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26

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" ...


17

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 ...


16

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 ...


16

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 ...


16

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 ...


13

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 ...


13

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 ...


10

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 ...


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 ...


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 ...


9

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 ...


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} ...


7

From Wikipedia (which cites the paywalled http://dx.doi.org/10.1006/icar.1996.0117), we get http://en.wikipedia.org/wiki/Tidal_locking#Timescale Now, a is the semi-major axis (or orbital radius) of the object (I'm not totally sure if the logic changes for elliptical orbits). Anyways, we can easily re-arrange the equation to express a (or the radius) in ...


7

That is an incorrect description of what happens. If the black hole is small, the gravitational forces pull you apart before you get anywhere near the horizon. On the other hand, if the black hole is large, you won't even notice anything as you cross the horizon, until you hit the singularity and then you'll be pulled apart. Remember, from your point of ...


6

This is, no doubt, one of the biggest challenges for realistic simulations: waves crashing, hair moving under wind and whatever other movement involving turbulence will be hard to solve. Though it is true that one can solve the equations of motion for each individual particle in a 'molecular dynamics' fashion, that is just infeasible for a system that goes ...


6

I think it is basically a coincidence at the current time. Earth's axis of rotation precesses with a period of about 26,000 years, and according to Wikipedia, its orbital axis precesses with a period of about 112,000 years. So the winter solstice and perihelion will have all possible relative phases over a long time period.


6

Because the differences in the strength of the gravitational field between each side of the glass are essentially 0. However, the difference in the strength of the gravitational field between the side of the Earth closest to the Moon and furthest to the Moon is enough to pull water more towards the side close to the Moon, causing high tide for that part of ...


6

A very sensitive device would be required to measure the minuscule change in the water depth along the glass walls, because the differences in the strength of the gravitational field between each side of the glass are essentially zero. Because of this the force exerted on the glass is the same, but due to the small volume of water in a glass as opposed to an ...


6

To essentially quote http://en.wikipedia.org/wiki/Tide: Energy of the Earth is not conserved while energy of the Earth-Moon system must exist. Energy from bodies of water are diminished (by about 3.75 TeraWatts) where about 98% of this energy loss is due to marine tidal movement. Because energy is lost in the water, this imposes a torque on the Earth ...


6

When an object comes within the Roche limit, it breaks up because of tidal stresses - the part closest to the earth feels a stronger gravitational attraction than the furthest part. Hence, the closest part will fall a little faster than the trailing parts. As a result, "disintegration" does not mean that the body will fly apart like a bomb. Instead, it ...


5

It would be better to think in terms of water trying to flow until it reaches a equal-potential surface. Like this. In the absence of competing forces from the sun and moon, the water doesn't rush down until the Earth stops pulling on it, it flows until there is nowhere for it to go because the other water is in the way. That happens when the surface of the ...


5

It has to do with the fact that the entire earth is in an accelerating reference from due to the attraction from the moon. Imagine 3 points on a line out from the moon. $N$ is the oceans near the moon, $C$ is the center of the earth, and $F$ is the far oceans. The attraction falls off with distance so the forces $F_N > F_C > F_F$ as you know. ...


5

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 ...


5

Yes, it is possible for objects to orbit 100 km away from the Earth. After all, lots of man-made satellites are doing so. It's also possible for a moon to be much closer than it is now. However, you must realize that the radius of the Moon is 1,700 km so 100 km can surely not be the distance between the Moon's center and Earth's surface. However, if the ...


5

It might affect climate, but not on the time scale of a month, and does not significantly affect the weather. The fact that the moon exists may significantly stabilise the inclination of the Earth relative to the Sun. This, in turn, affects climate in the long run. The debate is ongoing. For example, see long term axial tilt (Wikipedia): The Moon has ...


5

If the diameter would be increased by a factor of 2, then if the density stays the same, the mass would increase by a factor of $2^3 = 8$, since volume is proportional to the cube of the diameter. The main effects of doing this are: the Moon's pull on the Earth would be double of what it is now, since the gravitational force is proportional to the ...


4

You ask: How gravity of the moon creates tides only in water? This is wrong. Tides are created by the moon on all materials on earth that have some elasticity. The raising and falling of the ground has been measured at the beams in CERN, for example. The solid ground tides are called earth tides and their height can be 40cm. Is any things( other ...


4

You can see the graph used in context, in the paper 'The Habitable zones around main sequence stars 1993' written by James F. Kasting. The graph is on page 124. The locking formula he uses to calculate the graph comes from a book, a chapter written by Peale in 1977 called 'Rotation histories of the natural satellites' The formula is Tidal Locking Time in ...



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