See subject. Apologies if this an oft-told query.
Leo
Sure, but so tiny as to be negligible (and probably completely undetectable).
Yeah but you can counteract them by turning around 180 degrees.
By standing on your head? But then the tides would operate in the other direction, relative to your body.
Keep spinning - it’s what I do.
Tidal flow of the oceans is not really a direct pull on the water so no, not really.
Tides are more like a sideways pull, the lunar gravity causes what is a best thought as a slight “tilt” of the horizon, but water is very slick and runs “downhill”
Think of putting a cookie sheet perfectly level in the sand, now fill it to the brim with water. It does not take much of a tilt for the water to move quite a bit.
Add in the millions and millions of gallons in an ocean and the cumulative effect is the tide.
The liquid in your body is not as free to move as ocean water nor is it nearly as large of an amount, if you “tilt” the liquid does not move much.
If our oceans were smaller, lets say in the form of millions of lakes you would not notice a tide.
Here is a fairly good example that shows how it is the “flattening” of the earth that “tilts” it in a way that the water flows sideways.
http://scienceblogs.com/startswithabang/2010/02/how_tides_work.php
Also, the fact that we’re made mostly of water isn’t much of an issue: there are lunar tides in the earth’s crust!
It’s worth mentioning that tidal forces differ as the inverse cube of the distance, as opposed to the inverse square law of gravitational attraction. This is why the Moon is more influential than the Sun on earth’s tides: the Moon is so much closer.
However, if you had some way to breathe and stay warm but were heading straight into a black hole, the tides on your body would pull you apart. It should be the only thing you actually feel regarding the black hole.
And ditto it doesn’t much matter how much water we contain in this discussion. The only way it would matter is that free liquids slosh much better than solids do, so we notice tides in liquids generally. Flesh is closer to solid than liquid in this sense.
No, the real reason the Moon causes greater tides than the Sun is because it’s much denser. The Sun and Moon appear about the same size in the sky, even though the Sun is much larger because the Sun is far away. But the Sun is a mass of incandescent gas, while the Moon is made of rock. If the Sun were made of rock like the Moon, it would cause tides as strong as the Lunar tides, and if the Moon were a ball of gas the same density as the Sun it would cause tides as weak as the Solar tides.
Tides only affect big things, since the whole idea of tides is that the gravitational field is different at different distances. The distance from one side of the Earth to the other is small compared to the distance from the Earth to the Moon, but it’s not quite so small as to be completely negligible: There is a small difference in the Moon’s gravitational field at those distances, and so we see small tidal effects. The distance from your head to your toes, though, is much, much smaller than the distance to the Moon, so the difference in gravitational field is, in that case, negligible.
Not true. Tides are caused by a ***difference ***in gravitational pull between the center of the Earth and the oceans at the surface. The farther away you get from the Earth, the smaller the ***difference ***becomes.
It’s the proximity, not the mass/density that is the overwhelming factor that causes the tidal forces from the Moon to be so much greater than those from the Sun.
Actually, TerpBE, both you and Lemur866 are correct; you’re just looking at it in different ways. Lemur866 is correct in saying that, given that the Sun and the Moon have the same angular size in the sky, their tidal forces are proportional to their densities. But the catch is that the reason the Moon is the same angular size in the sky as the Sun, despite being much smaller, is that it’s also much closer, which brings it right back to distance being important.
Nay, nay, you must turn 90 degrees! Remember that there are tidal bulges at the points of greatest and least gravitational attraction, so if you spin all the way around your Personal Tidal Bulges (PTB’s) are in the same place.
And of course, spinning only works when the moon is on the horizon, when the tidal axis runs through your body horizontally. When the moon is overhead or below ground, you have to stand up and lie down for a 90-degree tilt.
Now get those PTB’s moving!
You beat me to it, so I’ll just elaborate a bit.
Tides are proportional to r1(M/R^3) - where r1 is the size of the object feeling the tides, M is the mass of the object causing the tide, and R is the distance to the object causing the tide.
But this can be reexpressed as r1(d 4/3pi r2^3/R^3) - where d is the density of the object causing the tides, and r2 is the radius of the object causing the tides (presumed to be spherical).
Rearranging again you get r1d4/3pi(r2/R)^3 - and r2/R is the angular radius of the object causing the tides, as viewed from the object feeling the tide.
So when comparing solar and lunar tides on a person (or mountain) on the Earth, r1 is the same in both cases (because we’re considering the same person or mountain in both cases), r2/R is the same because the angular radius of the Sun and the Moon are the same when seen from Earth; the only factor that is different is the density of the Sun/Moon.
So a steel ball-bearing held at a distance from you such that it exactly blocks the Sun from your view exerts larger tides on you than the Sun or the Moon would - because it’s denser than either.
??? Why would density matter? I thought that both for tides and for orbital attraction, a sphere could be represented by a point-source at its center. Newton worked that out for orbits; isn’t it true for tides?
Yes, it’s true that you can represent a sphere by a point mass.
Density doesn’t matter by itself - it’s just that you can express the magnitude of a tidal force in terms of density (as I showed above). You can do the same thing for orbits (to figure out the period of an orbit that just skims the surface of a spherical object, all you need to know is the density of that object) - sometimes it’s convenient to do so, and sometimes it’s not.
Yes, it is. Lemur866 is making the interesting point that if two equally dense objects appear equally large, the tidal forces associated with those objects will be the same, even if one of the two is in fact much larger and further away than the other.
In other words, while we might conventionally relate tidal forces to mass and distance, Lemur866 is pointing out that we can equivalently say that tidal forces depend on density and apparent size.
Except this isn’t true. The overall gravitational force will be the same, but the tidal forces won’t be. Tidal forces are caused by the difference between gravitational force from one side of the object to the other. The diameter of the earth compared to the distance from the earth to the moon is much larger than the diameter of the earth compared to the distance from the earth to the sun. So, the difference between gravitational force due to the moon from the point on the earth closest to the moon to the point on the earth furthest from the moon, is much greater than the difference in gravitational force due to the sun from the point on the earth closest to the sun to the point on the earth furthest from the sun. Even though the overall gravitational force may be the same (or would be if the densities of moon and sun were the same) the tidal effects will be much higher for the closer object.
I think Lemur866 is correct, though I’d never thought of it that way before. Gravitational force decreases as square of distance, but tidal force is the gradient of gravitational force, so it decreases as cube of distance. And if you compare different objects with the same apparent size and same density, the mass goes up as cube of distance. So different objects with the same density and same apparent size will have the same tidal force.
The apparent visual area of an object decreases with square of distance, not cube of distance.