Large astroid hitting earth

[piddlefoot]

''No offense, but where did you learn about E=mc^2? You seem to think it some some kind of miracle formula. ''

No , but its a law thats proven right and and can be applied to things for a guesstimate.

''K=m0 c^2(gamma-1)

where gamma=1/sqrt(1-v^2/c^2) ''

And how is this a well known law that can be applied , not even close to E=MC², to tell us about any or every atom in our universe ?
This is why E=MC² is E=MC², so we do have a general law , thats bound to the real physical world, and it can be used in many ways, but its not a miracle formular , though you could be mistaken for believing it is after making nuclea weopons from a 5 digit maths formulation realisation.
They teach E=MC² at school not K=m0 c^2(gamma-1)

where gamma=1/sqrt(1-v^2/c^2) , l cant ever remember anything on the above , as a law, but E=Mc² , so whats all that mean ?
We know what the E=MC² letters represent.

I dont think you understand , you can use E=MC² to predict an explosion even if total energy conversion isnt achieved, you can use it to tell through our spece instuments that stars have big planets spinning around them fast at close proximity and E=MC² tells us there aproximate mass, the list that E=MC² can be applied to is infinate.Supernova, E=MC² is the only thing here that makes sence. anyways......

 

[f95toli]

K is the relativistic kinetic energ
i.e.
K=mc^2-m0c^2

You can use E instead of K, but since it was the kinetic (and not the total) energy we were interested in I used K.

E=mc^2 is just a very small part of special relativity. It is the most famous part (since it looks good)but you want to actually calculate something you need to use the full theory.

Saying E=mc^2 explains everying is a bit like writing "H" and then saying "Look, I wrote down the Hamiltonian for the whole world, it explains everything!"
which would be true.
But unless you know what H is you can not use it to calculate anything.

 

[Woodchuck2000]

Originally posted by: piddlefoot

Originally posted by: Woodchuck2000

Originally posted by: piddlefoot
it can happen is the answer...

if its big enough and travelling fast enough , then yea a bloody big bang, E=MC².

That's simply doesn't follow. You're saying that if it's travelling fast enough, then E=MC^2... Except there's no velocity term in that equation.

Einstein's equation only applies with total conversion of mass into energy such as during nuclear fission. In this case the destruction happens due to a change in momentum. An asteroid weighing 2000 tonnes, travelling at 50 kilometers/second has kinetic energy (mv^2)/2 i.e. 5x10^15 joules or roughly 1 Megaton. That's gotta go somewhere on impact... And that's a small, slow asteroid.

E=MC² can be applied to everything/ explosion/ explosion prediction we ever see inside our universe. fact.Everything we have seen bar blackholes can have E=MC² applied to it for a mathematical answer, every explosion recorded in history so far has been inside the E=MC² law, yes it can tell us a total conversion and it can tell us so much more, and it tells us the faster something is moving effectivly the more mass it has, and the more mass something has the more ENERGY needed to stop or deflect the object, a basic of E=MC², and E=MC² can be applied to everything in our universe, every spec.

If it's a fact then find me a source. You won't be able to because you're totally and utterly wrong.

Einstein's equation describes an equivalence between mass and energy. It only applies when converting mass to energy or vice-versa. Try and formulate a mathematical explanation for a release of energy on impact using E=MC^2. Lets see.

Firstly, there's no velocity term. You stated that the faster an object moved the bigger the explosion. This is true but you can't explain that with Einstein's equation. Because it doesn't apply.

Secondly there's no net change in Mass or Energy. All the atoms there before the collision are there afterwards albeit somewhat scattered about. If you take your system as a closed spere about 0.1AUs in diameter around the earth, no change in energy within that system will result.

As I said, just try to find me a source for your claim. Einstein's equation is very simple but also very specific in its application and not a general cure-all as you state.

 

[Woodchuck2000]

No , but its a law thats proven right and and can be applied to things for a guesstimate.

Okay, please explain how to apply E=MC^2 to the asteroid case. And a law, applied to a situation, does not give a guesstimate. If it's applicable (which in this case it isn't) then it will give an exact answer.

