Editing Tidal force (section)

== Formulation ==
[[File:Inseparable galactic twins.jpg|thumb|Figure 7: Tidal force is responsible for the merge of galactic pair [[MRK 1034]].<ref>{{cite news|title=Inseparable galactic twins|url=http://www.spacetelescope.org/images/potw1325a/|access-date=12 July 2013|newspaper=ESA/Hubble Picture of the Week}}</ref> ]]
[[File:Tidal-forces.svg|thumb|right|Figure 8: Graphic of tidal forces. The top picture shows the gravity field of a body to the right (not shown); the lower shows their residual gravity once the field at the centre of the sphere is subtracted; this is the tidal force. For visualization purposes, the top arrows may be assumed as equal to 1 N, 2 N, and 3 N (from left to right); the resulting bottom arrows would equal, respectively, −1 N (negative, thus 180-degree rotated), 0 N (invisible), and 1 N. See Figure 2 for a more detailed version]]

For a given (externally generated) gravitational field, the '''tidal acceleration''' at a point with respect to a body is obtained by [[Euclidean vector#Addition and subtraction|vector subtraction]] of the gravitational acceleration at the center of the body (due to the given externally generated field) from the gravitational acceleration (due to the same field) at the given point. Correspondingly, the term ''tidal force'' is used to describe the forces due to tidal acceleration. Note that for these purposes the only gravitational field considered is the external one; the gravitational field of the body (as shown in the graphic) is not relevant. (In other words, the comparison is with the conditions at the given point as they would be if there were no externally generated field acting unequally at the given point and at the center of the reference body. The externally generated field is usually that produced by a perturbing third body, often the Sun or the Moon in the frequent example-cases of points on or above the Earth's surface in a geocentric reference frame.)

Tidal acceleration does not require rotation or orbiting bodies; for example, the body may be [[freefall]]ing in a straight line under the influence of a gravitational field while still being influenced by (changing) tidal acceleration.

By [[Newton's law of universal gravitation]] and laws of motion, a body of mass ''m'' at distance ''R'' from the center of a sphere of mass ''M'' feels a force <math display="inline">\vec{F}_g</math>,

<math display="block">\vec{F}_g = - \hat{r} ~ G ~ \frac{M m}{R^2}</math>

equivalent to an acceleration <math display="inline">\vec{a}_g</math>,

<math display="block">\vec{a}_g = - \hat{r} ~ G ~ \frac{M}{R^2}</math>

where <math display="inline">\hat{r}</math> is a [[unit vector]] pointing from the body ''M'' to the body ''m'' (here, acceleration from ''m'' towards ''M'' has negative sign).

Consider now the acceleration due to the sphere of mass ''M'' experienced by a particle in the vicinity of the body of mass ''m''. With ''R'' as the distance from the center of ''M'' to the center of ''m'', let ∆''r'' be the (relatively small) distance of the particle from the center of the body of mass ''m''. For simplicity, distances are first considered only in the direction pointing towards or away from the sphere of mass ''M''. If the body of mass ''m'' is itself a sphere of radius ∆''r'', then the new particle considered may be located on its surface, at a distance (''R'' ± ''∆r'') from the centre of the sphere of mass ''M'', and ''∆r'' may be taken as positive where the particle's distance from ''M'' is greater than ''R''. Leaving aside whatever gravitational acceleration may be experienced by the particle towards ''m'' on account of ''m''{{'}}s own mass, we have the acceleration on the particle due to gravitational force towards ''M'' as:

<math display="block">\vec{a}_g = - \hat{r} ~ G ~ \frac{M}{(R \pm \Delta r)^2}</math>

Pulling out the ''R''<sup>2</sup> term from the denominator gives:

<math display="block">\vec{a}_g = -\hat{r} ~ G ~ \frac{M}{R^2} ~ \frac{1}{\left(1 \pm \frac{\Delta r}{R}\right)^2}</math>

