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The acceleration of an asteroid (such as 2012DA14) as it approaches earth is proportional to the reciprocal of distance $r$ from earth center, squared. the derivative of the acceleration, or jerk, is then proportional to the cube of the reciprocal of distance $r$, and the speed $v$. Thus the jerk is rapidly increasing as the asteroid approaches. Will the asteroid experience stresses from the jerk that could possibly cause it to break up?

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3 Answers 3

An asteroid within the Roche Limit of a planet will experience a tidal stress tending to pull it apart that's stronger than the gravitational pull of the body's own gravity, tending to hold it together.

If the body is more than twice as dense as the planet, the Roche Limit is below the surface of the planet, and tides won't break it up before it collides (if it gets that close).

If the asteroid is a solid body, it's likely to survive tidal stress that exceeds its own gravity, as the mechanical strength of the material it's made of (metal and/or rock) is enough to hold it together. An astronaut standing on the surface of the asteroid might float away, along with any loose material on the surface, but the asteroid itself will remain in one piece.

On the other hand, some asteroids are thought to be mere "piles of rubble", loosely held together by gravity; such an asteroid is likely to be disrupted by a very close pass by a planet.

2012 DA14 didn't approach within Earth's Roche limit, which is generally a quite small multiple of the planet's radius. The recent meteorite that hit Russia did, but there probably wasn't enough time for it to be tidally disrupted before it hit the atmosphere.

Comet Shoemaker–Levy 9 is believed to have passed within Jupiter's Roche limit in 1992, causing to break up into multiple fragments. The fragments famously collided with Jupiter two years later.

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I think I understand the concept of tidal stress, although it is difficult for me to wrap my head around tidal stress in a relatively small body such as an asteroid, where the closest point to earth center and the farthest point may only differ by less than 100 yards. In other contexts, stress in a body is related to the time rate of change of acceleration, or "jerk". Thus jerk would be higher on an asteroid at say 17,000 miles from earth than at several hundred thousand miles. Or is this effect negligible? Or are we saying the same thing? –  Dave Feb 17 '13 at 14:05
    
Tide is the difference in (gravitational) acceleration over distance. It falls off with the cube of the distance from the primary, which is why the Moon's tidal effect on the Earth is stronger than the Sun's. For an asteroid near Earth, the (differential) force from the tide and the asteroid's own gravitational cohesion are both fairly small; within Roche's limit, the tide exceed's the asteroid's gravity. The net gravitational force on the asteroid is zero, since it's in free fall, so I don't think "jerk" comes into it, except perhaps as a second or third order effect. –  Keith Thompson Feb 17 '13 at 19:26

Possibly yes. However it would be from the tidal stresses that are on an object passing close to a larger body could be enough to break it up. It will depend on what the asteroid is actually made of and how close it passes to the Earth.

However, for an object the size of 2012DA14 on a flyby trajectory I wouldn't expect there to be enough tidal stress for it to break up this time. If it's orbit is changed so that it comes back across the Earth's path more frequently then the cumulative effect could be enough to break it up.

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The asteroid experiences no stress whatsoever, since it is in freefall. From the perspective of the asteroid, there's no difference whether it's in deep outer space or near a large source of gravity.

The only way it could break up is due to tidal forces, which are likely too minuscule for an object of its size.

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