Tidal acceleration

Tidal acceleration is an effect of the s between an orbiting (e.g. the ), and the primary  that it orbits (e.g. ).  The acceleration causes a gradual recession of a satellite in a  away from the primary, and a corresponding slowdown of the primary's rotation. The process eventually leads to, usually of the smaller first, and later the larger body. The Earth–Moon system is the best studied case.

The similar process of tidal deceleration occurs for satellites that have an orbital period that is shorter than the primary's rotational period, or that orbit in a retrograde direction.

The naming is somewhat confusing, because the speed of the satellite relative to the body it orbits is decreased as a result of tidal acceleration, and increased as a result of tidal deceleration.

Discovery history of the secular acceleration
was the first to suggest, in 1695, that the mean motion of the Moon was apparently getting faster, by comparison with ancient observations, but he gave no data. (It was not yet known in Halley's time that what is actually occurring includes a slowing-down of Earth's rate of rotation: see also . When measured as a function of  rather than uniform time, the effect appears as a positive acceleration.) In 1749  confirmed Halley's suspicion after re-examining ancient records, and produced the first quantitative estimate for the size of this apparent effect: a centurial rate of +10″ (arcseconds) in lunar longitude, which is a surprisingly accurate result for its time, not differing greatly from values assessed later, e.g. in 1786 by de Lalande, and to compare with values from about 10″ to nearly 13″ being derived about a century later.

produced in 1786 a theoretical analysis giving a basis on which the Moon's mean motion should accelerate in response to changes in the eccentricity of the orbit of Earth around the. Laplace's initial computation accounted for the whole effect, thus seeming to tie up the theory neatly with both modern and ancient observations.

However, in 1854, caused the question to be re-opened by finding an error in Laplace's computations: it turned out that only about half of the Moon's apparent acceleration could be accounted for on Laplace's basis by the change in Earth's orbital eccentricity. Adams's finding provoked a sharp astronomical controversy that lasted some years, but the correctness of his result, agreed upon by other mathematical astronomers including, was eventually accepted. The question depended on correct analysis of the lunar motions, and received a further complication with another discovery, around the same time, that another significant long-term perturbation that had been calculated for the Moon (supposedly due to the action of ) was also in error, was found on re-examination to be almost negligible, and practically had to disappear from the theory. A part of the answer was suggested independently in the 1860s by Delaunay and by : tidal retardation of Earth's rotation rate was lengthening the unit of time and causing a lunar acceleration that was only apparent.

It took some time for the astronomical community to accept the reality and the scale of tidal effects. But eventually it became clear that three effects are involved, when measured in terms of mean solar time. Beside the effects of perturbational changes in Earth's orbital eccentricity, as found by Laplace and corrected by Adams, there are two tidal effects (a combination first suggested by ). First there is a real retardation of the Moon's angular rate of orbital motion, due to tidal exchange of between Earth and Moon. This increases the Moon's angular momentum around Earth (and moves the Moon to a higher orbit with a lower ). Secondly there is an apparent increase in the Moon's angular rate of orbital motion (when measured in terms of mean solar time). This arises from Earth's loss of angular momentum and the consequent increase in length of day.

Effects of Moon's gravity
Because the 's mass is a considerable fraction of that of Earth (about 1:81), the two bodies can be regarded as a system, rather than as a planet with a satellite. The plane of the Moon's around Earth lies close to the plane of Earth's orbit around the Sun (the ), rather than in the plane perpendicular to the axis of rotation of Earth (the ) as is usually the case with planetary satellites. The mass of the Moon is sufficiently large, and it is sufficiently close, to raise s in the matter of Earth. In particular, the of the s bulges out towards and away from the Moon. The average tidal bulge is synchronized with the Moon's orbit, and Earth rotates under this tidal bulge in just over a. However, Earth's rotation drags the position of the tidal bulge ahead of the position directly under the Moon. As a consequence, there exists a substantial amount of mass in the bulge that is offset from the line through the centers of Earth and the Moon. Because of this offset, a portion of the gravitational pull between Earth's tidal bulges and the Moon is not parallel to the Earth–Moon line, i.e. there exists a between Earth and the Moon. Since the bulge nearer the moon pulls more strongly on it than the bulge further away, this torque boosts the Moon in its orbit and slows the rotation of Earth.

As a result of this process, the mean solar day, which is nominally 86,400 seconds long, is actually getting longer when measured in s with stable s. (The SI second, when adopted, was already a little shorter than the current value of the second of mean solar time.) The small difference accumulates over time, which leads to an increasing difference between our clock time  on the one hand, and  and  on the other hand: see. This led to the introduction of the in 1972  to compensate for differences in the bases for time standardization.

