Tides due to the moon vs. the sun?

[nautica]

Tides due to the moon vs. the sun?

The acceleration of gravity on the Earth due to the sun is 177 greater than gravity on Earth due to the moon.

Why are the tides predominantly due to the moon and not the sun, in spite of this number?

Nautica

 

[mathman]

I can't give you any numbers, but the moon is so much closer that its effect on the oceans is much larger than that of the sun.

 

[russ_watters]

Originally posted by mathman
I can't give you any numbers, but the moon is so much closer that its effect on the oceans is much larger than that of the sun.

To expand, tides are due to the DIFFERENCE in gravitational force between the sun/moon and the near and far sides of the earth. Since the sun is so far away, the near and far sides of the Earth are very close to the same distance from the sun - so not much tidal force.

 

[Janus]

Originally posted by russ_watters
To expand, tides are due to the DIFFERENCE in gravitational force between the sun/moon and the near and far sides of the earth. Since the sun is so far away, the near and far sides of the Earth are very close to the same distance from the sun - so not much tidal force.

To expand further, tidal force falls of by the cube of the distance while g-force falls off by the square of the distance.

Thus, the sun is 26730000 times more massive than the sun and 400 times the distance.

therefore its g-force on the Earth is 26730000/400² = 167 times that of the moon's, but its tidal effect is 26730000/400³ = 0.4 times, or a little less than 1/2 that of the moon's/

 

[nautica]

Originally posted by mathman
I can't give you any numbers, but the moon is so much closer that its effect on the oceans is much larger than that of the sun.

The sun's effect is only 1/2 that of the moon.

But the 177 x's is based on how close the moon is to the Earth relative to the distance of the sun to the Earth - distance has already been taken into consideration. So, I do not see how distance from the sun to the Earth vs the moon to the Earth makes a difference in the effect of the tides.

Thanks
Nautica

 

[Janus]

Originally posted by nautica
The sun's effect is only 1/2 that of the moon.

But the 177 x's is based on how close the moon is to the Earth relative to the distance of the sun to the Earth - distance has already been taken into consideration. So, I do not see how distance from the sun to the Earth vs the moon to the Earth makes a difference in the effect of the tides.

Thanks
Nautica

The near side of the Earth is 12756 km closer to both the sun and the moon than its far side.

The sun is 149,000,000 million miles away from the Earth so the distance difference is about .00856 %. This means that the difference in force across the Earth varies by about .017%

The moon is 384000 km away, so the % distance difference is 3.3%, and the force difference is about 6%

If the acceleration due to gravity of the sun is 167 time that of the moon, then its tidal effect (the difference in force between far and near side of the Earth) is 167 * .017% = 2.839%, when compared to the moon's, or again, about 1/2.

 

[Hurkyl]

Tidal forces aren't caused by the strength of gravitational attraction; they're caused by the differences in gravitational attraction at different points on an object.


Here's an exercise for you.

We know the mass of the sun is 1.99 * 10^30 kg.
The mass of the moon is 7.35 * 10^22 kg.
The Earth's radius is 6.38 * 10^6 m.

(assume all of these values are exact)

At a particular point in Earth's orbit, its center is 1.50 * 10^11 m from the center of the sun. Compute the acceleration due to the sun's gravity at the point of the Earth nearest to the sun, and the acceleration at the point of the Earth farthest from the sun.

The tidal stress on the Earth due to the sun is (proportional to) the difference of these two accelerations.


At a particular point in the moon's orbit, its center is 3.84 * 10^8 m from the center of the earth. Do the same calculation as you did for the sun. You'll find the tidal stress caused by the moon is greater!

 

[nautica]

Is that not what I did, when I determined that the pull of the sun was 177 times greater than that of the moon?

Thanks
Nautica

 

[Janus]

Originally posted by nautica
Is that not what I did, when I determined that the pull of the sun was 177 times greater than that of the moon?

Thanks
Nautica

No, all you did was calculate the pull of each body on the center of the Earth.

