So could this be the end of us all??? Deep Impact
#33
I would thought the problem would be the dust in the atmosphere rather than just the impact! so for UK to survive you weant it to hit New Zealand.. although the idea of moving to New Zealand and it hitting the UK seems more appealing.
Mind you if it hits the weather might be totally different everywhere...
JGM
Mind you if it hits the weather might be totally different everywhere...
JGM
#34
Here's a thought:
Assume the asteroid has a mean density somehwere near the mean density of earth (which isn't a wild assumption, as it probably formed out of the same material as earth, at about the same time) that is 5000kg/m^3. Its total volume (assuming it is spherical, which again isn't a bad thing to do) is given by:
4/3Pi(r)^3 = 4/3Pi(2000)^3 = 3 x 10^10 m^3
Therefore its mass is = density (rho) x volume = 1 x 10^14
I'm not sure of its velocity at impact, its orbital velocity (which is quoted at 28km/s I think) is not its impact velocity, but the MINIMUM velocity of impact will be 11.2Km/s (the escape velocity of Earth). The energy of the impact will then be given by:
1/2mass x velocity^2 = 1/2 x (1 x 10^14) x (11200)^2 = 6 x 10^21 Joules.....
It is worth noting that the amount of energy release by the bomb dropped on Hiroshima was around the order 1 x 10^13 joules. I.E this will be 10million times as big.
My advice if it is going to hit: Stick a paper bag on your head.
[Edited by Dan B - 7/25/2002 1:48:29 PM]
Assume the asteroid has a mean density somehwere near the mean density of earth (which isn't a wild assumption, as it probably formed out of the same material as earth, at about the same time) that is 5000kg/m^3. Its total volume (assuming it is spherical, which again isn't a bad thing to do) is given by:
4/3Pi(r)^3 = 4/3Pi(2000)^3 = 3 x 10^10 m^3
Therefore its mass is = density (rho) x volume = 1 x 10^14
I'm not sure of its velocity at impact, its orbital velocity (which is quoted at 28km/s I think) is not its impact velocity, but the MINIMUM velocity of impact will be 11.2Km/s (the escape velocity of Earth). The energy of the impact will then be given by:
1/2mass x velocity^2 = 1/2 x (1 x 10^14) x (11200)^2 = 6 x 10^21 Joules.....
It is worth noting that the amount of energy release by the bomb dropped on Hiroshima was around the order 1 x 10^13 joules. I.E this will be 10million times as big.
My advice if it is going to hit: Stick a paper bag on your head.
[Edited by Dan B - 7/25/2002 1:48:29 PM]
#40
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Its ok everyone my dad has spoken,
He says it's easy to knock a large 2km wide chunk of rock into tiny harmless pieces and we don't even need to land on it like Bruce Willis did.
Nice one dad, when did you start working for NASA?????
He says it's easy to knock a large 2km wide chunk of rock into tiny harmless pieces and we don't even need to land on it like Bruce Willis did.
Nice one dad, when did you start working for NASA?????
#43
Along way away.............................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .................................................. .............................................
#44
An impact on land is bad, but it's much worse if it hits the sea. It will cause tsunami in all costal regions around the impact. Boil millions of tons of water and vaporise the ocean floor allowing sea water to rush in and cause further boiling. All that water vapour in the atmostohere is not going to do us much good.
An ocean impact has much further reaching devistation than land impact.
Since the earth surface is 75% sea / 25% land it a good chance is doing in the drink.
I agree the chances of it actually hitting the earth are small, but this is the biggest known threat like this in recent times.
An ocean impact has much further reaching devistation than land impact.
Since the earth surface is 75% sea / 25% land it a good chance is doing in the drink.
I agree the chances of it actually hitting the earth are small, but this is the biggest known threat like this in recent times.
#46
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hey come on, if it was that bad and we were all gonna die do you think they would have told us???
I mean surely we would all run around in a mass panic?
well maybe in 17 years time
I mean surely we would all run around in a mass panic?
well maybe in 17 years time
#47
Dan,
I think there may be a few errors in your computations...