In addition to the effect of the ocean tides, there is also a tidal acceleration due to flexing of Earth's crust, but this accounts for only about 4% of the total effect when expressed in terms of heat dissipation.

If other effects were ignored, tidal acceleration would continue until the rotational period of Earth matched the orbital period of the Moon. At that time, the Moon would always be overhead of a single fixed place on Earth. Such a situation already exists in the – system. However, the slowdown of Earth's rotation is not occurring fast enough for the rotation to lengthen to a month before other effects make this irrelevant: About 1 to 1.5 billion years from now, the continual increase of the Sun's will likely cause Earth's oceans to vaporize, removing the bulk of the tidal friction and acceleration. Even without this, the slowdown to a month-long day would still not have been completed by 4.5 billion years from now when the Sun will probably evolve into a and likely destroy both Earth and the Moon.

Tidal acceleration is one of the few examples in the dynamics of the of a so-called secular perturbation of an orbit, i.e. a perturbation that continuously increases with time and is not periodic. Up to a high order of approximation, mutual perturbations between major or minor s only cause periodic variations in their orbits, that is, parameters oscillate between maximum and minimum values. The tidal effect gives rise to a quadratic term in the equations, which leads to unbounded growth. In the mathematical theories of the planetary orbits that form the basis of, quadratic and higher order secular terms do occur, but these are mostly of very long time periodic terms. The reason that tidal effects are different is that unlike distant gravitational perturbations, friction is an essential part of tidal acceleration, and leads to permanent loss of from the dynamic system in the form of. In other words, we do not have a here.

Angular momentum and energy
The gravitational torque between the Moon and the tidal bulge of Earth causes the Moon to be constantly promoted to a slightly higher orbit and Earth to be decelerated in its rotation. As in any physical process within an isolated system, total and  are conserved. Effectively, energy and angular momentum are transferred from the rotation of Earth to the orbital motion of the Moon (however, most of the energy lost by Earth (−3.321 TW) is converted to heat by frictional losses in the oceans and their interaction with the solid Earth, and only about 1/30th (+0.121 TW) is transferred to the Moon). The Moon moves farther away from Earth (+38.247±0.004 mm/y), so its (in Earth's ) increases. It stays in orbit, and from it follows that its  actually decreases, so the tidal action on the Moon actually causes an angular deceleration, i.e. a negative acceleration (−25.858±0.003"/century2) of its rotation around Earth. The actual speed of the Moon also decreases. Although its  decreases, its potential energy increases by a larger amount.

The rotational angular momentum of Earth decreases and consequently the length of the day increases. The net tide raised on Earth by the Moon is dragged ahead of the Moon by Earth's much faster rotation. Tidal friction is required to drag and maintain the bulge ahead of the Moon, and it dissipates the excess energy of the exchange of rotational and orbital energy between Earth and the Moon as heat. If the friction and heat dissipation were not present, the Moon's gravitational force on the tidal bulge would rapidly (within two days) bring the tide back into synchronization with the Moon, and the Moon would no longer recede. Most of the dissipation occurs in a turbulent bottom boundary layer in shallow seas such as the around the, the  off , and the.

The dissipation of energy by tidal friction averages about 3.75 terawatts, of which 2.5 terawatts are from the principal M$2$ lunar component and the remainder from other components, both lunar and solar.

An equilibrium tidal bulge does not really exist on Earth because the continents do not allow this mathematical solution to take place. Oceanic tides actually rotate around the ocean basins as vast s around several s where no tide exists. The Moon pulls on each individual undulation as Earth rotates—some undulations are ahead of the Moon, others are behind it, whereas still others are on either side. The "bulges" that actually do exist for the Moon to pull on (and which pull on the Moon) are the net result of integrating the actual undulations over all the world's oceans. Earth's net (or equivalent) equilibrium tide has an amplitude of only 3.23 cm, which is totally swamped by oceanic tides that can exceed one metre.

Historical evidence
This mechanism has been working for 4.5 billion years, since oceans first formed on Earth. There is geological and paleontological evidence that Earth rotated faster and that the Moon was closer to Earth in the remote past.  are alternating layers of sand and silt laid down offshore from having great tidal flows. Daily, monthly and seasonal cycles can be found in the deposits. This geological record is consistent with these conditions 620 million years ago: the day was 21.9±0.4 hours, and there were 13.1±0.1 synodic months/year and 400±7 solar days/year. The average recession rate of the Moon between then and now has been 2.17±0.31 cm/year, which is about half the present rate. The present high rate may be due to near between natural ocean frequencies and tidal frequencies.

