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Space Science

Neil deGrasse Tyson On How To Stop a Meteor Hitting the Earth 520

An anonymous reader writes "Astrophysicist Neil deGrasse Tyson talks stopping extinction-level meteor hits: '...Here in America, we're really good at blowing stuff up and less good at knowing where the pieces land, you know...So, people who have studied the problem generally – and I'm in this camp – see a deflection scenario is more sound and more controllable. So if this is the asteroid and it's sort of headed toward us, one way is you send up a space ship and they'll both feel each other. And the space ship hovers. And they'll both feel each other's gravity. And they want to sort of drift toward one another. But you don't let that happen. You set off little retro rockets that prevent it. And the act of doing so slowly tugs the asteroid into a new orbit.'"
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Neil deGrasse Tyson On How To Stop a Meteor Hitting the Earth

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  • They don't need to be thrusting directly at the asteroid. Think 3 or more at angles, so they cancel each others' sideways thrust and the overall thrust misses the asteroid, whilst providing net 'away' thrust. Yes, this reduces efficiency.

  • by Electricity Likes Me ( 1098643 ) on Sunday March 03, 2013 @11:59AM (#43061449)

    The asteroidmay not be solid rock. It could be a rubble-pile type, and there might not be anything solid-enough to apply force to in a consistent way. It might be two closely orbiting bodies of rock, in which case you can't push on one in any type of consistent direction.

    The benefit of the gravity-tug approach is that if you have a body of some concentrated mass moving at you, then if you have a spaceship sit away from it and maintain a constant position relative to a point other then the asteroid, then you can act on it's entire mass consistently.

    Find it early enough, and you can do this with high-efficiency ion thrusters, rather then needing inefficient chemical rockets.

    Re: reactive force from retrorockets - you fire them off-angle to the asteroid so exhaust doesn't hit them. You can easily mount orthogonal engines which would carefully cancel the attraction of the asteroid without directing any exhaust at it.

  • by hawguy ( 1600213 ) on Sunday March 03, 2013 @12:00PM (#43061455)

    The "pull" between a spaceship and an asteroid would be equal to the apparent weight of the spaceship on its surface, decreased by the square of the distance between the two objects. This would reduce the traction to a very limited amount.

    You'd get better results with a cable from the ship attached to the surface, but the problem would be the rotation of both objects.

    To do a decent job, the spaceship would need to collect a large quantity of mass before attempting to drag the asteroid.

    I think the point is that you don't know how fragile the asteroid is (it could just be a big pile of rubble held together by its own gravity), so anything you do to it through physically touching it, like attaching a cable, landing on it, etc, may break it up into smaller pieces with the result that instead of one large asteroid, you now have a dozen or maybe hundreds of smaller asteroids that you have to deflect. And the set of smaller asteroids will have the same effect on earth as the one large asteroid.

  • by bill_mcgonigle ( 4333 ) * on Sunday March 03, 2013 @12:08PM (#43061525) Homepage Journal

    No, because you use ion engines on the tug which are tremendously more efficient per launch weight than chemical thrusters.

    This isn't a strategy for an "OMG - it's going to wipe us out next week!" asteroid - it's for ones where the orbit shows a near hit of Earth fairly far into the future. Small gravitational tugs over a long period of time are all that's required.

    Now, ideally those asteroids can be brought into a useful orbit where they can be mined for more mass to deflect more and more asteroids. In the mid-term perhaps only the ion engines need to be sent up from Earth.

    Tyson isn't inventing this - it's a well-accepted strategy in the community that he's trying to explain to a larger audience.

  • by Vulch ( 221502 ) on Sunday March 03, 2013 @12:19PM (#43061587)
    the momentum of the propellant from the rocket pushes against the asteroid, countering the thrust of the rocket

    Only if you let it. The Gravity Tractor idea usually uses two ion engines aimed so the exhaust goes either side of the body being towed. The tractor stays in place and there's no unwanted momentum transfer.

  • by egcagrac0 ( 1410377 ) on Sunday March 03, 2013 @12:35PM (#43061691)

    No, it won't. The surface/mass ratio will be different (smaller pieces can burn up more readily), and if they're spread out enough, instead of all that mass hitting at once, we just get a few nights of falling stars of little consequence.

    We seem to survive the Leonids OK, and we've been surviving them for a long time.

  • by Anonymous Coward on Sunday March 03, 2013 @12:46PM (#43061777)

    Actually, they died by a mysterious virus spread through their filthy phones unless I'm mistaken.

  • by Joce640k ( 829181 ) on Sunday March 03, 2013 @12:51PM (#43061815) Homepage

    Golgafrincham entered into a period of exceptional peace and prosperity.

    Um, no. They all died from a virulent disease contracted from a dirty telephone.

  • by K. S. Kyosuke ( 729550 ) on Sunday March 03, 2013 @01:36PM (#43062213)

    The whole idea is conceptually idiotic. You spend a strong force of reaction mass ejection to maintain a weak force of gravity at a constant distance from the target mass producing a microscopic tug on the object.

