New answers tagged

4

My simple answer to this is that high and low tides are a bit arbitrary; they're the turning points of a continuous function over time. At the places you've mentioned, the (usually most significant) bi-daily tidal harmonic is "swamped out" by the (usually less significant) daily harmonic. It's still there – it's almost certainly non zero, but it's just not ...


21

The tides are a result of the response of the Earth's oceans to the tidal forces exerted on the water by the Moon and the Sun. The responses are vastly complicated by the Earth's rotation about its axis, by the physical geography of the Earth, and by the nature of the orbits of these bodies. Of key interest with regard to this question are the inclination of ...


9

OK, here's a theory: It's all (when I say all I mean mostly) got to do with the fact that the moon's plane of revolution is inclined about 20 degrees or so relative to the earth's equator. That causes differences in the frequency of tides in various latitudes. In order to see how that happens, here's a simplified example: Draw a cross, so that you have 4 ...


0

I'm no expert on this, and the detail of tides is complex, but basically the situation is a =s follows. The moon (and sun to a lesser extent) provide a gravitational forcing with a period of just over 12 hours. Then the resultant ocean movement interacts with the local sea floor geography, which modifies the response. In most places the result is the ...


2

First, let me explain the question a little further. The Hulse-Taylor binary is a binary system composed of two neutron stars orbiting each other. Each star is an extended body, and is in the gravitational field of the other, so should experience tidal forces, because one part of the star is closer than another to the opposite star, so the gravitational ...


1

It's worth thinking about why the tidal bulges1 are not lined up with the line between the bodies (which is where you would naively expect them to) and then thinking about how that affects the gravitational interaction between them. Because the moon takes about one month to orbit and the Earth takes one day to turn, the naive location of the tidal bulges ...


3

I think the two big factors would be that the Earth would 'want to' become tidally-locked to the Moon and the Sun. The Moon would win here, which is easy to see because tides are caused more strongly by the Moon than by the Sun. So in due course the Earth would end up tidally-locked to the Moon with a rotation period which would be the same as a lunar ...


0

In an isotropic space (which means it's equal in all directions), Noether's theorem tells us that angular momentum is conserved. For the Earth this means that it keeps the angular momentum that it has now. This can be calculated from $$L = I \omega$$ where $I$ is the moment of inertia and $\omega$ the angular speed. You'll see that if $L$ remains constant ...


2

The angular momentum is conserved in central force motion (like what we have in the case of Earth-Moon system). In such a case, the force $\vec{F}$ and the radius vector $\vec{r}$ are parallel so that the resultant torque is zero $$\vec{\tau}=\vec{r}\times\vec{F}=o$$ This means the angular momentum ($\vec{L}$) of the of the Moon about the center is a ...


0

The stress-energy tensor is a stand-in for whatever forces you want to account for. For example, electromagnetic waves carry energy (as well as momentum, angular momentum, etc.) and you can choose to include these energies in the stress-energy tensor if you wish. Please see the Wikipedia article about the stress-energy tensor for E&M: ...


1

That's fairly small for an object. It wouldn't have any significant gravitational effect on the moon or the earth. Tidal effects go as the cube of the distance. So the sun has about half the tidal effects of the moon. If this object were in low earth orbit (400km altitude), then the relative tidal effects on the surface when it is overhead would be about ...



Top 50 recent answers are included