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Hi all! I was just wondering if anyone can explain me why the tides (created by moon effect) are created on opposite sides.
(image is on attachment)
Thank you.
Since that is an insignificantly small fraction of what is going on (not sure how small, but certainly less than 1%), it is best not to bring it up when someone is looking for the broad overview of tides....centrifugal force
What centrifugal force? In what rotating frame? Centrifugal forces only arise in rotating reference frames. You should be able to explain this phenomenon from any reference frame, and there is no centrifugal force in a non-rotating frame. Some phenomena (e.g., hurricanes) are much easier to explain in terms of a rotating frame than a non-rotating frame. This is not one of them.gravity and centrifugal force
In what rotating frame? Whenever you invoke centrifugal force you need to specify what your frame of reference is. There are no centrifugal forces in a non-rotating frame.gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit). there is an excess gravitational force on the moon side of the Earth and an excess centrifugal force on the opposite side of the earth.
at the center of the Earth the 2 forces are balanced.
This is the explanation I have found to be the most clarifying.The hump nearest the moon is caused by that ocean being pulled away from the earth. The hump furthest from the moon is caused by the Earth being pulled away from that ocean.
The Earth and Moon don't orbit in the rotating frame used to explain tidal effects from the perspective of centrifugal force. They are stationary in such a frame.gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit/rotate).
I don't know if you're beating him up excessively.The Earth and Moon don't orbit in the rotating frame used to explain tidal effects from the perspective of centrifugal force. They are stationary in such a frame.
I don't know if you're beating him up excessively.
That effect is so small as to be unimportant. Perhaps you could actually calculate its magnitude and compare it to the magnitude of the tidal force...gravity attracts the bodies toward one another and centrifugal force pushes them away from their center of mass (around which they orbit). there is an excess gravitational force on the moon side of the Earth and an excess centrifugal force on the opposite side of the earth.
at the center of the Earth the 2 forces are balanced.
The fact that someone in another forum said it means exactly the same as when someone (you) in this forum said it: they got it wrong. Not sure what the point of the other link was.http://www.scienceforums.net/forum/s...ad.php?t=36766 [Broken]
No, it is not an acceptable explanation. In the swinging bucket, if the bucket isn't swinging, the water falls to the ground. In the tides example, if the Earth and moon are not in orbit around each other, the tides are still there. What he is describing is not tidal force. It is a different effect entirely, and a miniscule one at that.I don't know if you're beating him up excessively.
Inasmuch as centrifugal force can explain to the layperson why water stays in a swung bucket on a string, surely this is also an acceptable answer for the tides.
Not that I think it's the best explanation...
If they do, they are wrong - as I said before, the tides exist even in a non-rotating system. What granpa is describing is not tidal force. Are there terms in the tidal force equation for location of the barycenter and period of rotation?I know some textbooks explain tides by centrifugal force with mass center of the two bodies.
No, it is not an acceptable explanation. In the swinging bucket, if the bucket isn't swinging, the water falls to the ground. In the tides example, if the Earth and moon are not in orbit around each other, the tides are still there. What he is describing is not tidal force. It is a different effect entirely, and a miniscule one at that.
So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?
The key is, if they were not rotating around each other, they would not be stationary relative to each other. They would accelerate toward each other. What would happen to the oceans in this case?So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?
Russ is exactly right. The tides exist, so one had better be able to explain them from any frame of reference. Some choices are good in the sense that the explanation is clear and concise. Other choices, such as a frame with origin at the center of Mars and rotating with Mars' rotation rate (1 revolution per 1.026 days), are not so good in the sense that the explanation is anything but clear and concise.If they do, they are wrong - as I said before, the tides exist even in a non-rotating system.I know some textbooks explain tides by centrifugal force with mass center of the two bodies.
