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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 ...
47
Comet Shoemaker-Levy 9 (SL9 for short) is a great example of a moon in a highly eccentric orbit that passes through the Roche limit at periapsis.
SL9 was discovered in 1993, but is thought to have been orbiting Jupiter for 20-30 years prior to discovery. It passed through Roche limit of the Jupiter/Comet pair and is thought to have broken apart in July of ...
47
Tidal forces generated are proportional to $m/r^3$, where $m$ is the mass of the Earth satellite and $r$ is the semi-major axis (we assume circular orbits for simplicity). The derivation of this relationship is performed nicely by Butikov.
So $m_m/r_m^3 = m_c/r_c^3$, where
$m_m \approx 7.3 \times 10^{22} \rm\, kg$ is the mass of the Moon,
$m_c \approx 9.1 \...
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 ...
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 ...
22
As is nicely put on the Wikipedia page about tidal forces, the tidal force is given by
$$T=Gm\frac{2r}{d^3}$$
where $T$ is the tidal force (see below), $G=6.67\cdot 10^{-11}\rm\,\frac{m^3}{kg s^2}$ is the gravitational constant, $r$ is the radius of the Earth, and $d$ is the distance between the centers of the two objects. This is not force in correct sense (...
21
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. (...
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, ...
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 ...
17
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 ...
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 ...
14
"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 (...
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
$$K_a\equiv\frac{3M_*R_p^3}{5\bar{\rho}a^3},\quad K_g\...
13
As PM 2Ring mentioned, you seem to have a misunderstanding that spin is involved in creating an orbit. What matters for gravitational attraction is the mass of the bodies and their distance, the effects of spin are more subtle and only an issue when bodies are non-spherical. Also, one body doesn't need to be smaller than the other, they can be the same size.
...
12
A belated answer, but neither of the existing answers properly explain this.
The proper explanation is simple. In Newtonian mechanics, tidal influences make all objects in retrograde orbits and those objects in prograde orbits below the equivalent of geosynchronous radius spiral inward. Only objects orbiting prograde above the equivalent of geosynchronous ...
12
Synchronous rotation is when the orbit of a body has the same period as its spin. If the inclination and obliquity are the same, then the same face of the body will always point towards the barycenter. Here is an animation of Pluto and Charon Credit: Stephanie Hoover Wikimedi commons.
Tidal forces impart torque to orbiting body rotations. An orbiting body ...
9
Pluto and its largest moon Charon are tidally locked to each other.
Charon and Pluto revolve about each other every 6.387 days. The two objects are both gravitationally locked to the other, so each keeps the same face towards the other. This is a mutual case of tidal locking . . .
Because of Charon's large size compared to Pluto, and because its ...
9
The basic tidal acceleration felt by a test mass near the surface of the Sun, due to a body of mass $m$ at a distance $r$ is given by
$$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
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
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 ...
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 ...
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
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 ...
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
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 ...
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