The [[Maclaurin series]] of <math display="inline">1/(1 \pm x)^2</math> is <math display="inline">1 \mp 2x + 3x^2 \mp \cdots</math> which gives a series expansion of:

<math display="block">\vec{a}_g = - \hat{r} ~ G ~ \frac{M}{R^2} \pm \hat{r} ~ G ~ \frac{2 M }{R^2} ~ \frac{\Delta r}{R} + \cdots </math>

The first term is the gravitational acceleration due to ''M'' at the center of the reference body <math display="inline">m</math>, i.e., at the point where <math display="inline">\Delta r</math> is zero. This term does not affect the observed acceleration of particles on the surface of ''m'' because with respect to ''M'', ''m'' (and everything on its surface) is in free fall. When the force on the far particle is subtracted from the force on the near particle, this first term cancels, as do all other even-order terms. The remaining (residual) terms represent the difference mentioned above and are tidal force (acceleration) terms. When ∆''r'' is small compared to ''R'', the terms after the first residual term are very small and can be neglected, giving the approximate tidal acceleration <math display="inline">\vec{a}_{t,\text{axial}}</math> for the distances ∆''r'' considered, along the axis joining the centers of ''m'' and ''M'':

<math display="block">\vec{a}_{t,\text{axial}} \approx \pm \hat{r} ~ 2 \Delta r ~ G ~ \frac{M}{R^3} </math>

When calculated in this way for the case where ∆''r'' is a distance along the axis joining the centers of ''m'' and ''M'', <math display="inline">\vec{a}_t</math> is directed outwards from to the center of ''m'' (where ∆''r'' is zero).

Tidal accelerations can also be calculated away from the axis connecting the bodies ''m'' and ''M'', requiring a [[Euclidean vector|vector]] calculation. In the plane perpendicular to that axis, the tidal acceleration is directed inwards (towards the center where ∆''r'' is zero), and its magnitude is <math display="inline">\frac{1}{2}\left|\vec{a}_{t,\text{axial}} \right|</math> in linear approximation as in Figure 2.

The tidal accelerations at the surfaces of planets in the Solar System are generally very small. For example, the lunar tidal acceleration at the Earth's surface along the Moon–Earth axis is about {{val|1.1|e=-7|u=''g''}}, while the solar tidal acceleration at the Earth's surface along the Sun–Earth axis is about {{val|0.52|e=-7|u=''g''}}, where ''g'' is the [[standard gravity|gravitational acceleration]] at the Earth's surface. Hence the tide-raising force (acceleration) due to the Sun is about 45% of that due to the Moon.<ref>
{{cite book
 | title=Admiralty manual of navigation
 | volume=1
 | author=The Admiralty
 | publisher=[[The Stationery Office]]
 | date=1987
 | isbn=978-0-11-772880-6
 | page=277
 | url=https://books.google.com/books?id=GCgXCxG4VLcC
}}, [https://books.google.com/books?id=GCgXCxG4VLcC&pg=PA277 Chapter 11, p. 277]
</ref> The solar tidal acceleration at the Earth's surface was first given by Newton in the ''[[Philosophiæ Naturalis Principia Mathematica|Principia]]''.<ref>
{{cite book
 | title=The mathematical principles of natural philosophy
 | volume=2
 | first=Isaac | last=Newton
 | date=1729
 | isbn=978-0-11-772880-6
 | page=307
 | url=https://books.google.com/books?id=6EqxPav3vIsC
}}, [https://archive.org/details/bub_gb_6EqxPav3vIsC/page/307 <!-- pg=307 --> Book 3, Proposition 36, Page 307] Newton put the force to depress the sea at places 90 degrees distant from the Sun at "1 to 38604600" (in terms of ''g''), and wrote that the force to raise the sea along the Sun-Earth axis is "twice as great" (i.e., 2 to 38604600) which comes to about 0.52 × 10<sup>−7</sup> ''g'' as expressed in the text.
</ref>