The whole Bill O’Reilly thing with the tides has provoked a question I haven’t thought about in a while. So I know that the moon and the sun, mostly the moon, cause the tides on earth. I have no problems with that. But, what I never got is why there are two bulges with the tides. The Wikipedia has not been very helpful. So, why are there two bulges? And, now that I think about it, why does the tidal force vary with the cube of distance and not the square of distance, like regular gravity? What’s the deal with that?
Consider this: the gravitational force of the sun on the earth is greater than that of the moon. However, the tidal force caused by the moon’s gravitational field is greater than that of the sun. How can this be?
Tidal force is the difference between the gravitational force on one side of an object and that on the other. The diameter of the earth is quite small relative to its distance from the sun, so the gravitational force of the sun is not that different between the near side and the far side of the earth.
By comparison, the diameter of the earth is an appreciable fraction of its distance from the moon, resulting in a bigger difference between the gravitational force experienced on the near and far sides of the earth.
Tidal forces tend to stretch objects in the direction radial to the gravitational source. If you imagine the earth as a big blob of liquid in the moon’s gravitational field, it would be elongated along an axis towards (and away from) the moon.
Even though the earth is mostly solid, the same principle holds true.
The Sun and the Moon appear to be almost exactly the same size, viewed from Earth’s surface. The reason the Sun doesn’t create the same tides is because the Sun is a lot less dense than the Moon. The Moon is a ball of rock. The Sun is a mass of incandescent gas plasma. Gas is less dense than rock, so even though the Sun as the same apparent diameter it has a lot less gravitational pull.
Not so. If I’ve got my sums right the gravitational pull of the sun on the earth (and vice versa!) is 3.5E22 N and the gravitational pull of the moon on the earth is 2E20 N.
In other words, the sun’s gravity is over a hundred times stronger than that of the moon.
This is how I had it explained to me, and it made so much sense it’s stuck with me all this time:
Imagine the earth as one big solid lump, evenly covered with a thin layer of water. Now put the moon in place. The water closest to the moon is going to be pulled toward the moon (slightly, of course, but let’s ignore size of the effect for now). This causes the water to bulge up right under the moon. With me so far? That explains one tide.
Now, however, you have to consider the fact that the moon is also pulling on the earth. The earth itself is tugged toward the moon. The tug isn’t as hard as the tug on the “close” water, because the distance from the moon to the close water is less than the distance from the moon to the center of the earth. And, most importantly, the earth is pulled harder than is the water on the far side of the earth for the same reason. The net result is that the earth is pulled out from under the water on the far side, leaving a similar bulge over there. Hence two tides.
An illustration would be very helpful here, but I don’t have one.
First I’ve heard of it, but here’s my guess. What is the rate of change of the gravitational pull with respect to distance? It’s the derivative of something proportional to 1/x^2, so it’s proportional to 1/x^3.
Lemur is correct, if you interpret “gravitational pull” in his last sentence to mean “tides”. The tidal force exerted by an object is proportional to the cube of its angular size times its density (which you can confirm yourself by working through the math). Since the Sun and Moon have the same angular size, the ratio of their tidal forces is the same as the ratio of their densities.
So, let me rehash this to see if I’m getting it right. The moon is pulling on the earth, and the two would crash into each other except that they’re orbiting each other. Thus the whole earth is affected by the moon’s gravity. But, because the near side of the earth to the moon has that much more gravity, the near side is affected more, and because the far side of the earth has less gravity, it kind of lags the earth as a whole.
Here’s my attempt at a crappy diagram:
O −> o
But with regard to the whole, the “sides” of the earth get different gravity, so we get
−O+ −> o
which leads to
(O) −> o.
And the same thing is happening on the moon, too, I presume. Except the moon’s bulge is roughly fixed since the same side of the moon always faces the earth. Is this also due to tides?
Ah. I was wondering exactly that, and it makes sense. Does this mean that the bulge on the opposite of the earth is a bit smaller than the one closest to the moon? My math seems to say yes, but I’m not sure. It would seem then that the difference in the strength of gravitational forces is exactly what the tidal force is.
The tidal distortion is not completely symmetrical: The side facing the Moon is slightly “pointier” than the side away from the Moon (in the extreme case, the bulge would have a conical cusp on the side facing the other body). In practice, though, the tides in the Earth-Moon system are small enough that the asymmetries are negligible.
It’s easier to picture why there are 2 tidal bulges if you picture it from the moons point of view first. There are 2 forces acting on the moon: gravity pulling the moon towards the Earth, and centripetal acceleration (Centrifugal force if you prefer) pulling the moon away from the Earth. The only point on the moon where these 2 forces are in harmony is the moons center of mass. The portion of the moon that is closer to the Earth than the center of mass experiences a greater pull towards the Earth than the rest of the moon, therefore it has a tidal bulge facing the Earth. The portion of the moon further from the Earth than the center of mass experiences greater centripetal acceleration pulling on the moon, therefore it bulges away from the Earth.
The Earth / Moon system are both orbiting around a central point called the barycentre. The Earth also experiences a gravitational pull from the moon which is what causes a tidal bulge facing the moon, and centripetal acceleration is what causes the tidal bulge on the opposite side of the planet. The Earth moves a lot less than the moon, but it’s still moving and subject to the same forces.
And note the same forces are slowing down the rotation of the Earth with respect to the orbit of the moon. Since the Earth is much more massive than the moon this will take quite some time to finish.
Of course, the Moon currently has no tides, even though the Earth’s gravity is much stronger than the Moon’s. Or rather, the Moon has much larger tidal bulges, on the exact center of the near side of the Moon and the exact center of the far side of the Moon, it’s just that those points are stationary on the Moon (except for tiny wobbles). The immense tidal force the Earth puts on the Moon has slowed its rotation until it rotates exactly once every revolution. The Moon’s tidal force is slowing the rotation of the Earth too, it’s just that the tidal force is a lot lower and the Earth has a lot more angular momentum. Back when the Earth was first formed, the day was only about 9 hours and has slowed to 24 hours over 4 billion years.
Given enough time the Earth’s rotation could slow so much that a day lasts a month, and then the Moon would be stationary in the sky. However, the Moon is receding from the Earth, which means that its tidal force on the Earth is lowering. So long before a day equals a month, the Sun will have expanded to a red giant and the Earth-Moon system will be swallowed up.
As an aside, you know how, when you were first learning astronomy, you learned that there was no significance to the “empty focal point” of an elliptical orbit? Well, a tidally-locked object like the Moon doesn’t actually keep the same face pointed at the Earth: It actually keeps the same face pointed at that empty focal point, due to those small wobbles.