Quantitative description of the Earth–Moon case
The motion of the Moon can be followed with an accuracy of a few centimeters by (LLR). Laser pulses are bounced off mirrors on the surface of the moon, emplaced during the missions of 1969 to 1972 and by  2 in 1973. Measuring the return time of the pulse yields a very accurate measure of the distance. These measurements are fitted to the equations of motion. This yields numerical values for the Moon's secular deceleration, i.e. negative acceleration, in longitude and the rate of change of the semimajor axis of the Earth–Moon ellipse. From the period 1970–2012, the results are:


 * −25.82±0.03 arcsecond/century2 in ecliptic longitude
 * +38.08±0.04 mm/yr in the mean Earth–Moon distance

This is consistent with results from (SLR), a similar technique applied to artificial satellites orbiting Earth, which yields a model for the gravitational field of Earth, including that of the tides. The model accurately predicts the changes in the motion of the Moon.

Finally, ancient observations of solar s give fairly accurate positions for the Moon at those moments. Studies of these observations give results consistent with the value quoted above.

The other consequence of tidal acceleration is the deceleration of the rotation of Earth. The rotation of Earth is somewhat erratic on all time scales (from hours to centuries) due to various causes. The small tidal effect cannot be observed in a short period, but the cumulative effect on Earth's rotation as measured with a stable clock of a shortfall of even a few milliseconds every day becomes readily noticeable in a few centuries. Since some event in the remote past, more days and hours have passed (as measured in full rotations of Earth) than would be measured by stable clocks calibrated to the present, longer length of the day (ephemeris time). This is known as. Recent values can be obtained from the (IERS). A table of the actual length of the day in the past few centuries is also available.

From the observed change in the Moon's orbit, the corresponding change in the length of the day can be computed:


 * +2.3 ms/century

However, from historical records over the past 2700 years the following average value is found:


 * +1.70 ± 0.05 ms/century

The corresponding cumulative value is a parabola having a coefficient of T2 (time in centuries squared) of:


 * ΔT = +31 s/century2

Opposing the tidal deceleration of Earth is a mechanism that is in fact accelerating the rotation. Earth is not a sphere, but rather an ellipsoid that is flattened at the poles. SLR has shown that this flattening is decreasing. The explanation is that during the large masses of ice collected at the poles, and depressed the underlying rocks. The ice mass started disappearing over 10000 years ago, but Earth's crust is still not in hydrostatic equilibrium and is still rebounding (the relaxation time is estimated to be about 4000 years). As a consequence, the polar diameter of Earth increases, and the equatorial diameter decreases (Earth's volume must remain the same). This means that mass moves closer to the rotation axis of Earth, and that Earth's moment of inertia decreases. This process alone leads to an increase of the rotation rate (phenomenon of a spinning figure skater who spins ever faster as they retract their arms). From the observed change in the moment of inertia the acceleration of rotation can be computed: the average value over the historical period must have been about −0.6 ms/century. This largely explains the historical observations.

Other cases of tidal acceleration
Most natural satellites of the planets undergo tidal acceleration to some degree (usually small), except for the two classes of tidally decelerated bodies. In most cases, however, the effect is small enough that even after billions of years most satellites will not actually be lost. The effect is probably most pronounced for Mars's second moon, which may become an Earth-crossing asteroid after it leaks out of Mars's grip. The effect also arises between different components in a.

Tidal deceleration
This comes in two varieties:

and are believed to have no satellites chiefly because any hypothetical satellite would have suffered deceleration long ago and crashed into the planets due to the very slow rotation speeds of both planets; in addition, Venus also has retrograde rotation.

Size of the tidal bulge
Neglecting, the tidal force a satellite (such as the moon) exerts on a planet (such as earth) can be described by the variation of its gravitational force over the distance from it, when this force is considered as applied to a unit mass $$dm$$:
 * $$\frac{\delta F}{\delta r} = -2\frac{G\,m\,dm}{r^3} $$

where G is the, m is the satellite mass and r is the distance between the satellite and the planet.

Thus the satellite creates a disturbing potential on the planet, whose difference between the planet center and the closest (or farthest) point to the satellite is:
 * $$\Delta V = 2\frac{G\,m\,dm\,A^2}{r^3} $$

where A is the planet radius.

The size of the tidal bulge created on the planet can be estimated as roughly the ratio between this disturbing potential and the planet surface gravity:
 * $$H \approx \frac{\Delta V}{G\,M\,dm/A^2} = 2\frac{m\,A^4}{M\,r^3} $$

A more exact calculation gives:
 * $$H = \frac{15}{8}\frac{m\,A^4}{M\,r^3} $$

assuming we neglect a second order effect due to of the planet material.

For the moon-earth system (m=7.3x1022 kg, M=6x1024 kg, A=6.4x106 m, r=3.8x108), this gives 0.7 meters, close to the true value for ocean tides height (roughly one meter).