The near side to the sun is 149,000,000 - 6378 km = 148993622 from the sun, and the far side is 149,000,000 + 6378 km form the sun. You need to calculate the acceleration due to gravity at each of these distances and the sun's mass, and then take the difference between the two to get the net force acting across the Earth which causes solar tides.

For the moon, the distances are 384000 - 6378 = 377622 km and 384000+6378 = 390378 km.

If you calculate the net difference for these distances, using the moon's mass, you will find that it is twice as great as that of the Sun.

 

[Loren Booda]

Tidal forces are important in relativity's spacetime, where gravitation acts not only radially but angularly between finite masses.

 

[nautica]

Nice, I am a little slow, but have now figured out what you are saying.

Thanks
Nautica

 

[nautica]

I did what you said and it appears that the difference of gravitational pull on each side due to the moon is 1.068678428 x's stronger on the near side than the far side.

The sun is 1.000170496 greater on the near side than the far side. So how does this account for the fact that the sun only has half of the effect when the numbers show it should have 93% of the affect.

Thanks
Nautica

 

[russ_watters]

You didn't account for the difference in gravitational force - the number you posted in the beginning which got you confused in the first place. That number is how you relate the actual tidal force caused by each.

So convert those two numbers to percent (chop off the 1 then multiply by 100%), then multiply the sun's by the difference in gravitational force.

 

[Hurkyl]

Remember I said compute the difference, not the ratio.

 

[nautica]

The difference in the pull of the moon is 2.20 x 10^-6

The difference in the pull of the sun is 1.01 x 10^-6

Which is .457 or roughly 1/2.

You guys are unbelievable.

Thanks
Nautica

 

[Anne-ke]

Thanks people!
This information was really useful to me!

BUT: WHY is it the DIFFERENCE between forces on the far and the near sides of the Earth that determines the tides??

Would there not be tides when these forces were (theoretically) equal?
I thought the tides are determined by the difference between the gravitational force and the centrifugal force. And that the differences between the far and the near sides of the Earth arise because of the difference in direction of the centrifugal force on both sides...

Can somebody explain me why??

 

[Kalimaa23]

I think this ties in with your question.

http://192.211.16.13/curricular/astro2001/projects/luna/tides/tidecause.htm

By the way, Anne-ke? Strange name, you wouldn't happen to be Belgian would you?

 

[russ_watters]

Originally posted by Anne-ke
BUT: WHY is it the DIFFERENCE between forces on the far and the near sides of the Earth that determines the tides??

Can somebody explain me why??

"Why" gets to be kinda a fuzzy question: there was a force that was useful and it needed a name.

Adding the forces does give you a useful number, its the total force on the object as a whole. But the difference is important too because the difference gives you the internal forces on the object. It may be hard to visualize how two forces pulling in the same direction can result in a force pulling an object apart, but the reason is the internal forces on an object must be consistent. So if you apply two forces to an object, there must be a resultant internal force.

Would there not be tides when these forces were (theoretically) equal?

Correct.

I thought the tides are determined by the difference between the gravitational force and the centrifugal force.

Since centrifugal force is equal on all sides (along equal lattitude lines), there is no force imbalance with which to create a tidal force. Centrifugal force does create the bulge at the equator, similar but not quite the same as a tidal bulge.

 

[Nereid]

Welcome to Physics Forums Anne-ke!

Assuming you're now clear on the magnitude, and cause, of tides, and are ready to dive into some more fascinating aspects of the Earth-Moon system ...

Consider:
- the Earth's rotational axis is tilted ~23.5^o to the plane of its orbital motion about the Sun
- the Moon's orbit in inclined ~5^o to the plane of the Earth's equator
- the Moon completes one revolution about the Earth in a time which is much greater than the time the Earth takes to spin on its axis (the Moon's 'year' is longer than the Earth's 'day')
- neither the Earth's orbit around the Sun (actually the Earth-Moon's centre of mass orbit) nor the Moon's orbit around the Earth are circles

... and you've got a lot of extra forces that the 'point masses in circular orbits' ideal case doesn't account for!