 

[Woodchuck2000]

Proof of my point.
This took 10 seconds to Google. If you still don't believe me that E=MC^2 doesn't apply, please do some research before you post.

 

[Cooler]

Yes it has happened before on the coast of Yucatan peninsula its called the Chicxulub Crater. It is 10 km in diameter, releasing an estimated 4.3×10x23 joules of energy and it is about 65 million years old. Now guess what it did.

 

[piddlefoot]

'''Yes it has happened before on the coast of Yucatan peninsula its called the Chicxulub Crater. It is 10 km in diameter, releasing an estimated 4.3×10x23 joules of energy and it is about 65 million years old. Now guess what it did. '''

What law was used to come to that conclusion ?

 

[Varun]

Originally posted by: piddlefoot
'''Yes it has happened before on the coast of Yucatan peninsula its called the Chicxulub Crater. It is 10 km in diameter, releasing an estimated 4.3×10x23 joules of energy and it is about 65 million years old. Now guess what it did. '''

What law was used to come to that conclusion ?

Well, it wiped out the dinosaurs!

I would guess you could calculate the energy using conservation of energy Ek = (1/2)mv^2

I'll use the estimated number from a previous post of 20km/s v.

4.3*10^23/(0.5*20000^2)= 2.15*10^15 kg

I would also agree that E=mc^2 is not a relevant formula as we are not converting mass to energy or vice versa.

 

[piddlefoot]

ok yall need to read better, never have i stated E=MC² was the precise mathamatical formula to work out an objects speed or impact.
All l did was generalise, and state that E=MC² tells me / energy and mass/ that an object moving FAST ,E=MC² tells me the faster something goes the heavyier it is, thus the more damage on impact, so in general l stated, that using E=MC² as our guide, for it is accurate, to ESTIMATE and object/ asteriod thats at an ESTIMATED SPEED, and with an ESTIMATE on its mass, which is worked out by its velocity via E=MC², we can ESTIMATE its potential energy release upon impact,
so E=MC² does tell me if its fast enough and heavy enough and hits EARTH , enough enegy would be released to kil l most life as we know it, what the precise amount of energy be released knowone can ever know, without the objects precice wieght , and theres objects moving so fast out there we cant track them.What the prisice amount of energy needed to wipe us out , l dont know. So what do you base any other maths on without a precise weight to get an accurate answer to start with ?
So for a generallisation that people can understand , like me some of em drink beer alot, and some are very young, and can understand or relate to E=MC² , K=mc^2-m0c^2 to me means squat if i saw it, so l used a known proven law , one we all know and my comments stand E=MC² tells me if its big enough and / or fast enough then yea bloody big bag. How is that technicaly wrong ?

Dont misunderstand read it for what it is.
A generalisation , estimation of the effects, using E=MC² as a guide nothing more.....

 

[f95toli]

But it is not a valid estimate.
While it is true that "the faster something goes the heavier it is" that is not relevant.
Since the velocity v of the asteroid is much smaller than c we just use the laws of Newton
and write K=mv^2/2 for the energy.
The effect of the asteroid becoming slightly heavier due to its speed is negligable.

You can of course use K=mc^2-m0c^2 instead (as I did above) but that will be almost
exactly equal to mv^2/2 in this case.

E=mc^2 is NOT correct because not only does that include the kinetic energy of the asteroid but also ithe energy that would result from the mass of the asteroid being converted into energy.
The fact objects traveling fast do more damage is NOT related to the fact that they become heavier, it is simply because they carry more kinetic energy.

If you want to be precise and include relativistic effects (that they become heavier) you need to use mc^2-m0c^2 for the energy, NOT E=mc^2

 

[Woodchuck2000]

ok yall need to read better, never have i stated E=MC² was the precise mathamatical formula to work out an objects speed or impact.

No, you said this:

E=MC² can be applied to everything/ explosion/ explosion prediction we ever see inside our universe. fact.Everything we have seen bar blackholes can have E=MC² applied to it for a mathematical answer, every explosion recorded in history so far has been inside the E=MC² law,

Which it can't. And you still havn't produced any kind of source for your bizzare interpretation of an astoundingly simple law. E=MC² cannot be applied to combustion or non-relativistic collisions in any meaningful way. Then you said:

All l did was generalise, and state that E=MC² tells me / energy and mass/ that an object moving FAST ,E=MC² tells me the faster something goes the heavyier it is, thus the more damage on impact, so in general l stated, that using E=MC² as our guide, for it is accurate, to ESTIMATE...