Note that two bulges are formed, one centered roughly around the point nearest to the satellite and the other centered roughly around the point farthest from it.

Torque
Due to planet rotation, the bulges lags somewhat behind the planet-satellite axis, which creates an angle $$\alpha$$ between the two. The size of this lag angle depends on inertia and (much more importantly) on dissipation forces (e.g. friction) exerted on the bulge.

The satellite applies different forces on the close and far bulges. The difference is roughly $$\delta F/\delta r$$ times the planet diameter, where we replace the unit mass in the calculation above with the approximate mass of each bulge, $$\pi\,\rho\,A^2\,H$$ (where &rho; is the mass density of the buldge):


 * $$\Delta F \approx \frac{\delta F}{\delta r}\cdot 2A \cdot \cos(\alpha) = 4\pi\frac{G\,m\,\rho\,A^3\,H}{r^3}\cos(\alpha)$$

where we took into consideration the effect of the lag angle $$\alpha$$.

In order to get a rough estimation for the torque exerted by the satellite on the planet, we need to multiply this difference with the lever length (which is the planet diameter), and with the sine of the lag angle, giving:
 * $$N \approx 8\pi\frac{G\,m\,\rho\,A^4\,H}{r^3}\cos(\alpha)\sin(\alpha)

=4\pi\frac{G\,m\,\rho\,A^4\,H}{r^3}\sin(2\alpha)$$

A more exact calculation adds a 2/5 factor due to the planet spherical form and gives:
 * $$N = \frac{8}{5}\pi\frac{G\,m\,\rho\,A^4\,H}{r^3}\sin(2\alpha)$$

Inserting the value of H found above this is:
 * $$N=3\pi\frac{G\,m^2\,\rho\,A^8}{M\,r^6}\sin(2\alpha)$$

This can be written as:
 * $$N = \frac{9}{4}k\frac{G\,m^2\,A^5}{r^6}\sin(2\alpha)$$

Where k is a factor related that can be expressed by, taking into considerations non-uniformity in the planet mass density; corrections due to planet rigidity, neglected above, also enter here. For earth, most of the bulge is made of sea water and has no correction for rigidity, but its mass density is 0.18 the average earth mass density (1 g/cm3 vs. 5.5 g/cm3), so $$k\approx 0.18$$. The literature uses a close value of 0.2 ($$=2k_2/3$$ )

A similar calculation can be done for the tides created on the planet by the sun. Here, m should be replaced by the mass sun, and r by the distance to the sun. Since &alpha; depends on the dissipation properties of earth, it is expected to be the same for both. The resulting torque is 20% that exerted by the moon.

Relation of the lag angle to energy dissipation
The work exerted by the satellite over the planet is created by a force F acting along the path of movement of a mass units moving in velocity u in the planet (in fact, in the bulge).

Forces and locations depend on the relative angle to the planet-satellite axis &theta;, that changes periodically with the angular momentum &Omega;. Since the force in the planet spherical coordinate system is symmetrical in the direction towards the satellite and in the opposite direction (it is outwards in both), the dependence is approximated as sinusoidal in 2&theta;. Thus the force exerted on a unit mass is of the form:
 * $$ dF(t) = dF_0 \cos(2\theta) = dF_0 \cos(2\Omega t)$$

and the translation projected on the same direction is of the form:
 * $$ \xi(t) = \xi_0 \cos(2(\theta - \alpha) = \xi_0 \cos(2(\Omega t - \alpha)$$

due to the lag angle. The velocity component in the direction of the force is therefore:
 * $$ u(t) = \frac{d\xi(t)}{dt} = -2\Omega \xi_0 \sin(2(\theta - \alpha)$$

And so the total work exerted over a unit mass during one cycle is:
 * $$ \int_0^{\pi/\Omega} \vec{dF}(t)\vec{u}(t)dt = -2\Omega dF_0 \xi_0 \int_0^{\pi/\Omega} \cos(2\Omega t) \sin(2(\Omega t - \alpha))dt = -\pi dF_0 \xi_0 \sin(2\alpha)$$

In fact, almost all of this is dissipated (e.g. as friction), as explained below.

Looking now at the total energy from the satellite potential in one of the bulges, this is equal to the total work performed on this in a quarter of the total angular range, i.e. from zero to maximal displacement:
 * $$E^* = \int_{-\pi/4\Omega+\alpha/\Omega}^{\alpha/\Omega} \vec{dF}(t)\vec{u}(t)dt = -2\Omega dF_0 \xi_0 \int_{-\pi/4\Omega+\alpha/\Omega}^{\alpha/\Omega} \cos(2\Omega t) \sin(2(\Omega t - \alpha))dt$$

-dF_0 \xi_0 \int_{-\pi/2}^{0} \cos(z+2\alpha) \sin(z)dz \approx \frac{1}{2}dF_0 \xi_0$$ where we have defined $$z=2(\Omega t - \alpha)$$, and approximated for small &alpha; in the last equality, thus neglecting it.

