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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 ...

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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 \... 34 How can Io be tidally heated while it is in tidal lock? It is tidally locked in a mean motion sense of "tidally locked". That Io is in an eccentric orbit rather than a circular orbit means that tidal stresses can and do build up. Lainey et al. claim that the global energy dissipation in a tidally-stressed moon is given by $$\dot E = -\frac{21}2 \... 22 As is nicely put on the Wikipedia page about tidal forces, the tidal force is given by$$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 (... 14 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 ... 13 That sentence on wikipedia continues "from friction generated within Io's interior as it is pulled between Jupiter and the other Galilean moons". There's an orbital resonance (with the other Galilean moons) that prevents Io's orbit from circularizing, and also prevents Io from migrating away from Jupiter. If the other moons didn't exist, Io would ... 8 Mercury has a 3:2 spin-orbit resonance, that has evolved as a form of tidal lock, but not synchronous rotation (a 1:1 resonance) "Tidal lock" could be used strictly to mean only the 1:1 resonance that develops in circular orbits, or it might also refer to other tidal resonances, such as Mercury's, that are possible with more elliptical orbits. 7 It can't cause tides of the same strength Though you could have Ceres exert the same force on the ocean it has to be at a much smaller orbit. A smaller orbit means a shorter orbital period. This means that the Ceres tides can't be same as the current ones. To calculate the wanted equations, we need to use sidereal periods: The Earth rotates once every 23.935 ... 6 I want to know why is this, what causes this parameter to have such a strange behaviour. TL;DR: Better observations. Measurements were absolutely lousy before the telescope era, and remained fairly lousy throughout much of the telescope era. It has only been the recent several decades where very long baseline interferometry (VLBI) measurements have made ... 6 The semantic difference is that "synchronous orbit" is simply the description of an observation, a phenomenon, while "tidally locked" describes a mechanism, an interaction. You asked because the mechanism typically results in the phenomenon; but they aren't perfectly congruent in the Venn diagram — there are tidal locks that don't result ... 6 For a planet like the Earth, it is reasonably straightforward to show that the tidal acceleration across the planet is around 1g at the innermost stable circular orbit (ISCO) of a \sim 10^8 M_\odot black hole (i.e. an orbit of 3 Schwarzschild radii in size). Since a person is about 1/6400000 of the extent of the planet, then stretching acceleration they ... 6 It may not be possible for Venus to become tidal locked I don't think we know if it's possible for Venus to become tidally locked. Correia et al. 2008 expect the equilibrium rotation to differ from the synchronous motion for planets like Venus with thick atmospheres relatively close to the Sun. This might be best illustrated with a graphic from Auclair-... 3 Wikipeida (quoting Icarus ) gives$$\frac{15}{8}\frac{mA^4}{Mr^3}$$Where$m$and$M$are the masses of the moon and planet, respectively;$r$is the orbit radius of the moon and$A$is the radius of the planet. For Earth this is a little less than a metre. Systems with multiple moons (or moon and sun, like on Earth) will have multiple bulges which can add ... 2 To answer this question, one must first define what "weight" is. There are two competing definitions. One, which some call "actual weight", is simply the vector sum of all of the gravitational forces acting on an object. The other, which some call "apparent weight" is what can be measured by a local experiment such as an ideal ... 2 On the leading or trailing points of IO, you are orbiting Jupiter at the same speed as the center of mass of Io, so you experience its gravity only On the equator, facing Jupiter, you are 1821.6 km closer to Jupiter, but still orbiting with the same period. So there is a bit of Jupiter's gravity that is not countered by Io's orbital speed. How much? Gravity ... 2 In short, you're understanding is correct in the essentials. I think the reason it is challenging to get a grasp on this is because the literature is not always careful it its usage of "fluid Love number" and "tidal Love number". I've seen many cases where these two terms are used somewhat interchangeably or without context and I've even ... 2$ \Delta T \$ is dependent upon the rotation of the Earth, which is affected by multiple factors. Some of these factors are known and can be calculated/predicted, such as the gravitational pull of the Moon, the Sun, the planets, etc., but some take place inside the Earth itself: mantle currents, for example, are main contributors. Since we can’t predict or ...

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If you were to swap the sun with a black hole of exactly the same mass, the only thing you’d feel different is cold. The Earth would go along its slightly elliptical orbits as if nothing happened, as the new black hole and the old sun would have the same gravitational effects.

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You are right. Highest monthly tides are during a new or full moon since the sun and moon combine their gravitational attraction in line with the earth. The highest tides of the year are middle of Northern Hemisphere winter because that's when the earth's orbit is closest to the sun. The usual simple modeling of tides assumes an oceanic earth with no ...

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