It all leads to some really good-fun maths ... and profound long-term effects on the Earth's climate (among other things).

 

[Loren Booda]

Nereid,

Do these take into account precession or nonlinearity?

 

[Anne-ke]

That's an interesting point!
I understand (more or less) your explanation about the internal force. I can accept that such a force exists and that it has an effect. But in which direction?

So this internal force is the driving force for the tides... Why is direct gravity from the moon not important? I could imagine that when there would be no difference in force between the 2 sides, the moon still exerts a force on the water. Apparently this force is not important (otherwise the sun would have more tidal influence...), but I don't understand why? Same with an Earth as an infifnite point with the same mass... the moon still pulls the water to one side, leading to tides, I thought?? Or do I think wrong??

Another thing that I do not understand is why poeple use the Law of Newton for gravity with R square, and for tides with R to the power 3? Where does that 3 come from?

About the centrifugal force: I thought it is not equal at all points on earth, because Earth is not spinning around its axis, but around the common center of mass of Earth and moon. Because this point is located in the Earth on the side of the moon, the centrifugal force is bigger on the far side of the earth, then on the near side. Adding the gravitational and the centrifugal forces on both far and near sides, gives en equal force on both sides and thus equilibrium. Is this correct??

Anne-ke

PS: I'm not Belgian, I'm Dutch!

 

[russ_watters]

Anne-ke wrote on 02-25-2004 08:47 AM:
I can accept that such a force exists and that it has an effect. But in which direction?

An internal force is the 2d equivalent of pressure: pressure at a point exists in all directions at the same value. An internal force in a beam (or an object with two forces along the same axis) exists in both directions and is simply either a tension or compression.

Trick question (demonstration) from a physics lab I had: Two kids are standing on skateboards, holding a rope between them. One is the biggest guy in the class, the other is the smallest girl. They are told to pull toward each other as hard as they can. Who is pulling harder? Answer: tension is tension and they are both pulling with the same force.

Why is direct gravity from the moon not important? I could imagine that when there would be no difference in force between the 2 sides, the moon still exerts a force on the water. Apparently this force is not important (otherwise the sun would have more tidal influence...), but I don't understand why?

The force is important - its what keeps the moon and Earth in orbit around each other. You're seeing that the moon exerts a force on the water, but missing that the moon is also exerting a force on the earth and those forces are doing exactly the same thing: accelerating the Earth toward the moon.

Perhaps another analogy: freefall. An astronaut doesn't fall toward the side of the space shuttle because the astronaut and the space shuttle are both in freefall and accelerating toward the Earth at the same rate. That's exactly the situation with the water on earth.

Another example: flying in formation in orbit. The water bulges on the near and far sides of the Earth are essentially flying in formation with each other but as a result, the water on one side is below orbital velocity and the other is above it.

Picture 3 space ships in orbit around the earth, all right next to each other. Two of them separate from the middle and fly exactly parallel to the center one, but several miles apart, one closer and one further from the earth. What happens? Tracing out a bigger circle, the further one out is now flying faster to keep up with the leader and flying in a smaller circle, the lower one is flying slower than the leader to keep in formation. But that's a problem: they are no longer flying at the right orbital speed. The outer one is flying too fast and unless it fires its engines constantly it'll spiral away from the earth. The inner one is flying below orbital speed for its altitude and will spiral into the Earth unless it fires its engines constantly. The force the engines must apply to keep the spaceships in orbit is the tidal force.

Another thing that I do not understand is why poeple use the Law of Newton for gravity with R square, and for tides with R to the power 3? Where does that 3 come from?

Its a mathematical relation resulting from the fact that it isn't just the distance that matters but the difference in distance. Basically, its two equations combined. The % difference in distance decreases linearly with average distance while the total force decreases as a square function of average distance. Combine a linear and a square function and you get a cubic one.