The fraction of energy dissipated in each cycle is represented by the effective specific dissipation function, denoted by $$Q^{-1}$$ and defined as the total dissipation in one cycle divided by $$2\pi E^*$$. This gives:
 * $$Q^{-1} = \sin(2\alpha)$$

The value of this is estimated as 1/13 for earth, where the bulge is mainly liquid, 10−1-10−2 for the other inner planets and the moon, where the bulge is mainly solid, and as 10−3-10−5 for the outer, mostly gaseous planets.

With this value for earth at hand, the torque can be calculated to be 4.4x1016 N m, only 13% above the measured value of 3.9x1016 N m.

Note that in the distant past, the value of $$k\cdot \sin(2\alpha)$$ for the earth - moon system was probably somewhat smaller.

Retardation of the planet's rotation
Again neglecting, The change over time in the planet L is equal to the torque. L in turn is the product of the &Omega; with the  I.

For a spherical planet of approximately uniform mass density, $$I=f\,M\,A^2$$, where f is a factor depending on the planet structure; a spherical planet of uniform density has f = 2/5 = 0.4. Since the angular momentum This gives:

$$\frac{d\Omega}{dt} = \frac{dL}{Idt} =\frac{N}{I} = \frac{45}{8}k\frac{G\,m^2\,A^3}{M\,r^6}\sin(2\alpha)$$

Since the earth density is larger at depth, its moment of inertia is somewhat smaller, with f = 0.33.

For the earth-moon system, taking $$\sin(2\alpha)$$ of 1/13 and k=0.2, we get 4.5x10−22sec−2. For a 24-hour day, this is equivalent to 17 seconds in 1 million years, or 1 hour (i.e. lengthening of the day by 1 hour) in 210 million years. Due to the additional 20% effect of the sun, the day lengthens by 1 hour in approximately 180 million years.

A similar calculation shows that the earth had exerted angular momentum through tidal friction on the moon self-rotation, before this became. At that period, one calculates the change in the moon angular momentum &omega; in the same manner as for &Omega; above, except that m and M should be is switched, and A should be replaced by the moon radius a = 1.7x106 meter. Taking $$\sin(2\alpha)$$ of 10−1-10−2 as for the solid planets and k=1, this gives d&omega;/dt = 3x10−17-3x10−18sec2. For a 29.5-day long rotation period, this is equivalent to 1.5–15 minutes in 1 year, or 1 day in 102-103 years. Thus in astronomical timescales, the moon became tidally locked very fast.

Effect on the satellite motion around the planet
Due to conservation of angular momentum, a torque of the same size as the one exerted by the satellite and of opposite direction is exerted by the planet on the satellite motion around the planet. Another effect, which will not be dealt with here, is the changes in the eccentricity and inclination of the orbit.

The moment of inertia of this motion is m r2. However now r itself depends on the angular velocity which we denote here n: according to :
 * $$r^3\,n^2 = GM $$

Thus the satellite orbit angular momentum, l, satisfies (neglecting ):
 * $$l = m\,r^2\,n = m \sqrt{GM}\,r^{1/2}$$
 * $$N = \frac{dL}{dt} = \frac{1}{2}m \sqrt{GM}\,r^{-1/2}\frac{dr}{dt}$$
 * $$\frac{dr}{dt} = \frac{2r^{1/2}}{m \sqrt{GM}}N

=\frac{9}{2}k\sqrt{\frac{G}{M}}\frac{\,m\,A^5}{r^{11/2}}\sin(2\alpha)$$ Additionally, since $$n = \sqrt{GM}r^{-3/2} $$, we have:
 * $$\frac{dn}{dt} = -\frac{3}{2}\sqrt{GM}r^{-5/2}\frac{dr}{dt}

=\frac{27}{4}kG\frac{m\,A^5}{r^8}\sin(2\alpha)$$

Note that assuming all rotations are on the same direction and &Omega;>omega;, as time passes, the angular momentum of the planet decreases and hence that of the satellite orbit increases. Due to its relation with the planet-satellite distance, the latter increases, so the angular velocity of the satellite orbit decreases. For the earth-moon system, dr/dt gives 1.2x10−9 meter per second, or 3.7 cm per year. This is a 1% increase in the earth-moon distance in 100 million years. dn/dt is 1.3x10−23sec−2, and for a period of 29.5 days is equivalent to 7 minutes in 1 million years, or 1 day (i.e. lengthening of the lunar period in 1 day) in 210 million years.