    I pity the cranially impoverished people who modded this up as "Insightful". Go back to high school, would you? The two forces you're referring to are exactly equal in size, as per Newton's third law. The probe gets positioned at a distance at which the thrust of the engine is equalized by the asteroid's gravity, and the probe consequently pulls the asteroid with identical force (modulo its sign) while keeping a stationary position above its surface.

    What you get here is exactly what you'd get by putting the probe onto the asteroid and pushing it, but you're avoiding the potentially dangerous contact with the asteroid. Moreover, the probe is likely to be powered using solar arrays, and asteroids sort of tend to rotate, which would severely complicate your attempts at creating a sustained thrust, not to mention the fact that your thrust vector would also rotate. Separating the probe from the asteroid and acting gravitationally upon it gives you constant insolation of the panels and the ability to exert constant thrust in a single direction.

  • by __aaltlg1547 ( 2541114 ) on Sunday March 03, 2013 @02:31PM (#43062557)

    That gets you around the exhaust problem but worsens the problem of gravitational force limiting. The closer your tractor ship to the asteroid, the more propellant you must use pushing in useless directions. The farther you put it, the more efficient the use of fuel but the amount of time you must pull to get the same delta P increases as the square of the distance to avoid the tractor escaping.

    The optima don't align at all. If you want to minimize the amount of time you must spend towing, you put the "tractor" very close, about 1.25 radii from the asteroid. (Actually maybe a little farther to ensure you don't hit the asteroid with any of the propellant.) That that reduces your efficiency to about 60% of what it would be in the limit and your force to about 38% of the weight of the tractor on the asteroid's (assumed spherical) surface. For non-spherical towed objects, it gets worse. But assuming efficiency isn't a consideration, you're still limited to less than the gravitational force between the asteroid and your tractor.

    It must have a lot of mass when it gets there because if it doesn't it won't have enough gravity to pull anything anywhere. The smallest objects we'd probably need to move are 100-meter asteroids that mass something like 3E9 kg. So the force you can apply this way is limited to less than .074 Nt/ton of tractor. Over a year of such pulling, you get a delta-v of about .0008 meters per second. How much do you have to change the velocity to miss the Earth? About by the diameter of the Earth. It turns out to do that in a year takes about a 2 ton tractor. A rock twice that big has 8 times the mass and would take 4 times the tractor to move it in the same time.

    This has to be compared in practicality to other methods. While it solves the problem of not having to physically land on the object, you still must match velocities exactly and must send a bunch of dead weight to pull your object with.

  • by CyprusBlue113 ( 1294000 ) on Sunday March 03, 2013 @02:47PM (#43062675)

    Taking a 30,000 kg ship at 20 km from the asteroid and the gravitational constant to be 6.7x!0^-11 m^3 kg^-1 s^-2, you would have a gravitational acceleration of about 5x10^-14 m/s^2. To move it 6,400,000 m (one earth radius) would require 1.6x10^10 seconds (about fifty years).

    Also, for two asteroids orbiting each other, if these are "small" asteroids, what gravity would obtain between them? Two asteroids of radius 20 kilometers each with 50 kilometers between their centers (and having earth's density for a mass of 5x10^13 kg each) would have a mutual gravitational acceleration of 1.3x10^-5 m/s^2. Would it be that hard to separate them?

    But you aren't talking about moving it X distance, you're talking about deflecting it X angle...

    Even 1 degree is a decent change at a few AU

  • by Hotawa Hawk-eye ( 976755 ) on Sunday March 03, 2013 @09:35PM (#43064507)

    I would normally agree but the whole thing sounds preposterous. The gravitational pull of a spaceship is negligible.

    The force of gravity between the asteroid and the ship is small, not necessarily negligible. Let's take some basic physics. The gravitational force between two objects is g (about 9.8 m/s/s) times the mass of the first object times the mass of the second object divided by the square of the distance between them.

    F = g*m1*m2/d^2

    The acceleration of an object imparted by a given force is the force divided by the mass of the object being accelerated.

    F = m*a or a = F/m

    So combining those two equations, the acceleration of the meteor being moved due to the gravity of the spaceship is g times the mass of the spaceship divided by the square of the distance between the meteor and the spaceship. The mass of the meteor cancels out.

    a = g*m_{ship}/d^2

    Now this is likely to be small, true. But just like interest in the financial world, the secret is time.If you can impart even a small acceleration over a long enough time frame, the thing you're accelerating ends up going at a pretty good velocity.

    The approach you'd need to use to react to a meteor depends on the size and how long in advance of its impact you can detect it. If you detect it a year or two ahead of time? At _best_ you might have time to prepare an Armageddon-style mission; more likely you'd want to find some way to record information about humanity's progress so the survivors can recover or aliens can learn about the extinct human race. Ten years would be better for that type of "plant a bomb on it" mission or to prepare to launch a nuclear missile at it. Fifty to a hundred years? A low, slow method like gravitational towing may be good enough to ensure that Earth and the meteor don't pass through the same point in Earth's orbit at the same time.

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