Tidal bulges?

[Gareth56]

I can understand why there is a tidal bulge on the side of the Earth that is facing the Moon because basically the Moon is pulling on the sea that is facing it, however what I cannot understand is why there is a there a similar bulge on the opposite side of the Earth as there is nothing on that side pulling on the sea.

 

Could someone put me out of my misery as this has confused me for years?

 

Thanks

[D H]

The Earth as a whole is accelerating toward the Moon:

 

[math]\boldsymbol{a}_{e\to m} = -\,\frac {GM_m}{||\boldsymbol{r}_{e\to m}||^3}\boldsymbol{r}_{e\to m}[/math]

 

The acceleration at some point pon the surface of the Earth is slightly different from this:

 

[math]\boldsymbol{a}_{p\to m} = -\,\frac {GM_m}{||\boldsymbol{r}_{p\to m}||^3}\boldsymbol{r}_{p\to m}[/math]

 

The tidal acceleration at that point is the difference between the gravitational acceleration toward the Moon at that point and the gravitational acceleration of the Earth as a whole toward that point. The acceleration at the point on the Earth's surface directly in line with the Moon's and Earth's center of mass is directed toward the Moon with magnitude approximately

 

[math]a_{tidal} \approx 2\frac {GM_m}{||\boldsymbol{r}_{p\to m}||^3}r_e[/math]

 

The tidal acceleration at the point on the opposite side of the Earth has nearly the same magnitude and is directed away from the Moon. Hence the bulges.

[Gareth56]

Sorry but all that "gubbins" is way above my head.

[Flashman]

It's to do with the gravitation differential. The earth does not block the moons gravity, therefore on the opposite side of the earth to where the moon is, the moon "under your feet" is adding to the earth's gravity at that point. This makes an area of higher gravity, and water runs "down" the gravitational gradient to seek equilibrium.

 

Then on the moon side, it's not so much that the moon is pulling the water, it's that the gravitational center of the earth moon system is displaced toward that side, making the area below the moon, the "lowest" point wrt to the center of gravity of the system, and again the water runs down the gravity gradient towards it.

 

At least that's how I understand it.

[Gareth56]

It's to do with the gravitation differential. The earth does not block the moons gravity, therefore on the opposite side of the earth to where the moon is, the moon "under your feet" is adding to the earth's gravity at that point. This makes an area of higher gravity, and water runs "down" the gravitational gradient to seek equilibrium.

 

Then on the moon side, it's not so much that the moon is pulling the water, it's that the gravitational center of the earth moon system is displaced toward that side, making the area below the moon, the "lowest" point wrt to the center of gravity of the system, and again the water runs down the gravity gradient towards it.

 

At least that's how I understand it.

 

I'm not too sure how the moon can be under my feet (emboldened above) also the water is being "pulled" away from the Moon on the opposite side of the earth to where the Moon so again I'm not sure how the water can run "down"

[iNow]

If you have a rubber ball, and one side gets pushed out, the rest of the ball will deform. Think of squashing the ball between your hand and the floor. The area around the equator gets expanded, and the area around the poles moves closer to the center.

 

Similar with earth. It's not so much that "one side is getting closer to the moon," it's that the earth bulges at the center and squashes in from the poles, hence the bulge is seen around the entire planet, not just the moon facing side.

[Gareth56]

So if I understand this correctly the "squashing" of the Earth, which is due to the rotation of the Earth, is primarily responsible for the tidal bulges we observe?

 

So the effect of the Moon is negligible when it comes to the tides?

[iNow]

So if I understand this correctly the "squashing" of the Earth, which is due to the rotation of the Earth, is primarily responsible for the tidal bulges we observe?

 

So the effect of the Moon is negligible when it comes to the tides?

 

No, that's not really what I intended. I think this link explains it better than I could:

 

 

http://www.space.com/scienceastronomy/moon_mechanics_0303018.html

Gravity is said to be the weakest of all the fundamental forces. But one aspect of it is very consequential: Gravity never goes away. It weakens with distance, but it is always at work. This fact is the primary driver of tides. The side of Earth nearest the Moon always gets tugged more than the other side, by about 6 percent.

 

Hey, you might say, there are two high tides on this planet at any given moment. True. And another far more complex set of phenomena explains this.

 

The Moon does not just go around the Earth. In reality, the two objects orbit about a common gravitational midpoint, called a barycenter. The mass of each object and the distance between them dictates that this barycenter is inside Earth, about three-fourths of the way out from the center.

 

So picture this: The center of the Earth actually orbits around this barycenter, once a month. The effect of this is very important. Think, for a second, of a spacecraft orbiting Earth. Its astronauts experience zero gravity. That's not because there's no gravity up there. It's because the ship and its occupants are constantly falling toward Earth while also moving sideways around the planet. This sets up a perpetual freefall, or zero-g.

 

Like the orbiting spaceship, the center of the Earth is in free-fall around the barycenter of the Earth-Moon system.