About the centrifugal force: I thought it is not equal at all point on earth, because Earth is not spinning around its axis, but around the common center of gravity of Earth and moon. Because this point is located in the Earth on the side of the moon, the centrifugal force is bigger on the far side of the earth, then on the near side. Adding the gravitational and the centrifugal forces on both far and near sides, gives en equal force on both sides and thus equilibrium. Is this correct??

The Earth rotates around its axis and revolves around the center of gravity (COG) of the earth-moon system. You don't want to mix those up because you can use the rotation of Earth separately to calculate the equatorial bulge of the earth. The tidal bulge is in addition to the bulge created by rotation.

 

[Loren Booda]

The distortion of solid Earth (by tidal forces) has some nonzero magnitude. Does this distortion stay in step over Earth's crust with that of its waters?

 

[russ_watters]

Originally posted by Loren Booda
The distortion of solid Earth (by tidal forces) has some nonzero magnitude. Does this distortion stay in step over Earth's crust with that of its waters?

Since water flows, the tides lag a little. And since the Earth is a solid, it doesn't actually stretch much due to tides. Is this what you were asking?

 

[Loren Booda]

Yes. I assume the lag is constant. What would its value be (in time or degrees), and does this difference show any significance in observable phenomena?

 

[Nereid]

To answer the earlier question - coastlines do matter, esp north-south ones, and bays. I don't recall how much they matter, in terms of deviations from a homogeneous Earth, or one with only oceans.

You can get an idea of how big the effect should be - OOM - by working out a) the average water tide, b) the average land tide, c) the difference in gravitational field (at a distance of the average orbit of the Moon) between an ocean covered Earth and one with no ocean, then d) put in various lags.

To see how big the lag is, consult your favourite tide table, and compare the times of local high and low tide with the times of meridian passage of the Moon. Of course, there are several adjustments that you'd need to make ...

 

[selfAdjoint]

In some places coastline effects are large enough to reduce the normal two high tides a day to one.

 

[Anne-ke]

Yes, semi-diurnal tides can be reduced to diurnal tides. For example in the South China Sea (see picture attach).

This is a very complex phenomenon, in which many factors like coastal configuration and ocean depth play a role. But what is the basic principle? Is it that the tidal bulge moves over the Earth (for example the Pasific Ocean), then meets a coast line (like the Philipines), and has to push itself through the straits as through a funnel, leading to reduced tides??

If that is the case, then why does not the whole of the South China Sea have diurnal tide? Why are there these little red spots where tides are semi-diurnal (see attach)? And why on that locations?

And, one last question about the forces: If centrifugal force is equal on all points on Earth and the gravitational force is different on the near and far side of the earth, why is then the resultant tidal force still equal on both sides, leading to tidal bulges with the same height (is that so? are they of equal height?)? Or is that the story of the boy and the girl on the skateboard? Are actually the gravitational forces equal on both sides?

 

[selfAdjoint]

And, one last question about the forces: If centrifugal force is equal on all points on earth

It's not. Centrifugal force is proportional to the distance from the axis of rotation. It's a maximum at the equator, zero at the poles, and has a unique value for each circle of latitude.

and the gravitational force is different on the near and far side of the earth, why is then the resultant tidal force still equal on both sides, leading to tidal bulges with the same height (is that so? are they of equal height?)?[/b]

Yes, the bulges are the same height. On the near side to the moon the water is closer to the moon than the Earth and is more strongly pulled, hence it is pulled up into a bulge. On the far side the Earth is closer to the moon than the water, so it is pulled away from the water, making a water bulge. Tidal forces are all about differential effects of gravitation at different distances.

 

[Sariaht]

Huston we have a problem. We have landed on a very big planet with very strong winds, They are called solarwinds. Have we forgotten something? I get a strong feeling we have. Perhaps the sun is a bit warm.

-Keep searching for intelligent life. Aaouch, sun of a star.

