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It’s because they are much smaller than Io. Tidal forces are differential forces, that is, they result from the difference in gravitational pull on one side of a body compared to the other. When an object is small, the difference in distance to the two sides of it is necessarily small as well.
According to Wikipedia, Amalthea, the largest of those four ...
23
"It's believed that the Earth was rotating about once every 5 hours
before the theorized collision with a Mars sized coorbiting object
referred to as Theia."
Almost. Theia did not have to be co-orbiting, just an intersecting orbit. We have no idea what the Earth's spin was before the collision, but it is theorized that the Earth rotation had a 5 hour ...
20
Some hypothesize that the Earth did have a subsurface ocean during the Cryogenian period, which lasted from 720 to 635 million years ago. The Cryogenian saw the two greatest known ice ages in the Earth's history, the Sturtian and Marinoan glaciations. There is some evidence that the Earth was completely covered with ice and snow during those glaciations. (...
17
There are four moons that are closer to Jupiter than Io with higher eccentricities, yet they don't seem to have any volcanism at their surface.
Only one of those innermost moons (Thebe) has an eccentricity higher than that of Io. The other three have lower eccentricities.
The reason they don't exhibit volcanism is because they are too small. The largest of ...
16
A rough back-of-the-envelope way of seeing what's going on...
The tidal force is due to the difference in gravitational force, so follows an inverse cube law:
$$F_\mathrm{tide} \propto M_\ast R^{-3}$$
where $M_\ast$ is the stellar mass and $R$ is the distance. So at the same distance from a less massive star, the tidal force will indeed be weaker. But ...
15
As the moon orbits Earth, tidal forces slow down the Earth's rotation by 2 milliseconds per century. Eventually, in tens of billions of years, the Earth and Moon would achieve a double tidal lock, where both are stuck with one side facing the other as they orbit the Earth-Moon barycenter. In 7.5 billion years, the Sun will expand past the Earth's current ...
14
Here is how the tides move the moon away from the Earth:
The moon orbits the earth, and there is a difference in gravitational force between the the side of the Earth nearest the moon, and the side far from the moon.
This difference in force tends to pull the Earth into a oval shape with its long axis pointing towards the moon.
But the Earth is also ...
14
Leconte et al. (2015) suggested that the presence of an atmosphere could prevent or at least slow tidal locking. The star should exert two separate torques: one on the atmosphere and one on the solid body of the planet:
$$T_a=-\frac{3}{2}K_ab_a(2\omega-2n),\quad T_g=-\frac{3}{2}K_gb_g(2\omega-2n)$$
where
$$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\...
13
"Protection" isn't the only effect of Earth. Here is a different POV: Earth may have accelerated impactors by gravity assist.
A different approch is the thinner crust, as suggested for the near side, which may have allowed asteroids to penetrate Moon's crust, such that lava could flow into the basins, or which may have favoured volcanism on the near side (...
13
As PM 2Ring mentioned, you seem to have a misunderstanding that spin is involved in creating an orbit. What matters for gravitational attraction is the mass of the bodies and their distance, the effects of spin are more subtle and only an issue when bodies are non-spherical. Also, one body doesn't need to be smaller than the other, they can be the same size.
...
11
A belated answer, but neither of the existing answers properly explain this.
The proper explanation is simple. In Newtonian mechanics, tidal influences make all objects in retrograde orbits and those objects in prograde orbits below the equivalent of geosynchronous radius spiral inward. Only objects orbiting prograde above the equivalent of geosynchronous ...
9
The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$a_{\rm tidal} = 2\frac{Gm R}{r^3},$$
where $R$ is the radius of the Sun.
You can work out that Jupiter has the largest tidal effect on the Sun, very closely followed by Venus, and both produce tidal forces that are only a ...
9
Both expressions are incorrect. The first should be
$$\frac{GM_{\text{moon}}}{(R_{\text{moon}}-r_{\text{planet}})^2} - \frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2}\tag{1b}$$
or
$$\frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2} - \frac{GM_{\text{moon}}}{(R_{\text{moon}}+r_{\text{planet}})^2}\tag{1a}$$
where $R_{\text{moon}}$ is the distance between the center of ...
8
Pluto and its largest moon Charon are tidally locked to each other.
Charon and Pluto revolve about each other every 6.387 days. The two objects are both gravitationally locked to the other, so each keeps the same face towards the other. This is a mutual case of tidal locking . . .
Because of Charon's large size compared to Pluto, and because its ...
8
Interesting question. I would say from an energy standpoint, it almost certainly it has no effect.