 

Here's the kicker:

On the side of Earth opposite the Moon, the force of the Moon's gravity is less than at the center of the Earth, because of the greater distance. It can actually be thought of as a negative force, in essence, pulling water away from the Moon and away from Earth's surface -- a second high tide.

 

Our planet rotates under these constantly shifting tides, which is why high and low tides are always moving about, rolling in and rolling out as far as observers on the shore are concerned.

 

The Sun, too, has a tidal effect on Earth, but because of its great distance it is responsible for only about one-third of the range in tides. When the Earth, Moon and Sun are aligned (at full or new Moon), tides can be unusually dramatic, on both the high and low ends. When the Moon is at a 90-degree angle to the Sun in our sky (at first quarter or last quarter) tides tend to be mellower.

[D H]

Sorry but all that "gubbins" is way above my head.

That is some fairly simple math.

 

In short, the tidal force is

• Inversely proportional to the cube of the distance (rather than the square of the distance, as is the case with gravity), which is why the Moon has a greater tidal effect than does the Sun,
  
• Directed away from the center of the Earth at the two points on the Earth's surface intersected by the line connecting the Earth and Moon centers of mass,
  
• Directed toward the center of the Earth anywhere on the great circle defined by the plane that contains the center of the Earth and normal to the Earth-Moon line.

 

 

 

If you have a rubber ball, and one side gets pushed out, the rest of the ball will deform. Think of squashing the ball between your hand and the floor. The area around the equator gets expanded, and the area around the poles moves closer to the center.

Kind of like that, but in reverse. Think of it as attaching strings to two opposite points on a rubber ball. Those opposite points will define a great circle on the ball -- the set of all points halfway between the two points. If you pull on the strings, and squeeze on the great circle you will get a picture of what the Moon is doing to the Earth.

[Gareth56]

I love these forums

 

I love these forums

 

If the maths isn't too complicated or the explanation not too involved what does this mean?

 

"It can actually be thought of as a negative force, in essence, pulling water away from the Moon and away from Earth's surface -- a second high tide.

[swansont]

If the maths isn't too complicated or the explanation not too involved what does this mean?

 

"It can actually be thought of as a negative force, in essence, pulling water away from the Moon and away from Earth's surface -- a second high tide.

 

Relative to the moon, the force on the near side (N) is greater than the force on the earth's center © , which is greater than the force on the far side (F).

 

[math]F_N>F_C>F_F[/math]

 

In our frame, we feel no force from the moon, so subtract F_C from everything.

 

Now F_N is positive and F_F is negative

[SkepticLance]

I have always thought of it in terms of orbits. Think of the Earth as orbiting a point in space a little in the direction of the moon. Now think of the Earth as 3 parts.

1. Inner bit facing the moon.

2. Middle bit, making up most of the Earth.

3. Outer bit facing away from the moon.

 

 

If you change the speed of something in orbit, it changes its orbital position. If a satellite orbiting the Earth is slowed down, it falls towards the Earth. If a satellite is sped up, it moves outwards.

 

The inner and outer part of the Earth tied together, and are forced by the Earth's gravity to travel at the same speed. This slows down the inner part, and speeds up the outer part, more than a natural orbit.

 

The fact that the inner part travels too slow means it 'falls' towards the point in space that the Earth orbits - just as a satellite that is slowed down falls towards the Earth. The outer part is going faster than it should in pure orbit, which creates a centrifugal action pulling it outwards, just like a satellite being sped up moving outwards.

Edited December 4, 2008 by SkepticLance

[granpa]

the earth orbits the center of mass of the earth moon system. the centrifugal force and the pull of the moon on the earth cancel exactly at the center of the earth and roughtly speaking at the poles. elsewhere they dont.

 

btw. the math isnt as hard as you might think. many factors cancel out.

[D H]

Relative to the moon, the force on the near side (N) is greater than the force on the earth's center © , which is greater than the force on the far side (F).

 

[math]F_N>F_C>F_F[/math]

 

In our frame, we feel no force from the moon, so subtract F_C from everything.

 

Now F_N is positive and F_F is negative

 

Just to add: Both of these relative forces, F_N-F_C and F_F-F_C, are directed away from the Earth.

 

========================

 

The high tides typically do not occur when the Moon is directly overhead (or directly underfoot). The ocean tides are fairly complex. They vary *a lot* based on terrain (e.g., the Bay of Fundy). Some places have diurnal tides, other places have just daily tides, and yet others have something in between (two high tides, but one is much higher). On average, the tidal bulge leads the Moon by a bit because the lunar day is 24 hours and 50 minutes long. This phase difference in turn is what causes the Earth rotation rate to slow down a bit (one solar day is now 86400.002 seconds long; it was 86400 seconds long in 1820 or so) and what causes the Moon to retreat from the Earth.

[granpa]

http://www.coas.oregonstate.edu/research/po/research/tide/index.html