How exactly are you applying it? There's no velocity term in the equation...

so E=MC² does tell me if its fast enough and heavy enough and hits EARTH , enough enegy would be released to kil l most life as we know it, what the precise amount of energy be released knowone can ever know, without the objects precice wieght , and theres objects moving so fast out there we cant track them.

How exactly does E=MC² tell you this? Assume a spherical asteroid 10Km across of density 8000Kg/M^3 travelling at 30Km/s. That's all the data you should need, show your working please.

E=MC² tells me if its big enough and / or fast enough then yea bloody big bag. How is that technicaly wrong?

That's not only technically wrong, it's very actually wrong. E=MC² simply does not tell you that. Or prove me wrong, and explain how (showing a source).

I challenge you to find some kind of source for your freakish claims.

Jeez, just admit when you're wrong...

 

[Armitage]

Originally posted by: scottish144
It also depends on its make up. If the asteroid is from the outer (Kuiper) Belt, or was a comet, than it would be mostley ice and a lot of it would burn up in the atmosphere.

Above a certain size & speed, the atmosphere isn't really relavent anymore. The velocity of a potential impactor starts at about 11Km/s, and 60+ Km/s isn't unreasonable. Let's pick 25 Km/s, and say it starts hitting appreciable atmosphere at about 150Km up - you've got about 6 seconds to melt that 1Km diameter ice cube. You can't transfer heat through it fast enough.

And even if it could, so what? All of that energy has to go somewhere, whether the impactor actually hits the ground or not. Dumping 327,249 megatons of energy (from the badastronomy link - excellent site by the way) into the earth is a bad idea no matter how you do it. Actually, I might argue that you'd be better off if a big chunk of it does impact and dump the energy directly into the earth, rather then all of it into the atmosphere. The area directly below the impact is toast regardless, might as well absorb as much energy there as possible.

If its mostly rock or metal, then we're screwed. Additionally, screw texas. An asteroid 1 km long would just about screw the world on its own. Consider an asteroid the size of a soccer ball can make a crater hundreds of feet deep and more than a mile wide. Scale up from there. Also, if a 1 km asteroid landed in the ocean, worldwide tsunami.

 

[AbsolutDealage]

Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

 

[Soccerman06]

Originally posted by: AbsolutDealage
Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

I hope you meen Ke=mv^2.

The thing is with asteroids, meteors, comets and whatnot is if one was going to hit Earth, and if we detected and launched some device as early as say 10 milion miles away, we would easily be able to deflect it out of the way. Even if you nudge it less than 1 degree, that is still way more than enough deflect it out of Earths path.

 

[f95toli]

I think he means K=mv^2/2 which the formula for the kinetic energy.

 

[Armitage]

Originally posted by: Soccerman06

Originally posted by: AbsolutDealage
Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

I hope you meen Ke=mv^2.

The thing is with asteroids, meteors, comets and whatnot is if one was going to hit Earth, and if we detected and launched some device as early as say 10 milion miles away

10 million miles isn't that far on this scale - only about 3.8 degrees of the earths orbital path. If that's the soonest we can intercept, I suspect we're toast. If you have less then the orbital period of the asteroid to work with before impact you want to do your manuever about 180 degrees away from the impact point for in-plane manuevers, or 90 degrees away for out of plane (inclination) manuevers. Out of plane takes alot more energy, so forget about it for the moment. In this in-plane case case you are hoping to raise or lower the orbit of the impactor by at least half the radius of the earth.

Let's pick a very simple case - an asteroid with a aphelion @ 300,000,000 Km and perihelion @ the earth - 150,000,000 Km. So let's say we want to raise the perihelion so the object no longer crosses earth orbit - make it an even 13,000 Km.

