Editing Tide (section)

=== Forces ===
{{main|Tidal force}}
The tidal force produced by a massive object (Moon, hereafter) on a small particle located on or in an extensive body (Earth, hereafter) is the vector difference between the gravitational force exerted by the Moon on the particle, and the gravitational force that would be exerted on the particle if it were located at the Earth's center of mass.

Whereas the [[gravitational force]] subjected by a celestial body on Earth varies inversely as the square of its distance to the Earth, the maximal tidal force varies inversely as, approximately, the cube of this distance.<ref>{{cite book |last=Young |first=C. A. |date=1889 |title=A Textbook of General Astronomy |url=https://www.gutenberg.org/files/37275/37275-pdf.pdf |page=288 |access-date=2018-08-13 |archive-date=2019-10-05 |archive-url=https://web.archive.org/web/20191005003655/http://www.gutenberg.org/files/37275/37275-pdf.pdf |url-status=live }}</ref> If the tidal force caused by each body were instead equal to its full gravitational force (which is not the case due to the [[free fall]] of the whole Earth, not only the oceans, towards these bodies) a different pattern of tidal forces would be observed, e.g. with a much stronger influence from the Sun than from the Moon: The solar gravitational force on the Earth is on average 179 times stronger than the lunar, but because the Sun is on average 389 times farther from the Earth, its field gradient is weaker. The overall proportionality is

: <math>\text{tidal force} \propto \frac{M}{d^3} \propto \rho\left(\frac{r}{d}\right)^3,</math>

where {{mvar|M}} is the mass of the heavenly body, {{mvar|d}} is its distance, {{mvar|ρ}} is its average density, and {{mvar|r}} is its radius. The ratio {{math|''r''/''d''}} is related to the angle subtended by the object in the sky. Since the Sun and the Moon have practically the same diameter in the sky, the tidal force of the Sun is less than that of the Moon because its average density is much less, and it is only 46% as large as the lunar,<!-- numbers double-checked by User:JEBrown87544, 03 Jan 2007 -->{{efn|According to [https://web.archive.org/web/19980224174548/http://imagine.gsfc.nasa.gov/docs/ask_astro/answers/961029b.html NASA] the lunar tidal force is 2.21 times larger than the solar.}} thus during a spring tide, the Moon contributes 69% while the Sun contributes 31%. More precisely, the lunar tidal acceleration (along the Moon–Earth axis, at the Earth's surface) is about 1.1{{e|−7}}&nbsp;''g'', while the solar tidal acceleration (along the Sun–Earth axis, at the Earth's surface) is about 0.52{{e|−7}}&nbsp;''g'', where ''g'' is the [[standard gravity|gravitational acceleration]] at the Earth's surface.{{efn|See [[Tidal force#Formulation|Tidal force – Mathematical treatment]] and sources cited there.}} The effects of the other planets vary as their distances from Earth vary. When Venus is closest to Earth, its effect is 0.000113 times the solar effect.<ref name="Science Mission Directorate 2000">{{cite web | title=Interplanetary Low Tide | website=Science Mission Directorate | date=2000-05-03 | url=https://science.nasa.gov/science-news/science-at-nasa/2000/ast04may_1m | access-date=2023-06-25 | archive-date=2023-06-04 | archive-url=https://web.archive.org/web/20230604014510/https://science.nasa.gov/science-news/science-at-nasa/2000/ast04may_1m | url-status=live }}</ref> At other times, Jupiter or Mars may have the most effect.

[[File:Tidal field and gravity field.svg|thumb|The lunar [[gravity]] residual [[Vector field|field]] at the Earth's surface is known as the ''[[tide-generating force]]''. This is the primary mechanism that drives tidal action and explains two simultaneous tidal bulges; Earth's rotation further accounts for two daily high waters at any location. The figure shows both the tidal field (thick red arrows) and the gravity field (thin blue arrows) exerted on Earth's surface and center (label O) by the Moon (label S).|alt=Diagram showing a circle with closely spaced arrows pointing away from the reader on the left and right sides, while pointing towards the user on the top and bottom.]]

The ocean's surface is approximated by a surface referred to as the [[geoid]], which takes into consideration the gravitational force exerted by the earth as well as [[centrifugal force]] due to rotation. Now consider the effect of massive external bodies such as the Moon and Sun. These bodies have strong gravitational fields that diminish with distance and cause the ocean's surface to deviate from the geoid. They establish a new equilibrium ocean surface which bulges toward the moon on one side and away from the moon on the other side. The earth's rotation relative to this shape causes the daily tidal cycle. The ocean surface tends toward this equilibrium shape, which is constantly changing, and never quite attains it. When the ocean surface is not aligned with it, it's as though the surface is sloping, and water accelerates in the down-slope direction.