# Tag Info

162

There is no tidal bulge. This was one of Newton's few mistakes. Newton did get the tidal forcing function correct, but the response to that forcing in the oceans: completely wrong. Newton's equilibrium theory of the tides with its two tidal bulges is falsified by observation. If this hypothesis was correct, high tide would occur when the Moon is at zenith ...

91

Yes gravity can kill you because as you approach something super dense like a black hole, the gravity will change with the square of the distance which means that eventually the gravity at your feet would become significantly larger than at your head. This gravitational gradient is referred to as tidal forces and is the same effect that keeps the same side ...

77

What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different is the force of gravity at a point on the Earth's surface near the sun relative to a point on the Earth's surface far from the sun. If you compare it with the moon, the result will be that the tidal force from the sun is ...

53

Imagine that we have a very massive object in space. At some distance away (call it ten units) we release three tennis balls in a row: The tennis balls all fall towards the massive object. But because gravity goes like distance squared, the nearer balls feel a stronger attraction than the farther balls, and they move apart from each other: You're ...

49

First we must understand a little what is meant by "tide." A tide is the difference of gravitational force an object feels across its volume from another object. In the Earth's case the side closest to the moon feels a stronger force pulling it towards the moon than the center of the Earth does, while the side opposite the moon feels a force weaker than ...

45

Suppose the Moon didn't orbit the Earth at all, so it just stayed at some fixed point while the Earth rotated underneath it: In this case every point on the equator would pass directly under the Moon every 24 hours, and we'd get a high tide every 24 hours. (There's another high tide when we're exactly on the opposite side of the earth to the moon, but let'...

43

Tides are caused by the gradient of the gravitational field - so the tidal "force" experienced drops with the third power of the distance. This means that the relative strength of the tides should go as \begin{align} \mathrm{ratio} & = \frac{M_\mathrm{moon} \cdot D_\mathrm{sun}^3}{M_\mathrm{sun} \cdot D_\mathrm{moon}^3} \\ & = \frac{7 \cdot 10^{...

35

It's fair to say that the "spaghettification" problem occurs where there's a strong enough gravitational gradient across a typical body-distance - but your question seems to ask about cases where the gravity is great - extreme, but uniform (at least in terms of human proportions/distances). Einstein’s principle of equivalence states that, as long as there's ...

34

The moon does work on the Earth because the Moon orbits around the Earth at a different speed than the Earth's rotation. So, as time passes, different parts of the Earth feel the maximum strength of the Moon's pull, leading to oscillations in the height of the Earth's surface, which can be exploited for energy extraction. The Earth has always been ...

31

This diagram shows the Earth rotating round the Sun at it's orbital velocity $v$. That is the centre of the Earth is orbiting around the Sun at velocity $v$. NB the scale is rather fanciful - don't take it literally! I'll also assume the orbit is circular, and for convenience I'll ignore the Earth's rotation i.e. assume it's tidally locked. To calculate ...

30

The relevant "100%" from which you should calculate the percentage isn't the depth of the ocean but the radius of the Earth $$R\sim 6,378,000\,{\rm m}$$ Multiply this $R$ by $10^{-7}$ and you will get $0.6$ meters, a reasonable estimate for average tides. You must understand that the surface of the ocean always tries to create an "equipotential surface" –...

28

The differential force of gravity on the atmosphere works the same as it does for the rest of the earth (the oceans etc). However, moving the equipotential surface by a few m will be almost undetectable on the atmosphere, since the density of the atmosphere decreases so gradually – over many km. Contrast this with the surface of the ocean, which ...

27

The Earth is free falling towards the Moon. Because gravity decays with distance, the side near the moon wants to fall faster than the center of the Earth, while the other side tends to fall slower. So observed on the Earth, the other side "lags behind" and therefore we have high tide there.

25

The Wikipedia article you linked states: Atomic clocks show that a modern day is longer by about 1.7 milliseconds than a century ago If we take this change of 1.7 ms/century and multiply by 2.5 million centuries (250 million years) then we get a change of 4,250 seconds or 1.18 hours. So 250 million years ago the day length would have been 22.82 hours. ...

