^2}$$

The acceleration of B will be:

$$\frac{GM_L}{(R+r_B)^2}-\omega^2c=0+\Delta\tag{15}$$
$$\Delta=-1.07\cdot10^{-6}\,\mathrm{m/s^2}$$

Again, the result is identical to the case 2. Maybe here it is easier to accept the existence of an uniform acceleration field because it is “transferred” to the RF.

The Moon’s gravitational acceleration on the Earth’s center equals the centrifugal acceleration: the net result is zero and the Earth appears stationary.

### Case 5: at the Moon’s center

In this last case we choose the RF in the center of the Moon and rotating along with the Earth. This seems a bizarre choice, without any practical interest. However, this situation offers a point of view in wich many fictitious forces arise but, once again – surprisingly – they compensate each other leaving the “usual” final result.

Kinematics

The Earth and the Moon are at rest, but the Earth rotates around its own axis with angular velocity ω.
All the points of the Earth’s surface undergo a centripetal acceleration whose intensity is $$\omega^2r$$.

Fig. 17 – Left, the situation seen from an inertial RF. In this fifth case the RF chosen is in the center of the Moon and is both rotating and translating (rototranslatory motion).
Right, are shown two fields of fictitious forces: one uniform and the other radial, centered on the Moon. In this representation Coriolis’ forces are not shown.

Dynamics

This reference system not only rotates on itself, but it also moves in a circular path around the CM. Thus it is in rototranslatory motion, that generates three fictitious force fields.

One is uniform, with acceleration having intensity $$\omega^2(R-c)$$ and is due to the translation of the center of the RF around the CM.
The other two are due to the rotation of the RF around its own origin. The rotation generates a centrifugal acceleration whose intensity is $$\omega^2x$$ where with x we have indicated the distance of the Moon to the point being analysed. It generates also a Coriolis’ force that – as we have already said – has intensity $$2\omega v=2\omega^2r$$ and its direction is given by the vector product between ω and v, inverted.

Let’s repeat the calculations of the accelerations for the A and B points, as the sum of:
– Moon’s gravitational acceleration
– uniform fictitious acceleration
– centrifugal fictitious acceleration
– fictitious Coriolis’ acceleration.

Paying attention to the vectors’ direction, the acceleration of A will be:

$$\frac{GM_L}{(R-r_A)^2}+\omega^2(R-c)-\omega^2(R-r_A)-2\omega^2r_A=-\omega^2r_A+\Delta\tag{16}$$

but all the terms $$\omega^2r_A$$ and $$\omega^2R$$ cancels out, then

$$\Delta=1.13\cdot10^{-6}\,\mathrm{m/s^2}$$

For B, recalling that vectors $$\vec{r_A}$$ and $$\vec{r_B}$$> are opposites, we will write,

$$\frac{GM_L}{(R+r_B)^2}+\omega^2(R-c)-\omega^2(R+r_B)+2\omega^2r_B=\omega^2r_B+\Delta\tag{17}$$

this time too all the terms $$\omega^2r_B$$ and $$\omega^2R$$ cancels, thus

$$\Delta=-1.07\cdot10^{-6}\,\mathrm{m/s^2}$$

Again, we got the same result as the previous cases.

### Conclusions

Let’s stop a moment to think about an important fact, emerged in all the different cases.

To the Moon’s attraction in the different points of the Earth’s surface, we always have to subtract an uniform field, whose value is equal to the intensity of the Moon’s attraction at the center of the Earth. Why?

The answer lies in the deep meaning of the condition “to be in orbit around a celestial body”. This condition is equivalent, in every case, to the free fall condition.
The Earth is in free fall towards the Moon, attracted by the gravitational force. This is why the Moon feels a uniform acceleration whose intensity is $$\frac{GM_L}{R^2}$$. But since the Earth is not actually falling along a straight line but is in orbit, this linear acceleration “turns into” a centripetal acceleration $$\omega^2c$$, of equal intensity.

The fact that the Earth rotates around the CM or on itself, unfortunately throws dust into our sight. Those effects do not produce any tidal deformation, they just confuse us.

I want to thank Donald E. Simanek, whose article inspired this work, Mino Saccone, Dieter Kreuer, Salvatore Ganci and Giampiero Barbieri for the useful discussions and Andrea del Monte for the suggestions about Flash animations.