The energy of the asteroid orbit is: E = -u/2a = 1.3271544e11/(2*225000000) = -294.9232 km^2/s^2

Orbital velocity is given by E = V^2/2 - u/r

For the initial orbit:
Vaphelion = 17.173328157 Km/s
Vperihelion = 34.34665631 Km/s

for the raised orbit:
Vaphelion = 17.173824254 Km/s
Vperihelion = 34.34467197 Km/s

So the dV needed at aphelion is 0.4961 meters/second

Let's assume a 1Km diameter spherical stone asteroid. Given a density of 2 g/cm^3 its mass will be approximately 8.3776e12 Kg

So the energy needed to give it that dV will be 1/2 M V^2 = 1.0309e12 kg m^2/s^2

This works out to 0.2464 kilotons

Quite a bit less then I was expecting - I better check my math . However, when you consider bombs, I suspect that setting off a bomb next to an asteroid will be a very inefficient & unpredictable way to change the velocity of a rock in space. If the asteroid has a lower perihelion, the energy is higher, so anything you do with it will take more energy as well.

Now if you have several revs to deal with it, you can get away with less dV. You don't have to raise the orbit by that amount, but simply alter the period enough so that even though it crosses earths orbit, the timing is changed and the earth isn't there at the same time as the asteroid. This is left as an exercise for the reader

, we would easily be able to deflect it out of the way. Even if you nudge it less than 1 degree, that is still way more than enough deflect it out of Earths path.

One degree of what?

 

[Gibsons]

One reason your number might be apparently low is because that's the amount of energy needed, which isn't the same as the size of the warhead you'd need. How big of a real world warhead would you need to actually deliver .25 kilotons to the target?

 

[Armitage]

Originally posted by: Gibsons
One reason your number might be apparently low is because that's the amount of energy needed, which isn't the same as the size of the warhead you'd need. How big of a real world warhead would you need to actually deliver .25 kilotons to the target?

Yep thtat's what I was getting at when I said that bombs are probably a pretty inefficient way to move an object in space. No atmosphere to transmit the shockwave & such. I figure a good bit of the impulse would come simply from boiling material off of the face that's toward the explosion.

Might look for some stuff on the Orion project and see how their math worked out.

 

[Soccerman06]

Originally posted by: Armitage

Originally posted by: Soccerman06

Originally posted by: AbsolutDealage
Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

I hope you meen Ke=mv^2.

The thing is with asteroids, meteors, comets and whatnot is if one was going to hit Earth, and if we detected and launched some device as early as say 10 milion miles away

10 million miles isn't that far on this scale - only about 3.8 degrees of the earths orbital path. If that's the soonest we can intercept, I suspect we're toast. If you have less then the orbital period of the asteroid to work with before impact you want to do your manuever about 180 degrees away from the impact point for in-plane manuevers, or 90 degrees away for out of plane (inclination) manuevers. Out of plane takes alot more energy, so forget about it for the moment. In this in-plane case case you are hoping to raise or lower the orbit of the impactor by at least half the radius of the earth.

Let's pick a very simple case - an asteroid with a aphelion @ 300,000,000 Km and perihelion @ the earth - 150,000,000 Km. So let's say we want to raise the perihelion so the object no longer crosses earth orbit - make it an even 13,000 Km.

The energy of the asteroid orbit is: E = -u/2a = 1.3271544e11/(2*225000000) = -294.9232 km^2/s^2

Orbital velocity is given by E = V^2/2 - u/r

For the initial orbit:
Vaphelion = 17.173328157 Km/s
Vperihelion = 34.34665631 Km/s

for the raised orbit:
Vaphelion = 17.173824254 Km/s
Vperihelion = 34.34467197 Km/s

So the dV needed at aphelion is 0.4961 meters/second

Let's assume a 1Km diameter spherical stone asteroid. Given a density of 2 g/cm^3 its mass will be approximately 8.3776e12 Kg

So the energy needed to give it that dV will be 1/2 M V^2 = 1.0309e12 kg m^2/s^2

This works out to 0.2464 kilotons

Quite a bit less then I was expecting - I better check my math . However, when you consider bombs, I suspect that setting off a bomb next to an asteroid will be a very inefficient & unpredictable way to change the velocity of a rock in space. If the asteroid has a lower perihelion, the energy is higher, so anything you do with it will take more energy as well.