 

[Shukie]

Yes, the bulges are the same height. On the near side to the moon the water is closer to the moon than the Earth and is more strongly pulled, hence it is pulled up into a bulge. On the far side the Earth is closer to the moon than the water, so it is pulled away from the water, making a water bulge. Tidal forces are all about differential effects of gravitation at different distances.

Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the Earth and moon revolving around their centre of mass?

 

[sylas]

Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the Earth and moon revolving around their centre of mass?

It's not centrifugal force; that has nothing to do with it.

The answer you have quoted is correct and it explains the bulges on both sides and why they are about equal. Read it again. It says: Tidal forces are all about differential effects of gravitation at different distances. It stretches out a fluid sphere into an ellipsoid shape. Relevant to the center of the sphere, you get bulges on each side. There's a difference between the Moon's gravity at the center of the Earth and the nearside bulge, and there's about the same difference between the Moon's gravity at the center of the Earth and the far side bulge.

 

[billiards]

I found this to be a very insightful and interesting undergraduate thesis on tides by a world-class student (currently I think a PhD student at Caltech).

http://www.gps.caltech.edu/~carltape/research/pubs/thesis/tides_CHT_thesis.pdf" (6.7 MB)

 

[D H]

Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the Earth and moon revolving around their centre of mass?

That is a common but completely erroneous explanation. First off, what centrifugal force?

Secondly, and more importantly, suppose an object of the Moon's mass is falling straight toward the Earth with zero tangential velocity. There is no centrifugal force in this case; the object is falling straight into the Earth. At the point in time that this object reaches the Moon's orbital distance the tidal forces will be exactly equal to those exerted by the Moon. The tidal forces exerted by the Moon has nothing per se to do with Moon's orbit. It is solely a function of where the Moon is.

So, how to explain that there are two bulges? The answer is simple. The Earth as a whole is accelerating gravitationally toward the Moon:

$$a_e = \frac {GM_m}{{R_m}^2}$$

where a_e is the acceleration of the Earth toward the Moon, M_m is the mass of the Moon, and R_m is the distance between the centers of the Earth and the Moon.

The distances between the center of the Moon and the points on the surface of the Earth directly opposite the Moon and directly between the Earth and Moon are R_m+r_e and R_m-r_e, respectively. Here r_e[/i] is the radius of the Earth

Because these points are a bit further from/closer to the Moon than the center of the Earth, the acceleration toward the Moon at these points will be slightly different than that of the Earth as a whole. In particular,

$$a_p = \frac {GM_m}{(R_m\pm r_e)^2}$$

It is the difference in acceleration that is important here. Calculating this,

$$a_{p,\text{rel}} = \frac {GM_m}{(R_m\pm r_e)^2} - \frac {GM_m}{{R_m}^2} \approx \mp\, 2\,\frac{GM_mr_e}{R_m^3}$$

The point on the surface of the Earth directly between the centers of the Earth and the Moon experiences a bit more acceleration toward the Moon that does the Earth as a whole, so this differential acceleration is directed toward the Moon -- that is, away from the center of the Earth. The point on the surface of the Earth directly opposite the Moon experiences a bit less acceleration toward the Moon that does the Earth as a whole, so this differential acceleration is directed away from the Moon -- which is, once more, away from the center of the Earth. The Moon pulls the submoon point and its antipode apart, like a piece of taffy. For the points on the surface of the Earth where the Moon is at the horizon, the action is to squeeze those points inward a tiny bit.

For the Moon these tidal forces are very, very small. For a neutron star or black hole they are not. There is a cool technical term for how black holes pull the extremes of some object apart and squeeze in at the middle: http://en.wikipedia.org/wiki/Spaghettification" .

 

[russ_watters]

Yeah, I know it's an old topic, but something bugged me about this answer. I have never heard this explanation for the second water bulge. I thought the second one was created due to the centrifugal force that arises from the Earth and moon revolving around their centre of mass?

The easiest way to deal with this is to get rid of the rotational motion altogether. Without the moon being in orbit and the Earth rotating, the tides would still be there (for a little while, anyway - until the moon crashed into the earth!).