asteroid strike 100% likely, but... (warning - physics post)

[[xeno]Julios]

but that's not what got me in this article:

NASA monitors 127 asteroids with the potential to collide with the planet and cautions there may be 20,000 more yet to be discovered.

Among the identified threats is Apophis, a 140-metre piece of rock with the potential to strike in 2036. If the asteroid crashed into the ocean near the U.S. coast, the resulting tidal wave would cause $400-billion in damage, according to Mr. Schweickart.

"That's the equivalent of six or seven or eight [Hurricane] Katrinas," he said in a telephone interview.

...

Scientists propose deflecting asteroids using unmanned space craft. Edward Lu, a former astronaut employed by the NASA Johnson Space Center, told the AAAS conference the gravity of a ship hovering close to the asteroid could exert a gentle tug on its trajectory, over time directing the rock away from the planet.

Ok, let's think this through.

First of all, Apophis is more than 140 metres long:

Based upon the observed brightness, Apophis's length was estimated at 415 m (1350 ft); a more refined estimate based on spectroscopic observations at NASA's Infrared Telescope Facility in Hawaii by Binzel, Rivkin, Bus, and Tokunaga (2005) is 320 m (1050 ft). Its mass is estimated to be 4.6×10^10 kg.

http://en.wikipedia.org/wiki/99942_Apophis

Now, the mass is 46 billion kilograms (let's assume it's only 1 billion for an "easier" asteroid.

Let's say we could get a payload of 1 billion kg into space (which is probably far beyond our current capacity)


Ok, now let's say we fly this ship 1 metre next to the asteroid:

The force between them will be:

GM1M2/R^2

= (G)(1X10^18)/1

= 66700000 newtons

Assuming my physics is correct, since the masses are equal, we can assume each body experiences half that force towards the other.

So the target asteroid is being pulled away from its trajectory by a force of 33350000 newtons (33 million newtons)

a=f/m

therefore a = 0.03335 m/s/s

let's say the asteroid was headed towards the centre of the earth.

In order to deviate its trajectory so it misses, we'd need to push it at least the radius of the earth away, PLUS whatever is needed for it to escape the gravitational pull of the earth.

But EVEN IF WE FORGET about the gravitational pull of the earth, we'd need to move it about 6 thousand kilometres, or 6 million metres.

d = 6 million metres
a = 0.03 m/s/s
v1 = 0


6 million metres = (1/2)(0.03)(t)(t)

t = 20000 seconds

= 5 hours of side by side travel

hm, ok it might work - i was expecting a much larger answer.


and i guess if the asteroid was headed towards the boundary of the gravitational "sink", you might need to push it by a lesser amount.

Am I doing the calculations right here?

[feedback]

Why would it have to be pulled 6 million meters? That would only be accurate if it was traveling in a completely straight line- which it isn't, because it's on an orbit around the galaxy.

[mjrpes]

Along with feedback... the farther back you mess with the asteroid's orbit, the less you have to move it, no? Like an arrow shot at a target: if you're off very slightly at launch, you'll end up way off.

[feedback]

I got a D+ in physics 201, julios

[mjrpes]

Maybe not along with feedback.

[[xeno]Julios]

the logic is that if it's aimed for the centre of the earth along one axis (let's call it x-axis), then you'd need to displace it by the radius of the earth along the y axis, by the time it closes the x-distance to the earth.

To be maximally effective, you'd position the heavy ship in such a way that the pertubation force was exactly orthogonal to the x-axis - so the pertubation force acts in the direction of the y axis.

since the axes are orthogonal, you can treat each independently, so if u find the acceleration along y-axis, you can calculate how much time is required for it to clear that distance.

[[xeno]Julios]

Why would it have to be pulled 6 million meters? That would only be accurate if it was traveling in a completely straight line- which it isn't, because it's on an orbit around the galaxy.

Let's assume that during its collision course it's essentially travelling in a straight line - the curve of a galactic orbit is so large that the approximation of a straight line is probably not too big an issue.

Of course, if we actually implemented a plan, we'd need more accurate models.


Either way though, you'd still need a net displacement of 6 million metres - without displacement we're assuming it hits centre of earth, and we're not taking earth's gravity into account.

We want an end result of the trajectory just skimming the edge of the planet, so the applied force has to displace the object by 6 million metres along the axis defined by the line from the centre of the earth to the edge at which it skims.

Now you're sorta right in that you mightn't need a big force to accomplish this, but it's a misconception to think that the effect is magnified - the key is to understand that you have a lot of time for this displacement to occur.

[[xeno]Julios]

oh and my calculations are assuming that the force is being applied for five hours.

In other words, you'd need a five hour nudge.

If you stopped nudging after 1 hour, the asteroid would still move along the y-axis, but by the time it got to the earth it wouldn't have moved as much.

I think a ring of nukes attached to the surface would do the trick, sorta like armageddon

it's only a few hundred metres - shouldn't be hard to crack

[feedback]

why not just ram the spacecraft into the asteroid, instead of flying parallel? Then you'll get your newtons in a fraction of the time. Hooray for impact!

[Eraser]

What we need to do is send a manned spaceship up there who drill a hole deep down to the core of the asteroid, place a nuclear explosive in there and push the 'splode!! button.

Then we're all safe.