Effect of the Sun
The sun-planet system has two tidal friction effects. One effect is that the sun creates a tidal friction in the planet, which decreases its spinning angular momentum and hence also increases its orbital angular momentum around the sun, hence increasing its distance and reducing its angular velocity (assuming the orbital angular velocity of the sun is smaller than that of the planet spinning; otherwise directions of change are opposite).

If MS is the sun mass and D is the distance to it, then the rate of change of D is given, similar to the above calculation, by:
 * $$\frac{dD}{dt} = \frac{2D^{1/2}}{M \sqrt{GM_S}}N_S

=\frac{9}{2}k\frac{\sqrt{G}\,M_S^{3/2}\,A^5}{M D^{11/2}}\sin(2\alpha) $$ The planet orbital angular velocity, &Omega;S, then changes as:
 * $$\frac{d\Omega_S}{dt} = -\frac{3}{2}\sqrt{GM_S}D^{-5/2}\frac{dD}{dt}

=\frac{27}{4}k\frac{G\,M_S^2\,A^5}{M D^8}\sin(2\alpha)$$

For the earth-sun system, this gives 1x10−13 meters per second, or 3 meters in 1 million years. This is a 1% increase in the earth-sun distance in half a billion years. The decrease in the earth orbital angular momentum is 2x10−31sec2, or equivalently for a 1-year period, 1 second in 1 billion years.

Another, relatively negligible, effect is that the planet creates tidal friction in the sun. This creates a change in the distance to the sun and the orbital angular velocity around it, as it does for the satellite in the satellite-planet system. Using the same equations but now for the planet-sun system, with AS standing for the sun radius (7x108 meter), we have:
 * $$\frac{dD}{dt}

=\frac{9}{2}k_S\sqrt{\frac{G}{M_S}}\frac{\,M\,{A_S}^5}{D^{11/2}}\sin(2\alpha_S)$$
 * $$\frac{d\Omega_S}{dt}

=\frac{27}{4}k_SG\frac{M\,A_S^5}{D^8}\sin(2\alpha_S)$$ where kS is a factor, presumably very small, due to the non-uniformity of mass densities of the sun. Assuming this factor times sin(2&alpha;S) to be not larger than what is found in the outer planets, i.e. 10−3-10−5, we have a negligible contribution from this effect.

Potential perturbation created by the moon on earth
The potential per mass unit that the moon creates on earth, whose center is located at distance r0 from the moon along the z-axis, in the earth - moon rotating, and in coordinates centered at the earth center, is:
 * $${\cal{W}} = -\frac{Gm}{|\vec(r)-r_0\hat{z}|} + \frac{1}{2}\omega^2|\vec{r}-r_1\hat{z}|^2 $$

where $$r_1$$ is the distance from the moon to the center of mass of the earth - moon system, &omega; is the angular velocity of the earth around this point (the same as the lunar orbital angular velocity). The second term is the effective potential due to the of the earth.

We expand the potential in around the point. The linear term must vanish (at least on average in time) since otherwise the force on the earth center would be non vanishing. Thus:


 * $${\cal{W}} = \frac{1}{2}\omega^2\left(x^2+y^2+(z-r_1)^2\right) - \frac{Gm}{\sqrt{x^2+y^2+(z-r_0)^2}} $$
 * $$ = const + \frac{1}{2}\omega^2(x^2+y^2+z^2) - \frac{Gm}{r_0^3}\left(z^2 - \frac{1}{2}(x^2+y^2)\right) +

\frac{Gm}{r_0^5}\left(\dots \right) + \dots$$

Moving to spherical coordinates this gives:


 * $$ {\cal{W}} = const + \frac{1}{2}\omega^2 r^2 - \frac{Gmr^2}{r_0^3}\frac{1}{2}\left(3 \cos^2(\theta) - 1\right) +

\frac{Gm}{r_0^5}\left(\dots \right) + \dots $$
 * $$ = const + \frac{1}{2}\omega^2 r^2 - Gm\sum_{n=2}^{\infty}\frac{r^n}{r_0^{n+1}}P_n(\cos \theta)$$

where $$P_n(\cos \theta)$$ are the.

The constant term has no mechanical importance, while the $$r^2$$ causes a fixed dilation, and is not directly involved in creating a torque.

Thus we focus on the other terms, whose sum we denote $${\cal{W}}_{2+}$$, and mainly on the $$P_2(\cos \theta)$$ term which is the largest, as $$\frac{r}{r_0}$$ is at most the ratio of the earth radius to its distance from the moon, which is less than 2%.

