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The answers to this question about encounters with Jupiter keep mentioning the Roche limit, inside which solid objects break up due to tidal gravity gradients.

Over what time scale does this effect actually occur? The Roche limit is usually discussed in the context of near-circular satellite orbits, implying that the time scale could be quite long - millions of years even.

For a collision trajectory such as discussed in the linked question, the time from entering the Roche limit to contact might be measured in minutes or hours. Would an Earth-sized body approaching Jupiter have time to break up between entering the Roche limit and hitting Jupiter?

The Roche limit in this case is about 53,000km (assuming Earth density for the intruder). If a collision's going to happen it will happen within no more than 35 minutes from crossing the Roche limit.

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The Roche limit is just the distance from Jupiter at which the "stretching" of the object by Jupiter's tidal force exceeds the internal gravitational force of the object.

So certainly in some sense yes, 35 minutes is long enough for the Earth to "break up". Anything loose on the surface, such as dust and litter and the atmosphere and pedestrians and the hydrosphere, which in astronomical terms is all "part of the Earth", will start to fall away from the surface of the Earth, starting at the Roche limit. Note that this happens on both sides, with things "falling" either towards or away from Jupiter, for much the same reason that there are two high tides per day.

However, the Earth has some tensile strength, which means that anything "nailed down" will need to be closer to Jupiter before it comes free of the surface. Over the course of 35 minutes, Earth probably isn't very ductile (or anyway its surface isn't), so it's not really a question of time, so much as the magnitude of the force. Rock won't slowly give over 35 minutes, like stretching gum with a constant force until it thins out and comes into two pieces. It will reach breaking point and shatter catastrophically or it won't.

Tidal force goes as the inverse cube of distance, so it will ramp up quite quickly over the course of that 35 minutes to impact. However, the limit you state is little more than 1/3 the diameter of Jupiter, so I suspect that most of the Earth will still be in one piece when it "impacts" (or "enters" if you prefer) the surface of Jupiter. A geologist or an architect might come along and contradict me, though, since rock doesn't have a whole lot of tensile strength compared with its weight. So it might only require that the tidal force reaches a small proportion above Earth-weight before great chunks of rock come loose (or, you know, continents).

Furthermore, I reckon that to anyone standing on the surface, it will feel like the planet is breaking up before it actually does. Any tidal deformation of the core will break up the less-fluid surface, even before it detaches.

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  • $\begingroup$ Doesn't the lack of rotation change the Roche limit at all? (Ie, aren't total Tidal forces different for a free-falling body than an orbiting body?) $\endgroup$ – Yakk Jul 21 '15 at 18:32
  • $\begingroup$ @Yakk: the simple Roche limit I think it makes no difference, but I don't actually know what goes into the calculation of the more realistic variants. $\endgroup$ – Steve Jessop Jul 21 '15 at 19:44
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    $\begingroup$ @dotancohen: it's not about spinning, it's that the centre of the Earth is deeper in Jupiter's gravity well than the far surface and therefore (for as long as the bulk of the Earth holds together), the Earth is accelerating faster than the dust/pedestrians. Relative to Earth, the dust is accelerating away from Jupiter. In the "incoming projectile" scenario, a perfectly symmetrical ball of unconnected dust held together only by gravity would stretch out rather than "break apart" as in the orbital scenario. $\endgroup$ – Steve Jessop Jul 22 '15 at 8:26
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    $\begingroup$ ... for really deep gravity wells (black holes as opposed to Jupiter) this is called "spaghettification" rather than "breaking apart", I suspect because black hole physicists over-estimate the ductility of humans for dramatic purposes ;-) $\endgroup$ – Steve Jessop Jul 22 '15 at 8:28
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    $\begingroup$ @dotancohen: hmm, so now I need to question whether what I believe to be the reason there are two high tides a day is correct. I think there would be two high tides on Earth whether it was rotating relative to the Moon or not. The reason we pass under both high tides in one day is the definition of "day" in terms of spinning! $\endgroup$ – Steve Jessop Jul 22 '15 at 8:30
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The Roche limit does not cause an object to instantly degrade, it has to orbit within the Roche limit to degrade. Phobos, larger and closer moon of Mars, is believed to orbit inside the Roche limit, and has not as of yet broken apart.

Thus, it isn't fair to say that an object intersecting Jupiter would break apart as a result of the Roche limit. However, once you cross the Roche limit, for a weakly held together object like a comet, it is increasingly likely it will break up. And the closer it is, the more likely that becomes.

Comet Shoemaker-Levy was in fact a loose collection of rock, and broke apart when it reached deep within the Roche limit of Jupiter, about 2 years prior to its impact on Jupiter. But if something is more strongly held together (Earth, for instance), it probably wouldn't break up, but it would cause strong atmospheric tides due to tidal stress and might lose it to accretion disk connecting with Jupiter. Otherwise, any object headed on a direct impact course would probably remain almost intact until impact due to stronger cohesive forces and own gravity than the amplitude of tidal stress within the Roche radius.

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