25

The picture of high tides on opposite sides of the Earth with a period of about 12 hours (actually 12 hours 25 minutes, due to the rotation of the Earth) is an oversimplification. It's just a starting point. Tides would behave this way in the limit of an all-water Earth with ocean depth so great that it had no effect on the surface wave. But the Earth has ...

23

incompressible: not able to be compressed. compressible: In thermodynamics and fluid mechanics, compressibility is a measure of the relative volume change of a fluid or solid as a response to a pressure (or mean stress) change. So water does not change its volume. The same volume of water can take many shapes. When no other gravitational force except ...

22

This is a gravitational phenomenon known as tidal lock. It is closely related to the phenomenon of tides on Earth, hence the name. Tidal locking is an effect caused by the gravitational gradient from the near side to the far side of the moon. (That is, the continuous variation of the gravitational field strength across the Moon.) The end result is that the ...

22

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 ...

19

There are a few things that keep Saturn's rings roughly the way they are. First, Saturn's D ring actually is "raining" down on Saturn currently. But, the phenomenon of shepherd moons prevents the vast majority of material from leaving the other rings: "The gravity of shepherd moons serves to maintain a sharply defined edge to the ring; material that ...

19

Ultimately, the short answer is yes. The moon is slowly moving away from the Earth, at a rate of 4 centimeters per year. This is due to it being in a slightly higher orbit than equilibrium. At the same time, the Earth, being a satellite of the sun, is having its rotation rate slowed, until it will eventually be tidally locked with the sun. Both of ...

18

In the ocean, it's called the tides, and it happens twice a day. You can see it if you're standing on the seashore, but on the deep ocean it happens so gradually you wouldn't even notice it. On land the tidal bulge only rises up a third of a meter or so (not seven meters, because the moon's gravity is much less than the earth's), and, again, it happens so ...

18

When we say that the Moon rotates, we don't mean relative to an observer on Earth, because we're also rotating. Maybe best is to think of it from the perspective of the Sun. If you were at the centre of the solar system, looking at the Earth, you'd see the Moon rotates once every 28 days or so. That also happens to be the amount of time it takes for the Moon ...

17

Begin by imagining that the moon isn't quite a perfect sphere. One side is just a little bigger than the other. As the moon rotates, the heavier face will swing around towards the earth a little faster, and it will swing away from the earth a little slower, since it feels a stronger gravitational attraction via its larger mass. Since gravity is a ...

17

I suspect that what has confused you is the difference between remaining a fixed distance from the black hole and falling freely into it. Let me attempt an analogy to illustrate what I mean. Suppose you are carrying a large and heavy backpack. You can feel the gravitational force of the backpack weighing you down. However this only happens because you're ...

15

If you could take from orbital energy, then it would decrease, until at some point in the future it would zero. Hence, it can't be perpetual.

14

This says it concisely, when describing the effect of tides: Gravitational coupling between the Moon and the tidal bulge nearest the Moon acts as a torque on the Earth's rotation, draining angular momentum and rotational kinetic energy from the Earth's spin. In turn, angular momentum is added to the Moon's orbit, accelerating it, which lifts the Moon ...

14

The Roche limit applies when the astronomical body in question is held together by gravity rather than electromagnetic forces. This is the case for bodies with a diameter larger than around 500km. Obviously for smaller bodies, like humans, we can get arbitrarily close to the surface, but i suspect this isn't what you're asking about. For moons much smaller ...

14

You probably got voted down cause this can easily be google searched, but the simplest way to explain it is that a tide happens because the lunar tug on one side of the ocean is measurably more than on the other side of the ocean and as the earth rotates the tidal "bump" follows the moon so you get 2 high tides and 2 low tides a day. A tide is effectively ...

12

In principle, yes, the ultimate source of energy for a tidal power plant is Earth's rotational energy, so these plants are slowing down the Earth's rotation. By conservation of angular momentum, that means they are pushing the Moon further away as well, although I wouldn't phrase it as being due to "waves in the gravitational field," as that expression ...

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