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110

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 ...


62

What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different the gravity is at a point on the surface near the sun relative to a point on the other side. Because the Sun is far away, gravity doesn't change much between the two extremes on earth. However, if you compare it with the ...


46

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 ...


44

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 ...


36

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 $$ratio = \frac{M_{moon} \cdot D_{sun}^3}{M_{sun} \cdot D_{moon}^3}\\ =\frac{7 \cdot 10^{22}\cdot (1.5 \cdot 10^{11})^3}{2\cdot 10^{30}\cdot (3.7\cdot ...


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" ...


29

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 ...


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.


22

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. ...


20

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 ...


19

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 ...


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 ...


18

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 ...


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

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 ...


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 ...


13

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 ...


11

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 ...


11

As stated in other answers it is how much the gravitational force is different by on opposite sides of the earth that creates the tides. You can still show this using $a=\frac{GM}{d^2}$ but you need to consider the difference, not the absolute force on the earth. The sun while much more massive is just far enough away that it is getting to a much flatter ...


10

Increasing the diameter and distance of the Moon by a factor 2 would lead to a number of very subtle differences. I will list the ones I came up with: Apparent size If $R_m$ is the radius of the Moon and $D_m$ its geocentric distance (that is, the distance between the centre of the Moon and the centre of the Earth), then its geocentric angular diameter is ...


10

Let's simplify. Let's eliminate the Moon. Let's get rid of the Sun temporarily. Let's replace the Earth with an equivalent mass-and-density perfect sphere of iron that is neither moving linearly nor spinning or revolving in any way. We place two 1KG iron test masses on opposite sides of the Iron Earth, suspended 1 M above the surface by identical ...


10

I think the reasoning has an error. It assumes $v$ is constant, but instead we ought to assume the angular momentum is constant. By dimensional analysis that leads to $r \propto \frac{L^2}{GM}$ so as $M$ decreases, $r$ increases (the original post had $r \propto M$, not $r \propto 1/M$. On the other hand, assuming a circular orbit seems dubious. As ...


9

The mistake you're making is that you're looking at the full acceleration when you should look at the relative one. At distance $R=1\mathrm{au}$ from the sun, the gravitational acceleration is given by $$ a_0 = \frac{GM_\odot}{R^2} $$ Assuming a spherical cow earth (in vacuum), at midday at the equator, we're one earth-radius $r$ closer to the sun, ie $$ ...


9

Yes your weight will change. The moon will have a bigger impact than the sun, so you need to look at the position of the moon to decide when you will be heaviest (basically - you are lighter when the moon is overhead, or on the opposite side of the earth; and heaviest when it is on the horizon. So a full moon rising makes you fat...) The effect (the ...


8

If you experience such a uniform force, e.g. when an astronaut on a space walk near the ISS (just earth's gravity), you don't experience any forces at all. That's freefall. Even with 10G, you'd experience a rapid freefall, but that is still harmless. It's the hitting the ground which kills you - that's not a uniform force.


8

The highly upvoted answer is right but to make things much simpler: Tides are based on the change in gravity, not the gravity. That means they drop off at the cube of the distance rather than the square of the distance like gravity itself does. Thus the object with the most gravity isn't necessarily the one that causes the most tide.


8

You should do some hard assumptions, as no atmosphere, but the real problem is the so called Roche Limit, that states that at this limit, the tidal forces , difference of the gravitational potential between the face facing the Earth and the one opposite, is so large that rips the body (Moon) apart.Wiki With the Wikipedia formula: $ d = 2.44 R_{Earth} ...


8

In the summer of 2010, I had the opportunity to attend a presentation by Reiner Rummel, involved in the GOCE satellite containing a very precise gradiometer. The presentation can be found on the ESA website. It contains a table with orders of magnitude for the accelerations they could measure when still in the lab. The gravitational acceleration in the lab ...



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