The volume of a sphere is 4/3 pi r cubed. If it has a 2 km diameter than the radius is 1km, not 2000 as you have shown. This reduces the volume by a factor of eight down to 4.189e09 cubic metres. Assuming that your average density is correct then that is 2.09e13 kilogrammes.
Your estimation of impact velocity is also curious; there is no reason for it to be travelling at least at escape velocity, it isn't trying to escape. Given that it is estimated to be travelling at 28,000 m/s just before it hits, it stands to reason that ALL the kinetic energy would need to be absorbed by the planet (which includes the atmosphere). That makes the energy 6.157e21 joules.
The two errors pretty much cancel out and thus the net effect is that the energy is roughly the same (certainly within the same order of magnitude).
Given your energy figure for the Hiroshima bomb, 1e13 joules, this being 6e21, it would make it six hundred million times larger.
Deflection at the earliest time is generally the most effective means. Other influences will also affect its trjectory, such as gravitational influences on space-time, variances in the ISM and heliosphere, collisions with other stray matter (for example from the asteroid belt). These factors need to be taken into account, so it would equally be foolish to act too soon as the original computations may be affected in a yet to be determined way which could alter the trajectory in an unfavourable way.
Cheers,
Pat.
I think there may be a few errors in your computations...
The volume of a sphere is 4/3 pi r cubed. If it has a 2 km diameter than the radius is 1km, not 2000 as you have shown. This reduces the volume by a factor of eight down to 4.189e09 cubic metres. Assuming that your average density is correct then that is 2.09e13 kilogrammes.
Your estimation of impact velocity is also curious; there is no reason for it to be travelling at least at escape velocity, it isn't trying to escape. Given that it is estimated to be travelling at 28,000 m/s just before it hits, it stands to reason that ALL the kinetic energy would need to be absorbed by the planet (which includes the atmosphere). That makes the energy 6.157e21 joules.
The two errors pretty much cancel out and thus the net effect is that the energy is roughly the same (certainly within the same order of magnitude).
Given your energy figure for the Hiroshima bomb, 1e13 joules, this being 6e21, it would make it six hundred million times larger.
Deflection at the earliest time is generally the most effective means. Other influences will also affect its trjectory, such as gravitational influences on space-time, variances in the ISM and heliosphere, collisions with other stray matter (for example from the asteroid belt). These factors need to be taken into account, so it would equally be foolish to act too soon as the original computations may be affected in a yet to be determined way which could alter the trajectory in an unfavourable way.
Cheers,
Pat.
#48
Ta Pat: Thought is was 2km radius..... ooops.
There are a few reasons why I didn't take the impact velocity to be 28km/s. Earths velocity, asteroids velocity, acc^ns due to the thousands of gravitation bodies it will pass, but a MINIMUM impact velcity of 11.2km/s is valid. Escape velocity is derived from equating the gravitational potential energy to the kinetic energy and solving for v. If the earth were the only body in the universe, and the asteroid started at infinity, then it would hit the earth at 11.2km/s.....
The point is that we would be screwed if it did hit.
I agree, deflection is the best solution we have, lucky we can now land on comets and stuff (ref: Giotto) so we could apply some force to one if it were on its way.
Personally I would extend the warp field of the enterprise to reduced the asteriods effective mass, and then use its tractor beam...... Or just get Q to do it
There are a few reasons why I didn't take the impact velocity to be 28km/s. Earths velocity, asteroids velocity, acc^ns due to the thousands of gravitation bodies it will pass, but a MINIMUM impact velcity of 11.2km/s is valid. Escape velocity is derived from equating the gravitational potential energy to the kinetic energy and solving for v. If the earth were the only body in the universe, and the asteroid started at infinity, then it would hit the earth at 11.2km/s.....
The point is that we would be screwed if it did hit.
I agree, deflection is the best solution we have, lucky we can now land on comets and stuff (ref: Giotto) so we could apply some force to one if it were on its way.
Personally I would extend the warp field of the enterprise to reduced the asteriods effective mass, and then use its tractor beam...... Or just get Q to do it
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