Of course, the extreme case is Io, one of the Galilean moons whose heat source comes from the gravitational tidal stretching as it orbits very closely to the planet Jupiter. The heat that sustains the core of the Earth, however, is left over from its ...
8
The terrestrial planets are Mercury, Venus, Earth and Mars. Mercury and Venus are too hot for liquid water to exist at any level, Mars has lost nearly all its water and Earth has a surface ocean, not a subsurface one. The inner planets lost most of their volatiles (including water) as they formed, the water on Earth was provided by later icy asteroid ...
7
Tidal locking occurs because the planet deforms the satellite into an oval, with long axis pointing towards the planet. If the satellite is rotating the long axis will move away from being pointing towards the planet, and the gravity of the planet will tend to pull it back, slowing the rotation until one face is permanently facing the planet. Tidal locking ...
7
Yes - the earth and sun do have tidal forces like the moon and earth. There are two main reasons this is happening.
The sun is always losing mass due to nuclear reactions, the sun is always converting a tiny amount of its mass into energy. This means the pull on the earth is weakened.
The tidal forces that happen between the earth and moon also happen ...
7
The recession rate of the Moon from the Earth is given as 38.04 mm per year, due to tidal forces, according to Wikipedia. A good description (with diagrams) about how this occurs can be found here.
The orbital decay rate due to gravitational radiation can be determined by $$\frac{\mathrm{d}r}{\mathrm{d}t} = - \frac{64}{5}\, \frac{G^3}{c^5}\, \frac{(m_1m_2)...
7
Yes: It has a companion planet or an excessively large moon, with the two bodies orbiting their common center of mass (much like the Earth and the Moon). They could be tidally-locked to each other, but they cannot be tidally-locked to their star.
6
It depends what you mean by "very strong".
The announced first detection (14th September 2015) was considered a very strong signal and that still only stretched the 4 km detector arms by less than the width of a proton (a tiny particle inside the nucleus of an atom). This wave would have no effect on a person.
You could hypothesis a stronger wave that ...
6
That the Earth and Moon orbit about their center of mass from the perspective of an inertial frame of reference is a bit irrelevant. One thing that is quite relevant is that gravitational force is undetectable by a local measuring device. For example, people standing still on the surface of the Earth do not feel gravity. They instead feel the normal force ...
6
There are a series of resources on the leapsecond.com site which specifically talk about detecting tides with pendulum clocks. It has been done by e.g. the Shortt pendulum clocks which had a master pendulum in a vacuum to remove disturbances synchronized to a slave pendulum. From the first link in the resources at 1, the order of magnitude of the change in ...
6
When we do mathematical calculations, sometimes terms cancel and we get to draw a line through them.
But in the real world cancellations and "total cancellations" don't really happen.
Forces just add. Often there's a big force and one or more little forces, and as long as the big force substantially dominates the little forces, we tend to notice only the ...
6
tl;dr
The inner moons are much to small (and too stiff) to experience significant tidal work. They also cool down much faster though I don't think this is relevant in this case.
There're two things to look at, here:
The total amount of tidal work that is done on a moon
The amount of energy that is radiated away over time
When looking at the list of Jupiter'...
5
Well, I wasn't sure if tidal forces between the Earth and Sun were strong
enough to have any effect on the matter. Truth to be told, the article
barrycarter linked clarifies that tidal forces have neglible effect
relative to the Sun's nuclear fusion mass loss.
That's correct.
Some more details on this.
Tidal "tugging" is a two step process. ...
5
You're right that density is the important thing here. The Roche limit is the distance from the main body $d$ such that
$$d=1.26R_M\left(\frac{\rho_M}{\rho_m}\right)^{\frac{1}{3}}$$
where $_M$ denotes the main body and $_m$ denotes the satellite.
As you can see from the chart on the Wikipedia page, Pandora and Prometheus are both at least one and a half ...
5
First it will be important to consider the term 'relativistic speed'. If by that you mean something like 0.1c, it will only change the colour of the stars as you mentioned in the bounty description. However, if it means something with higher Lorentz Gammas (like 0.9c or 0.99c), all other relativistic effects come into play. There's relativistic beaming and ...
5
Other answers are right at explaining why tidal forces move Earth and Moon apart but they don't move apart a pair of black holes. However, I think it is also needed to explain why the phenomenons making two black holes spiral inward don't make the Moon spiral inward to the Earth.
In fact, every pair of rotating masses radiate gravitational waves. What makes ...
5
The more likely case is actually a spin-orbit resonance that is not 1:1 but a half odd multiple, like the 3:2 case of our own Mercury. Having eccentricity in the orbit encourages this situation.
I’ve been meaning to write this up on the Worldbuilding.SE but I have not re-found enough references. But see this video.
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