# Tag Info

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 ...

24

Roche limit happens where the gravity of the object, trying to pull the object together, becomes smaller than the tidal force (trying to pull the object apart). But the astronaut is bound by not gravity, rather by the electromagnetic interaction between his/her atoms. The own gravity of the astronaut is negligible, compared to the electromagnetic interaction....

14

The Roche limit is where the tidal forces exerted on an orbiting object are sufficient to overcome the self-gravity of that object. The "self-gravity" of an astronaut is tiny. We can estimate it as something like $$F_{\rm grav} \sim \frac{Gm^2}{(h/2)^2},$$ where $m$ is the mass of the astronaut (+ equipment) and $h$ is their size (height). Assuming $m=100$ ...

10

I consider small stars to mean main sequence stars (or brown dwarfs) with mass less than the Sun, and then also consider compact stellar remnants - white dwarfs and neutron stars. Main sequence stars and brown dwarfs What you need is a mass-luminosity relation combined with an expression for the tidal radius in terms of the stellar mass. The latter also ...

9

No. The Roche limit refers to the radius at which a satellite which is held together only by its own self gravity is disrupted by tidal forces. Spacecraft are not such satellites, but are much harder to tear apart.

7

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 ...

7

The Roche limit is defined in classical physics, dealing with materials that behave classically. That said, of course people have tried to estimate the counterpart in relativistic conditions. It turns out that it behaves roughly like the classical case when considering a liquid body orbiting a black hole. For the neutron star case and a more massive stellar ...

7

The Roche limit applies when a smaller body that would be held together by its own self-gravity is in the gravitational field of another, such that the tidal forces of the latter are stronger than the self-gravity of the latter, thus destroying the smaller body. However, the gravitational tidal forces of a black hole are always finite, except at the ...

7

It's different. Black holes are not objects, like planets or stars. Rather, they are powerful distortions of spacetime, maintained by the concentration of mass/energy therein (which itself is kept there prisoner by the spacetime distortion - a vicious cycle broken only slowly by the Hawking radiation). As such, they cannot be "ripped apart", since there is ...

7

The simple answer is yes, an asteroid can have rings. The current known example is 10199 Chariklo, whose rings were discovered in 2014 (see Braga-Ribas et al. (2014)). It actually has two rings, at distances of 391 km and 405 km, 7 km and 3km in width. As the discovery paper notes, they may be the (young) remnants of a debris disk around Chariklo, and have ...

5

You have asked two different questions, Muze. The title of your question asks whether the Roche limit is dangerous to spacecraft, but the body asks whether the Roche limit has been observed having an effect on an object. The answers to these two very different questions are "no" and "yes". The main question first: The Roche limit applies to objects that are ...

5

Expanding on Peterh's answer, we could try to find how should be an astronomical object for the tidal forces be felt by an astronaut orbiting it. I don't have any reliable data on how strong need the tidal force to be felt. However, with a big simplification, we can very roughly model the upper and lower body of an astronaut as two masses placed about 1 ...

5

Slightly long for a comment, so I'll put it here. A comparison could be drawn to the equatorial bulge. A planet that rotates is fatter at the equator than the poles. Nothing "lifts" off the surface because the outward force from the rotation is generally much weaker than the gravity, but the planet still bulges because the forces get added together and ...

4

A spaghettified object may escape; indeed, this is fairly common. The reason is that most objects approaching a black hole do it freely falling along hyperbolic orbits that swing by and then go back out to infinity (unless the closest distance is just a few Schwarzschild-radii out - then it may plunge in). When it passes the tidal radius it begins to be ...

4

First things first, but in this case, second things first. When an asteroid or moon, passes the Roche Limit, it breaks apart and forms rings around the primary body. This is a widely believed misconception of the Roche limit. The Roche limit pertains to objects that are held together by gravitation only. Once chemical bonds come into play the Roche limit ...

3

tl;dr Maybe it made rings, but they’re certainly not around today. The whole reason the Roche limit exists is because of tidal forces, and the whole reason tidal forces exist is because we choose to work in non-inertial reference frames. If you’re not orbiting an object, there’s no tidal forces from that object. Just as a meteor or asteroid can pass the ...

