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22

That is because the moon attracts both the water and the earth. The gravity of the moon reduces with distance (by the inverse square law). So, the moon's gravity is Greatest at the point nearest to the moon (called sub-moon point) Is lesser on the earth's center, and' Is the least in the opposite side of the earth. Assume that earth is falling to wards ...


19

The following diagram from the wikipedia article on the tidal force shows the tidal force that results from a moon. Note that the tidal force is directed away from the center of the planet when the moon (satellite) is directly overhead or underfoot but is directed toward the center of the planet when the moon is on the horizon. You are right that these very ...


19

Everything in the universe has a gravitational influence on everything else in the universe. It isn't a question of the strongest gravitational pull winning out and all the others doing nothing. The Earth is the strongest pull on the oceans, but the Moon and the Sun both have easily measurable effect in addition to the Earth's. Other bodies (Venus, Jupiter, ...


8

There will be tides, but they will be not very large. It's pretty easy to get a good estimate of their size. There are two things that control tides on Earth and both will be present anywhere else: The tidal forces from celestial bodies, and the size of the body of liquid in which the tides are raised along with resonances int he body of liquid. Taking ...


7

The possibility you mention is feasible if the moon is in the equivalent of a geostationary orbit, where the orbital period of the moon is exactly the rotation period of the planet and it orbits in the equatorial plane of the planet. Such a moon would always keep the same face to the planet if it were close enough to the planet (which in turn depends on the ...


7

Neptune's largest moon, Triton, is about a third the mass of Earth's moon, and roughly the same distance away from its primary, so it will indeed have noticeable tidal effect on Neptune's atmosphere and oceans. Neptune's other moons are all tiny (all of them together being about 0.5% of Triton's mass), so while they do exert tidal forces, those effects ...


7

I am probably going to get slammed for this, because it violates everything we were taught about tidal forces, but the antipodal tide is caused by the centrifugal force created by the Earth's rotation about the earth/moon barycenter, not differential gravitational forces. While the moon's gravity is less on the side of the earth furthest from it, that force ...


6

You're right; the effect of solar tides on the Earth would be to increase the Earth's (or Earth+Moon's) orbital angular momentum around the Sun. But it would be difficult to compute this properly, and probably impossible to measure the effect, because of all the other small alterations to the Earth's orbit going on. A simple way to see the effect (which I ...


6

There are two main gravitational causes of tides: the Moon, and to a lesser extent the Sun. When the moon is full or the moon is new, the Earth, Moon and Sun are roughly aligned, and the Lunar tide combines with the Solar tide to give a "Spring tide" that has a larger range. When the moon is at first quarter, or third quarter (ie a half moon), the solar ...


6

Yes Tidal locking is a mutual process, and both the satellite and the parent body will eventually permanently face each other. However, while tidal locking of a satellite can happen relatively quickly (that is, within the life span of the solar system), tidal locking of the parent body is extremely slow and will in a practical sense never happen (the ...


5

Tidal circularization is indeed a plausible mechanism for why Triton's orbital eccentricity is so low, and I believe it is in fact the prevailing theory, at the moment. In short, tidal forces from Neptune dissipate energy in Triton's interior, essentially "squeezing" the moon. This results in a loss of orbital kinetic energy. The exact calculations that lead ...


4

Solid-body tides on the Moon There are various good models for solid-body tides on the Earth. Are there any recommendations for tides on the Moon? This answer starts with simplistic explanations and links with trivial calculations and progresses to links that, along with their references, provide exacting and lengthy approaches. @scb, you might want to skip ...


4

Nope. (Sorry, the other answer beginning with yes, I wanted to begin one with no), but both answers are correct from a certain point of view. Jupiter's rotation is speeding up because it's contracting. See here. Jupiter does slow down due to it's tidal bulge and tugging from it's moons, but the bigger factor is that Jupiter loses more heat than it ...


4

Let us begin with the basics. If an exterior perturber generates the potential W, then an originally spherical planet gets distorted. Owing to this distortion, its potential acquires an increment U which is usually called the additional tidal potential or, simply, the tidal potential. Now let us fix a point on the surface of the planet. Let $\gamma$ be the ...