From the link, a bit out of order,Here is what one of my profs has to say:
http://www.math.nus.edu.sg/aslaksen/teaching/tides.html
Personally I like tidal gravity (or gravity gradient) as the explanation. The explanation pops out clear and concise from the math, and this also explains why the squeeze at the middle. Words don't do as much for me as does mathematics.But i myself agree with " The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean"
Sorry, I'm not sure if I said it(i was thinking it) but the Earth and moon can't be held stationary with respect to each other: if a hypothetical support was constructed to keep them there, the regular gravitational force would swamp the tidal force and all of the water would rush to the side of the Earth that the moon is on, so it isn't good for visualizing the issue. So consider instead the scenario I described before:So, if the Earth and Moon were not rotating around a common center of gravity, and were stationary relative to one another, the oceans would still bulge out on the far side? What would cause this bulge?
There are two tidal bulges.Me said:The Earth has no moon, but a moon-sized asteroid is on a direct collision course with earth. Are there two tidal bulges or one?
Briefly, but not too briefly - if the object were traveling at 30,000 km/hr (reasonable for an object on a collision course with us), it would take hours for it to get here from a distance of the moon's orbit (it would accelerate during that time of course, so i don't want to try to calculate how long exactly). Plenty of time to observe and measure the tidal bulges as they both grow prior to the collision.You'd still get the bulge, on both sides. Briefly. Then the ocean would be vaporized in the collision.
From the link:
Which is what I said and I honestly intend to do the calculation if I become less lazy in the future... I'm curious to know, because this is a common misunderstanding and it has come up before.In fact, a simple computation shows that the centrifugal force caused by the Earth's rotation around the Earth-Moon center of mass is tiny compared to the gravitational differential.
I know some textbooks explain tides by centrifugal force with mass center of the two bodies.
But i myself agree with " The moon's gravity pulls the near-side ocean away from the Earth and the Earth away from the far-side ocean"
Unfortunately, you are wrong. For example, seeI'm quite sure textbooks don't call this "centrifugal force", rather they will work in the frame which is co-moving with the Earth's center of mass.
Simple. Why invoke a rotating frame, and why invoke any frame but an Earth-centered frame? The goal, after all, is to compute the acceleration relative to the Earth.In this reference frame, the Earth's surface is stationary (we don't need to consider the Earth's rotation around its axis) because the Earth is (to good approximation) a rigid body. The gravitational force plus the fictitious force yileds the acceleration in this frame of reference.
On the side of the Moon, the sum of the gravitational force and the fictitious force points in the direction of the Moon, on the opposite side it points away from the Moon.
From the book,But the explanation given in thet book is correct, the only thing is that he calls the fictitious force the "centrifugal force."
Here is what one of my profs has to say:
http://www.math.nus.edu.sg/aslaksen/teaching/tides.html
Rotation of an object means angular motion about an axis that passes through an axis internal to the object in question. The Earth rotates once per sidereal day. If you want to attack the issue of tides from the perspective of a rotating Earth, have at it. We ignore the daily rotation of the Earth to get a first-order approximation of what makes the tides arise.why would we 'ignore the rotation of the earth'?
Wrong. The shape of the Earth is, ignoring mountains and valleys, an equipotential surface, not a constant force surface. Mean sea level is an equipotential surface.as the Earth spins on its axis it produces an outward centrifugal force but over millions of years the whole Earth itself (not just the water) has bulged at the equator thereby cancelling out this force.
Wrong again. This bulge has nothing to do with orbital motion. As Russ mentioned earlier, an asteroid headed straight toward the Earth with zero angular momentum would still raise tides -- and on both sides of the Earth. The tides result from the gradient in the gravitational force, not from centrifugal force.most of the rotation can therefore be safely ignored. but not all of it. the fraction of its rotation equal to 1 rotation per month does not produce a regular equatorial bulge. it produces a bulge on the side facing the moon and a bulge on the opposite side away from the moon.
In between there is the opposite of a bulge. Tidal forces are directed inward at the points where the Moon is on the horizon.in between there is no bulge.
This is nonsense.hence to a first approximation when calculating tidal force we can treat the Earth as though it rotated once per month. and the centrifugal force (due to the Earth and moon orbiting each other) is proportional to distance from the earth-moon center of mass (around which they orbit). the centrifugal force at the center of the Earth (and roughly speaking at the poles) is equal to the gravitational force. to a first approximation the gravitational force can be considered to be a constant.