3

Sometime before now and "tens of millions of years". Phobos currently orbits at a radius of about 9400km The theoretical Roche limit is different for rigid and fluid bodies. If Phobos were a fluid body, it would already have passed the Roche limit at 10500km. This is because fluid bodies will deform into an ellipse pointing towards the planet, ...

3

You've answered you own question. If there are cohesive forces beyond that of simple self-gravitation, then objects can survive intact inside the self-gravitating Roche limit - as does every solid item on the surface of the Earth for example. Saturn's rings are made of ice, not rocks. The tensile strength of ice is about $10^6$ N m$^{-2}$. For a self-...

3

It may also help your mental picture to consider a phenomenon many people see every day. The Earth orbits the Moon quite a distance outside the Moon’s Roche limit. Nevertheless, it is distorted by this. The distortion is a bit less than a metre: that is, a perfectly fluid frictionless inertialess material would settle into an ellipsoidal shape pointing ...

3

Because the density of the matter at the center of a black hole is infinite (or nearly so) the Roche Radius is 0 (or nearly so). Two black holes in orbit spiral in towards each other because they radiate gravity waves and form a mutual event horizon without any disturbance to the matter at their centers.

3

I think there are two key aspects to the answer. 1) Solid/rocky bodies should tend to collide before they reach the Roche limit. 2) When gaseous bodies reach the Roche limit (and undergo 'Roche-Lobe Overflow'), the dynamics are basically those of test-bodies and are fairly straightforward and well understood from binary stellar dynamics. To expand on both:...

3

(Yes, this should be a comment. It's too big, though.) To address Sidney's comment to Ed Shaya's answer: The Roche limit equation can be expressed as $$1.26 \times R_{\text{secondary}} \times \sqrt[3]{\frac{\text{Primary}}{\text{Secondary}}}$$ Since the radius of the secondary is zero and zero times anything is zero, the Roche limit is likewise zero. ...

3

A large moon (the size of a planet) would be modeled by the fluid model, meaning friction and tensile strength are too weak to modify the shape of the moon significantly, shape is determined by rotation, vortices, self-gravitation, and tidal forces. Smaller moons, like Phobos would likely follow a rubble pile model. On circular orbits tidal heating doesn't ...

3

The roche limit is a calculation of when pieces of a satellite will be pulled off the satellite. Any force that affects the whole satellite equally can be ignored, as this will affect the motion of the satellite as a whole but not (directly) whether pieces are pulled off. If you consider an object which maintains a constant orientation, but rotates about ...

2

When a satellite is in a circular orbit around a planet you may consider its motion in a rotating frame of reference; the frame rotates at the same rate as the satellite orbits. In this frame the satellite doesn't move. But because this is a rotating frame the formula $F=ma$ is doesn't work, you need to add extra terms for the centrifugal force and the ...

2

There are actually two main ways to form rings, which I'll call the bottom-up and top-down approaches. In the bottom-up approach, a moon or other body moves inside the planet's Roche limit, and is torn apart by tidal forces. The resulting particles then settle down into a plane and spread around the planet until a ring is formed, of roughly uniform density....

2

There are a few ways that a gigantic ring system outside the Roche limit might be possible. Ring systems can extend well past the Roche limit if they are very faint or very young. A faint/light ring system doesn't have enough mass to coalesce and a young system might not have enough time. Young small moons with lots internal heat of formation can be ...

2

Bodies typically progress outward rather than inward. (See Why is the Moon receding from the Earth due to tides? Is this typical for other moons? .) The only orbiting bodies that might approach are ones that orbit faster than the main object spins, IOW, closer than synchronous orbit. Even then they could recede if locked into resonance w/ other bodies, eg, ...

2

According to wikipedia, the roche limit for an object with the same density as the Moon would be around 10,000 kilometres. Since the density of the Earth is about 166% of that of the Moon, if you don't take the atmosphere, or oceans into account, then the binary planets could be much closer to each other. However, when you take into account the oceans, and ...

Only top voted, non community-wiki answers of a minimum length are eligible