Form of the bulge I: response to a
We treat the potential created by the moon as a perturbation to the earth's gravitational potential. Thus the height on earth at angles $$\theta$$, $$\varphi$$ is:
 * $$r(\theta,\varphi) = A + \delta(\theta,\varphi)$$

where $$\delta\ll A$$, and the amplitude of &delta; is proportional to the perturbation. We expand &delta; in Legendre polynomials, where the constant term (which stands for dilation) will be ignored as we are not interested in it. Thus:
 * $$\delta(\theta,\varphi) = \sum_{n=1}^{\infty}\delta_n P_n(\cos \theta)$$

where &delta;n are unknown constants we would like to find.

We assume for the moment total equilibrium, as well as no rigidity on earth (e.g. as in a liquid earth). Therefore, its surface is, and so $$V_E\left(r(\theta,\varphi)\right)+W_{2+}\left(r(\theta,\varphi)\right)$$ is constant, where $$V_E(r)$$ is the earth potential per unit mass. Since &delta; is proportional to $$W_{2+}$$, which is much smaller than VE, This can be expanded in &delta;. Dropping non-linear terms we have:


 * $$const = V_E\left(r(\theta,\varphi)\right)+{\cal{W}}_{2+}\left(r(\theta,\varphi)\right) \approx

V_E(A) + V_E^{\prime}(A) \delta(\theta,\varphi) + {\cal{W}}_{2+}(A)$$
 * $$const = V_E^{\prime}(A) \sum_{n=1}^{\infty}\delta_n P_n(\cos \theta) - Gm\sum_{n=2}^{\infty}\frac{A^n}{r_0^{n+1}}P_n(\cos \theta) $$

Note that $$V_E^{\prime}(r)\equiv dV_E(r)/dr$$ is the force per unit mass from earth's gravity, i.e. $$V_E^{\prime}(A)$$ is just the gravitational acceleration g.

Since the Legendre polynomials are, we may equate their coefficients n both sides of the equation, giving:
 * $$\delta_1 = 0$$
 * $$\delta_n = \frac{GmA^n/r_0^{n+1}}{V_E^{\prime}(A)} = \frac{GmA^n/r_0^{n+1}}{GM/A^2}= \frac{mA^{n+2}}{Mr_0^{n+1}}$$, $$n\ge 2$$

Thus the height is the ratio between the perturbation potential and the force from the perturbated potential.

Form of the bulge II: the deformation creating a perturbative potential
So far we have neglected the fact that the deformation itself creates a perturbative potential. In order to account for this, we may calculate this perturbative potential, re-calculate the deformation and continue so iteratively.

Let us assume the mass density is uniform. Since &delta; is much smaller than A, the deformation can be treated as a thin shell added to the mass of the earth, where the shell has a surface mass density &rho; &delta; (and can also be negative), with &rho; being the mass density (if mass density is not uniform, then the change of shape of the planet creates differences in mass distribution in all depth, and this has to be taken into account as well). Since the gravitation potential has the same form as the electric potential, this is a simple problem in. For the analogous electrostatic problem, the potential created by the shell has the form:
 * $$\sum_{n=0}^{\infty}a_n r^n P_n(\cos\theta)$$, $$r\le A$$
 * $$\sum_{n=0}^{\infty}a_n \frac{A^{2n+1}}{r^{n+1}} P_n(\cos\theta)$$, $$r\ge A$$

Where the surface charge density is proportional to the discontinuity in the gradient of the potential:
 * $$\sigma(\theta) = \varepsilon_0\sum_{n=0}^{\infty}(2n+1) a_n A^{n-1} P_n(\cos\theta)$$

$$\varepsilon_0$$ is the, a constant relevant to electrostatics, related to the equation $$U(r) = \frac{1}{4\pi\varepsilon_0}\frac{q^2}{r}$$. The analogous equation in gravity is $$U(r) = G\frac{m^2}{r}$$, so if charge density is replaced with mass density, $$\varepsilon_0$$ should be replaced with $$\frac{1}{4\pi G}$$.

Thus in the gravitational problem we have:
 * $$\rho \sum_n \delta_n P_n(\cos \theta) = \frac{1}{4\pi G}\sum_n (2n+1) a_n A^{n-1} P_n(\cos\theta)$$

So that, again due to the orthogonality of Legendre polynomials:
 * $$a_n = 4\pi \frac{G \rho}{(2n+1)A^{n-1}} \delta_n $$

Thus the perturbative potential per mass unit for $$r\ge A$$ is:
 * $$4\pi G \rho A^{n+2} \sum_{n=0}^{\infty} \frac{1}{(2n+1)r^{n+1}} \delta_n P_n(\cos\theta)$$
 * $$= 3 \frac{G M}{A^2} \sum_n \frac{A^{n+1}}{(2n+1)r^{n+1}} \delta_n P_n(\cos\theta)$$

Note that since earth's mass density is in fact not uniform, this result must be multiplied by a factor that is roughly the ratio of the bulge mass density and the average earth mass, approximately 0.18. The actual factor is somewhat larger, since there is some deformation in the deeper solid layers of earth as well. Let us denote this factor by x. Rigidity also lowers x, though this is less relevant for most of the bulge, made of sea water.

