Tides.

[Gareth56]

This has always perplexed me with regard to tides. I understand that the gravitational attraction of the Moon causes the oceans to bulge out in the direction of the Moon and another bulge occurs on the opposite side because the Earth is also being pulled toward the Moon (and away from the water on the far side of the Earth). My question is why isn't the water (on the far side of the Earth) also pulled toward the Moon because if the Moon can pull the far side of the Earth towards it why can't it also pull the water that's also on the far side towards it as well?

 

Thanks

[swansont]

This has always perplexed me with regard to tides. I understand that the gravitational attraction of the Moon causes the oceans to bulge out in the direction of the Moon and another bulge occurs on the opposite side because the Earth is also being pulled toward the Moon (and away from the water on the far side of the Earth). My question is why isn't the water (on the far side of the Earth) also pulled toward the Moon because if the Moon can pull the far side of the Earth towards it why can't it also pull the water that's also on the far side towards it as well?

 

Thanks

 

It does, but the acceleration is smaller because it's further away. Hence the bulge.

[Gareth56]

It does, but the acceleration is smaller because it's further away. Hence the bulge.

 

 

Presumably the acceleration of the water is less on the opposite side of the Earth because it's overall mass is less than that of the earth at the same location, so the Earth accelerates at a faster rate towards the Moon than the water does?

[swansont]

Presumably the acceleration of the water is less on the opposite side of the Earth because it's overall mass is less than that of the earth at the same location, so the Earth accelerates at a faster rate towards the Moon than the water does?

 

No, mass does not change that — the acceleration is independent of mass. F = ma = GMm/r^2, so m cancels and the acceleration depends on the sun's mass only, as well as r. It is because r is larger that you get the smaller acceleration, and thus the tidal bulge

[ewmon]

Also, IIRC, Earth is also relatively solid and tends to move as one piece, and the oceans are liquid and tend to deform more easily.

[D H]

Treating the Earth and Moon as spherical bodies and ignoring tiny relativistic effects, the acceleration of the Earth as a whole toward the Moon in some inertial frame is given by Newton's law of gravitation

 

[math]\vec a_e = \frac{GM_m}{||\vec r_{e\to m}||^3} \vec r_{e\to m}[/math]

 

where [math]\vec r_{e\to m}[/math] is the vector from the Earth's center of mass to the Moon's center of mass.

 

The acceleration at some point on the Earth's surface is similarly given by

 

[math]\vec a_p = \frac{GM_m}{||\vec r_{p\to m}||^3} \vec r_{p\to m}[/math]

 

where [math]\vec r_{p\to m}[/math] is the vector from the point in question to the Moon's center of mass. The tides result from the difference between these accelerations:

 

[math]\vec a_{p,\text{rel}} = \vec a_p - \vec a_e= GM_m\left(\frac {\vec r_{p\to m}}{||\vec r_{p\to m}||^3} - \frac {\vec r_{e\to m}}{||\vec r_{e\to m}||^3}\right)[/math]

 

 

 

For the point p₁ on the surface of the Earth directly between the Earth's and Moon's centers of mass, this becomes

 

[math]a_{p_1,\text{rel}} = GM_m\left(\frac {1}{(R_m-r_e)^2} - \frac{1}{R_m^2}\right)\hat m\approx \frac{2GM_mr_e}{R_m^3}\hat m = \frac{2GM_mr_e}{R_m^3}\hat p_1[/math]

 

where [math]R_m[/math] is the distance between the Earth and the Moon, [math]r_e[/math] is the radius of the Earth, [math]\hat m[/math] is the unit vector from the center of the Earth toward the Moon, and [math]\hat p_1[/math] is the unit vector from the center of the Earth toward the point p₁. Note that [math]\hat p_1=\hat m[/math].

 

Now consider the point p₂ on the surface of the Earth that is diametrically opposed to p₁. Note that the unit vector from the center of the Earth toward p₂, [math]\hat p_2[/math], is directed away from the Moon (the Moon is underfoot). For this point, the relative acceleration becomes

 

[math]a_{p_2,\text{rel}} = GM_m\left(\frac {1}{(R_m+r_e)^2} - \frac{1}{R_m^2}\right)\hat m\approx \frac{-2GM_mr_e}{R_m^3}\hat m = \frac{2GM_mr_e}{R_m^3}\hat p_2[/math]

 

So once again the acceleration is directed outward.

 

 

Edits

16 Dec 2010: Fixed a LaTeX error: The forward slash in [math]/hat m[/math] should have been a backslash, resulting in [math]\hat m[/math].

