The classic tide diagram looks like this. Why doesn’t it look like this?
The force from the moon is weaker on the farther side of the earth, but that doesn’t mean that there is an *opposite *force. Why doesn’t the moon’s gravity also pull the water on the far side towards it at least somewhat? It seems that bulge is even in defiance of earth’s own gravity.
I’m on vacation watching the tides in the St. Lawrence river and got to wondering about the physic of how tides work. I knew the bare-bones basic but I started doing some searching on it to find more detail. I have read a couple of descriptions of how tidal forces work but haven’t seen a good explanation of this aspect. IANAP but have a good understanding of mechanics and can read an equation.
Yes it’s an interesting question and I suspect most people don’t even know there is a tidal bulge on the opposite side of the Earth.
The answer is that the sea on the side opposite to the Moon has inertia. Accordingly it doesn’t want to carry on circling the Earth but instead tries to move in a straight line away into space.
Earth’s gravity is too strong to allow the sea to escape but it has a darned good try and bulges slightly.
On the other side, the sea is dragged into a bulge by the Moon’s gravity which is relatively weak. But again, the Moon has a good try at getting the Earths water so a bulge occurs.
This answers your question, in detail, with animations, but you may have to be an astrophysicist to understand it. In contrast, Cecil’s article doesn’t seem to cover it adequately.
The short answer is that, in effect there is an opposite force on the water, despite your categorical statement that there is not. If you take the force at the center of the earth as a reference, the much weaker pull at the “far side” amounts to a net force (when you subtract out the average at the earth’s center) away from the moon or sun.
Tidal forces are all due to variations in gravitational force with distance, the gradient of the force. This gradient gets larger the closer you get to an object, so the tidal force due to the relatively small moon are comparable to the tidal forces from the (much larger, but much farther away) sun.
This explanation is also given in my undergraduate classical mechanics textbook. So you only need about two or three years of undergraduate physics education to understand it.
Edit: I think my textbook’s calculations are clearer than this website’s. If interest persists, I can try to write up an explanation tonight. However, there are some actual physicists on this board, so I hope that one of them will give an adequate explanation and spare me the humiliation of forgetting a minus sign in my explanation.
Try it this way. The Moon does not orbit the Earth, they both orbit their common center of mass. Like a hammer thrower and the ball is connected by a chain, the Earth and the Moon are connected by gravity and they both wobble around. But unlike the chain gravity works everywhere and it works differently depending on distance so all object so bound are deformed. The softer they are, the more they are deformed, so the hard Earth and Moon are deformed just a little, while the soft surface oceans are deformed a lot. The Moon pulls more on the near side oceans than it pulls on the Earth, and more on the Earth than the far side oceans, so we get a bulge on both sides.
As the Earth rotates the bulge stands “still”, and we get high tide twice a day in the open ocean. And by “still” I mean, rotates with the orbiting moon completing a circuit of a hypothetical non-rotating Earth once a month.
I don’t think this works. If you held the Earth & Moon stationary with respect to each other (thus removing any inertial forces on the sea) you’d still have both tidal bulges.
Quite right, though it’s probably worth noting that this common center of mass (the barycenter) lies within the Earth (about 73% of the way from the center to the surface).
Depends on how you’re holding them in place. In fact I challenge you to come up with a way to hold them stationary that gives you the same tidal bulges.
If, however, they are free falling towards each other and you measure the bulges just as they pass the separation they have now …
Have you heard of the concept of objects being “spaghettified” as they get very close to a massive object such as a black hole? This is an extreme example of tidal forces at work. Tidal forces cause objects to stretch out in the direction of the gravitational field.
Now imagine an “earth” composed entirely of water? Can you imagine this watery sphere being stretched out by the gravitational field of the moon?
The effect is similar if you imagine a small rocky core at the heart of this water-world. The rocky core will be remain at the centre, so that the water forms a “bulge” on both sides. And that’s still true if you expand the rocky core until you have 8000 miles of rock and only one mile of water.
I’m sure **naita **knows that. The point is that if you somehow rigidly held them in place, you would only get one bulge. The reason you get two bulges is because the Earth and Moon are in mutual freefall. This requires (even in thought-experiment land) that they are orbiting each other (or that you get a one-shot measurement just before they collide).
This appears to be the key concept, and one that for some reason I am having trouble grasping, even after reading the excellent treatment that Xema linked.
Yes, but I always thought you would get ripped apart standing on a black hole because it was pulling on your feet so much harder than your head, not because your head was being pushed in the opposite direction.
I don’t think this is true - why would it be? Under the influence of a nearby decently massive body, another body is elongated (ovalized?) by tidal forces. I see no reason why this would not be the case if they were held at a certain distance from each other, in an inertial frame of reference.
Picture a hammer thrower with a big weight in front (The Hammer and the Sea) spinning around with a small weight (Just the Sea) on the back. Both weights get thrown out.
Fluids move such that they form an isosurface with gravitational potential.
Suppose you fix a rigid, gravitating sphere in place, and provide another distant gravitational force with 1% of the strength of the first. Then, objects on the “near” surface will feel 99% gravity (since they cancel), while objects on the far side feel 101% gravity.
If there were a bulge on the far side, there would be a problem–a small unit of water on the surface (say, of a 1 km depth ocean) has more potential energy than one on the near side. Just like a ball on a hill, it should “roll” downward, converting potential to kinetic (and ultimately heat) energy.
In math terms, we have:
PE=mgh
mg[sub]near[/sub]h[sub]near[/sub] = mg[sub]far[/sub]h[sub]far[/sub]
g[sub]near[/sub]h[sub]near[/sub] = g[sub]far[/sub]h[sub]far[/sub]
(9.8 m/s^2 * 99%) * 1010 =~ (9.8 m/s^2 * 101%) * 990 m
So in the steady state, the depth on the near side will be (approximately) 990 meters and on the far side 1010 meters. Between those two points we get a smooth interpolation.
The water is still “ovalized”. But the planet is closer to one end than the other. That’s not the case with real tides, where the CG of the planet follows a freefall trajectory.