Saw it myself in a movie.

[mjrpes]

this thread needs like charts and stuff.

[[xeno]Julios]

hm more details here:

http://www.metimes.com/storyview.php?St ... 1601-2872r

NASA astronaut Edward Lu said Hollywood-style solutions such as detonating a nuclear bomb in outer space to destroy an oncoming asteroid, could increase the chances of a hit on Earth.

"There is a random element to them. Things like hitting them with a bomb or flying a spacecraft into them - you just do not know what the results of that are going to be. It could make things worse."

Lu said a small "tractor spacecraft," similar in size to those used in the Apollo missions, would need to be deployed to deflect rogue asteroids.

"It would be positioned hovering in front or behind, with the intention to drag the asteroid off its trajectory with gravity," Lu said. "You can move an aircraft carrier with a tiny tug if you pull long enough."

Ok, so an apollo size craft - based on the specs here:

http://en.wikipedia.org/wiki/Apollo_spacecraft

looks like a total of 50 thousand kg max

That's 20 thousand times lighter than the one used in my calculations, and those calculations scaled down the mass of the asteroid 50 times...

I'm guessing they're thinking of tugging for months...

gonna try and email Edward Lu

[Denz]

Missing: speed of asteroid and the distance from the earth you start to pull the rock away from it's target.

[[xeno]Julios]

Missing: speed of asteroid and the distance from the earth you start to pull the rock away from it's target.

the analysis I used doesn't need that.

All I've done is show, using certain assumptions, how much time is required to displace the asteroid safely.

The shorter the distance from the earth the body is, and the faster it's going, the less time you have.

All i've done is show the minimum amount of time needed for the job.

If we're given the distance and speed, we can determine whether it's possible, since we'll know the time allowed.

[Denz]

Yeah but if you're actually going to apply your theory then you would also need: Speed and distance. Think about it, if the asteroid is the furthest distance from the earth it would take less of a pull to pull it off course then if it were closer to the earth. So therefore it would take less mass to nudge. Speed plays a part because of the time it would take to make the nudge, assuming that the rock is heading strait for the earth and not on an elliptical orbit.

[o'dium]

if>"big rock"="1" then "shit self"; else "get yhour hass to to mhas"

[JB]

if this bumper car spaceship was to explode on impact there would almost definetely be a hugely significant push in the right direction (no pun intended), greater than that of a simple bounce.

[[xeno]Julios]

Yeah but if you're actually going to apply your theory then you would also need: Speed and distance. Think about it, if the asteroid is the furthest distance from the earth it would take less of a pull to pull it off course then if it were closer to the earth. So therefore it would take less mass to nudge. Speed plays a part because of the time it would take to make the nudge, assuming that the rock is heading strait for the earth and not on an elliptical orbit.

yes - we're not disagreeing.

All I'm saying is that my analysis simply shows the minimum amount of time necessary to get the job done. It doesn't by itself say whether the job is possible.

As you say, you need the distance and speed so you actually have the time allowed.

All my analysis does is give the time required, and this is not affected by distance and speed of asteroid.

[JB]

why even calculate time? just make sure you have enough force to get the job done as fast as is humanly possible

[[xeno]Julios]

why even calculate time? j

because if you can know in advance how much time is required to displace an asteroid of a given size, you can get a sense of how feasible it is.

When beginning my calculations, I was expecting to come up with a time of a few years at least.

If that had been the case, it's hard to see how such a mission could be carried out.

I was surprised to discover that it only took five hours

(of course, that's with a MASSIVE spacecraft)

Just make sure you have enough force to get the job done as fast as is humanly possible

We're working within the constraints of gravitational tugging here, so the only way to increase the force is to make the spacecraft heavier.

[JB]

why tug when you can bounce? Then why bounce when you can detonate a rocket or two on the face of it?

[Denz]

why tug when you can bounce? Then why bounce when you can detonate a rocket or two on the face of it?

Detonation on the surface would do no good other than putting a blemish on the surface. I believe drilling a hole then a detonation is better. Where is Bruce Willis when you need him?

[[xeno]Julios]

why tug when you can bounce? Then why bounce when you can detonate a rocket or two on the face of it?

read the first post in the thread - the whole point was to critique the tugging approach.

As for your specific question, read a few posts back:

NASA astronaut Edward Lu said Hollywood-style solutions such as detonating a nuclear bomb in outer space to destroy an oncoming asteroid, could increase the chances of a hit on Earth.

"There is a random element to them. Things like hitting them with a bomb or flying a spacecraft into them - you just do not know what the results of that are going to be. It could make things worse."

Lu said a small "tractor spacecraft," similar in size to those used in the Apollo missions, would need to be deployed to deflect rogue asteroids.

"It would be positioned hovering in front or behind, with the intention to drag the asteroid off its trajectory with gravity," Lu said. "You can move an aircraft carrier with a tiny tug if you pull long enough."

[[xeno]Julios]

Detonation on the surface would do no good other than putting a blemish on the surface. I believe drilling a hole then a detonation is better. Where is Bruce Willis when you need him?

a 200 metre sized rock would probably experience significant effects with nukes. It's the unpredictability which is the problem. The tugging approach allows for fine tuning with precision.

[Denz]

Why not use a suction cup and some string to grab it and pull it from the appending doom of the earth?