4

There's two parts to this question, the first part is relatively easy. The 2nd part, more tricky. How are shepherd moons such as Pan and Daphnis able to exist without being disintegrated by Saturn's tidal forces despite literally orbiting within the rings themselves? I'm going to focus on Pan because it's larger and closer. It's unusual, ravioli ...


4

The short answer which may or may not be an "Aha!" answer is that what is plotted is what's left over after a much larger, uniform force is subtracted. The uniform force is the Force from the Moon evaluated at the center of the Earth, and the arrows show the deviation of the actual force from that average. Why do we do it that way? When looking at the ...


3

In the first case, the planet's rotation will carry the tidal bulge ahead of the satellite, this will tend to accelerate the satellite, causing its orbit to recede. In the other cases, the planet's rotation will pull the bulge behind the satellite, causing the orbit to decay. It doesn't matter if the period is greater or smaller in the case of a ...


3

Only a syzygial tide during/near an equinox is the strongest. This means, there must be either new or full moon. In general, syzygial tides are strong because three bodies (Earth, Moon, Sun) align near one line, and tidal effects of Moon and Sun on Earth become (nearly) collinear and sum to the maximal possible magnitude. The Moon’s orbit is inclined to the ...


3

I think it's a somewhat poorly written page and Xakarus Alldredge's answer is very good. I thought I'd add some pictures. The Tidal force the Earth exerts on the Moon is much stronger than the one the Moon exerts on the Earth and it's technically not a force either, it's a secondary effect from the force of gravity. But the tidal "force" on the moon is ...


2

As stated in the quoted book excerpt, the tidal forces that arise from the Moon's gravity create a bulge in Earth's oceans on the side of the planet facing the Moon and on the side facing opposite the Moon. Now imagine that you are on a shoreline of some arbitrary continent. As the Earth spins, the shoreline you are on is carried into the oceanic bulge ...


2

'Neap' tides when there is the least difference between high and low tide, generally occur around the time of the first and third quarter moons. 'Spring' tides when there is the largest difference between high and low tide, generally occur around the time of the New and Full moons. The exact timing is complicated by your position on earth, the shape / size ...


2

Gravity is an acceleration, and force = mass x acceleration. Thus, if we want to detect the Moon's gravitational pull, we need either a giant mass or a really precise instrument. Since the oceans are a giant mass, the force on the ocean is thus correspondingly large. The fluidness of the ocean also allows it to move, even if the force per volume is small. ...


2

To expand on Keith's comment, the equation for the tidal force is based upon the difference between the acceleration due to the orbiting object, and the gravity of the body itself, i.e. the difference between how hard the Moon pulls on you compared to how hard the Earth pulls. This results in an overall equation of: $$|a_{tidal}| = 2\Delta r G \frac{M}{R^3} ...


1

While gravity itself is weak relative to the other three known fundamental forces, that should not be construed to mean that the moon's gravitational pull on earth is necessarily weak. To say something is "weak," you really need some sort of benchmark against which to make that determination (e.g., I'm weak because I can only benchpress 50 kg, ...


1

I think it's good to start with the last question: How would the tides be affected without the Moon? The answer is simple: There would still be tides twice a day caused by the Sun, but they would be smaller. There are lots of sources online that talk about how tides from the Sun and Moon combine; without the Moon, we would just have the tides from the Sun. ...


1

The following is a plausibility argument based on the symmetry of an assumed circular orbit of the earth around the sun. The earth's rotational axis is inclined to the ecliptic plane by an angle of about 23.4°. For this reason idealized tidal bulges induced by the sun would travel around the earth's equator at the vernal and autumnal equinox whereas they ...


1

Over enough time there is a tendency of all orbiting objects to fall into resonance with the object they're orbiting. 1:1 being is most common. 1:1 usually called Tidal Locking, is quite common. The Moon to the Earth. Pluto and Charon to each other. All four of the Galilean moons are tidally locked to Jupiter, likely others. Mercury's 3:2 ratio is ...


1

What are the positions of the Earth, Moon, and Sun at a time when there is the least difference between high and low tides? TL;DR response: The simple answer to this question is twice per month, when the Moon is half full. A not so simple answer is roughly every 8.85 years, when the Moon is half full and the Moon is close to apogee. A full answer: Who knows?...


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