The deformation was created by the a perturbative potential of size $${\cal{W}}_{2+}(A) = \frac{G M}{A^2}\delta_n P_n(\cos\theta)$$. Thus for each coefficient of $$P_n(\cos\theta)$$, the ratio of the original perturbative potential to that secondarily created by the deformation is:
 * $$c_n\equiv \frac{3 x}{2n+1} $$

with x=1 for perfectly a non-rigid uniform planet.

This secondary perturbative potential creates another deformation which again creates a perturbative potential and so on ad infinitum, so that the total deformation is of the size:
 * $$\sum_n \sum_{k=0}^{\infty} c_n^k h_n P_n(\cos \theta)= \sum_n \frac{1}{1-c_n} \delta_n P_n(\cos \theta)$$

For each mode, the ratio to &delta;n, the naive estimation of the deformation, is $$\frac{1}{1-c_n}$$ and is denoted as $$h_n$$. For a perfectly a non-rigid uniform planet (e.g. a liquid earth of non-compressible liquid), this is equal to $$\frac{2n+1}{2}$$, and for the main mode of n=2, it is 5/2.

Similarly, n-th mode of the tidal perturbative potential per unit mass created by earth at r = A is the kn times the corresponding term in the original lunar tidal perturbative potential, where for a uniform mass density, zero rigidity planet kn is:
 * $$k_n = \sum_{k=1}^{\infty} c_n^k = \frac{c_n}{1-c_n}$$

For a perfectly a non-rigid uniform planet (e.g. a liquid earth of non-compressible liquid), this is equal to 3/2. In fact, for the main mode of n=2, the real value for earth is a fifth of it, namely k2 = 0.3 (which fits c2 = 0.23 or x = 0.38, roughly twice the density ratios of 0.18).

Calculation of the torque
Instead of calculating the torque exerted by the moon on the earth deformation, we calculate the reciprocal torque exerted by the earth deformation on the moon; both must be equal.

The potential created by the earth bulge is the perturbative potential we have discussed above. Per unit mass, for r = A, this is the same as the lunar perturbative potential creating the bulge, with each mode multiplied by kn, with the n=2 mode far dominating the potential. Thus at r=A the bulge perturbative potential per unit mass is:


 * $${\cal{U}(r=A, \theta)} = - Gm\sum_{n=2}^{\infty}k_n\frac{A^n}{r_0^{n+1}}P_n(\cos \theta)$$

since the n-the mode it drops off as r-(n+1) for r>A, we have outside earth:
 * $${\cal{U}(r, \theta)} = - Gm\sum_{n=2}^{\infty}k_n\frac{A^n}{r_0^{n+1}}\frac{A^n+1}{r^{n+1}}P_n(\cos \theta)$$

However, the bulge actually lags at an angle &alpha; with respect to the direction to the moon due to earth's rotation. Thus we have:
 * $${\cal{U}} = - Gm\sum_{n=2}^{\infty}k_n\frac{A^{2n+1}}{r_0^{n+1}r^{n+1}}P_n\left(\cos (\theta-\alpha)\right)$$

The moon is at r=r0, &theta;=0. Thus the potential per unit mass at the moon is:
 * $${\cal{U}(r=r_0, \theta=0)} = - Gm\sum_{n=2}^{\infty}k_n\frac{A^{2n+1}}{r_0^{2n+2}}P_n\left(\cos (\alpha)\right)

\approx - Gm k_2\frac{A^5}{r_0^6}P_2(\cos \alpha)$$
 * $$= - Gm k_2\frac{A^5}{r_0^6}\cdot\frac{1}{2}(3 \cos^2 \alpha - 1)$$

Neglecting eccentricity and axial tilt, We get the torque exerted by the bulge on the moon by multiplying :$${\cal{U}}$$ with the moon's mass m, and differentiating with respect to &theta; at the moon location. This is equivalent to differentiating $$m{\cal{U}(r=r_0, \theta=0)} $$ with respect to &alpha;, and gives:
 * $$N = m\frac{d{\cal{U}(r=r_0, \theta=0)}}{d\alpha} = \frac{3}{2}Gm^2 k_2\frac{A^5}{r_0^6}\cdot \sin(2\alpha)$$

This is the same formula used, with r=r0 and k there defined as 2k2/3.