Edited December 16, 2010 by D H

[between3and26characterslon]

simples

[pioneer]

Are there also tides in the atmosphere, due to the moon? I am not trying to be funny. It seems reasonable to assume that since the atmosphere is even closer to the moon, than the oceans, one would expect the moon's gravity would create tidal effects within the atmosphere.

Edited November 23, 2010 by pioneer

[D H]

Also, IIRC, Earth is also relatively solid and tends to move as one piece, and the oceans are liquid and tend to deform more easily.

Yes, but there are Earth tides. As the Earth's k2 Love Number is about 0.3, these Earth tides are a bit less than 1/3 the size of the ocean tides. However, since the Earth tides involve all of the Earth rather than a thin shell of water that covers 78% of the Earth's surface, the impact on the orbits of satellites is about 10 times those of the ocean tides.

 

Are there also tides in the atmosphere, due to the moon?

Yes, there are also atmospheric tides, and the Moon does contribute to these tides. However, unlike the ocean tides and Earth tides where the dominant feature is the ~12.4 hour long lunar semidiurnal, the dominant feature in the atmospheric tides is the diurnal bulge caused by the Sun.

Edited November 23, 2010 by D H

[ewmon]

the dominant feature in the atmospheric tides is the diurnal bulge caused by the Sun.

Wow! It gives a new meaning to "High Noon" (aka 1952 western starring Gary Cooper).

Edited November 25, 2010 by ewmon

[pioneer]

Is it possible that the lunar atmospheric tides are lumped into something else? It seems illogical to have this lunar gap.

 

One possible explanation is connected to the observation that the ocean tidal changes are not the same at all places on earth.

 

The World’s Highest Tides Ecozone presents an extraordinary tidal landscape. This includes the upper basins of the Bay of Fundy, where the peak tidal range is around 15 m (50 ft) — five times higher than typical tides on the rest of the Atlantic coast!

 

 

Gravity should be relatively uniform and not 5 times higher at the Bay of Fundy. This suggests the tidal effect can concentrate using other factors.

 

This might suggest the lunar atmospheric tides are concentrated and not uniformly distributed. Other factors cause the lunar tides to focus more energy in certain places which may be lumped with other dominant effects that are too strong to explain with gravity alone; bay of Fundy effect.

Edited December 4, 2010 by pioneer

[TonyMcC]

Other factors certainly do have a huge impact on the height of tides. In some river estuaries the narrowing of the river and its reducing depth result in water "piling up". In the UK the estuary of the river Severn gives the second highest tidal range in the world because of this phenomenon. This concentration of the water also sets up a tidal wave known as the Severn Bore which travels miles up the river Severn and on which people like to surf. There is plenty of information if you google" Severn bore". I will just mention I spent some months in the Maldives where the tidal range was only about 1 metre. Just as well really as the islands of the Maldives are only a metre or two above sea level!

Edited December 4, 2010 by TonyMcC

[swansont]

!

Moderator Note

matterdoc's contributions and the resulting discussion have been moved http://www.scienceforums.net/topic/53309-tides/

[D H]

For the point on the surface of the Earth directly between the Earth's and Moon's centers of mass, this becomes

 

[math]a \approx \frac{2GM_mr_e}{{R_m}^3}[/math]

 

Playing with this expression yields another explanation for why lunar tides are greater than solar tides. Designating [math]r_m[/math], [math]\rho_m[/math], and [math]\delta_m[/math] as the radius, density, and ngular diameter of the Moon, the above expression for the tidal force becomes

 

 

[math]a \approx \frac {\pi\,G\,r_e} 3 \, \rho_m \, {\delta_m}^3[/math]

 

There is nothing special about the Moon here. The same expression applies for Jupiter, Saturn, and the Sun. The tidal force at the points on the surface of the Earth along line connecting the center of the Earth and the perturbing object (Moon, Sun, etc) is proportional to the product of the object's density and the cube of the object's angular diameter.

 

Note hat the angular diameters of the Moon and the Sun as seen from the Earth are nearly equal (0.518° for the Moon versus 0.533° for the Sun). This means the relative strength of tidal forces from these two objects depends primarily their average densities, which is about 3.34 grams per cc for the Moon but only 1.41 grams per cc for the Sun.

 

 

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Is it possible that the lunar atmospheric tides are lumped into something else? It seems illogical to have this lunar gap.

What "lunar gap"?

 

Gravity should be relatively uniform and not 5 times higher at the Bay of Fundy. This suggests the tidal effect can concentrate using other factors.

Gravity is not 5 times higher at the Bay of Fundy. Have you ever sung in the shower? Your vocal chords are not 5 times stronger when you are in a shower.