Now if you have several revs to deal with it, you can get away with less dV. You don't have to raise the orbit by that amount, but simply alter the period enough so that even though it crosses earths orbit, the timing is changed and the earth isn't there at the same time as the asteroid. This is left as an exercise for the reader

, we would easily be able to deflect it out of the way. Even if you nudge it less than 1 degree, that is still way more than enough deflect it out of Earths path.

One degree of what?

1 degree if you think about the earth and the meteor/asteroid on an xy axis.

But I ment 10 billion sorry bout that.

There is another thing you can do besides using nukes. Basically launch many rockets into space and put them on an intercept course with the meteor/asteroid. Once they have gotten into the proper position, they would shoot out a tow hook which would ancor into the asteroid. Then powerful chemical jets would propel the rock out of our way. There would have to be many rockets depending on which direction you would want the rock to go, but the farther away, the less required. This solution could only work if the rock was not rotating at a major speed.

 

[Armitage]

Originally posted by: Soccerman06

Originally posted by: Armitage

Originally posted by: Soccerman06

Originally posted by: AbsolutDealage
Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

I hope you meen Ke=mv^2.

The thing is with asteroids, meteors, comets and whatnot is if one was going to hit Earth, and if we detected and launched some device as early as say 10 milion miles away

10 million miles isn't that far on this scale - only about 3.8 degrees of the earths orbital path. If that's the soonest we can intercept, I suspect we're toast. If you have less then the orbital period of the asteroid to work with before impact you want to do your manuever about 180 degrees away from the impact point for in-plane manuevers, or 90 degrees away for out of plane (inclination) manuevers. Out of plane takes alot more energy, so forget about it for the moment. In this in-plane case case you are hoping to raise or lower the orbit of the impactor by at least half the radius of the earth.

Let's pick a very simple case - an asteroid with a aphelion @ 300,000,000 Km and perihelion @ the earth - 150,000,000 Km. So let's say we want to raise the perihelion so the object no longer crosses earth orbit - make it an even 13,000 Km.

The energy of the asteroid orbit is: E = -u/2a = 1.3271544e11/(2*225000000) = -294.9232 km^2/s^2

Orbital velocity is given by E = V^2/2 - u/r

For the initial orbit:
Vaphelion = 17.173328157 Km/s
Vperihelion = 34.34665631 Km/s

for the raised orbit:
Vaphelion = 17.173824254 Km/s
Vperihelion = 34.34467197 Km/s

So the dV needed at aphelion is 0.4961 meters/second

Let's assume a 1Km diameter spherical stone asteroid. Given a density of 2 g/cm^3 its mass will be approximately 8.3776e12 Kg

So the energy needed to give it that dV will be 1/2 M V^2 = 1.0309e12 kg m^2/s^2

This works out to 0.2464 kilotons

Quite a bit less then I was expecting - I better check my math . However, when you consider bombs, I suspect that setting off a bomb next to an asteroid will be a very inefficient & unpredictable way to change the velocity of a rock in space. If the asteroid has a lower perihelion, the energy is higher, so anything you do with it will take more energy as well.

Now if you have several revs to deal with it, you can get away with less dV. You don't have to raise the orbit by that amount, but simply alter the period enough so that even though it crosses earths orbit, the timing is changed and the earth isn't there at the same time as the asteroid. This is left as an exercise for the reader

, we would easily be able to deflect it out of the way. Even if you nudge it less than 1 degree, that is still way more than enough deflect it out of Earths path.

One degree of what?

1 degree if you think about the earth and the meteor/asteroid on an xy axis.

But I ment 10 billion sorry bout that.

Sure, what's a few orders of magnitude between friends! But how do you plan to spot an earth impacting asteroid or comet at a distance more then double the radius of Neptunes orbit?

There is another thing you can do besides using nukes. Basically launch many rockets into space and put them on an intercept course with the meteor/asteroid. Once they have gotten into the proper position, they would shoot out a tow hook which would ancor into the asteroid. Then powerful chemical jets would propel the rock out of our way. There would have to be many rockets depending on which direction you would want the rock to go, but the farther away, the less required. This solution could only work if the rock was not rotating at a major speed.

Sure, that'll work.
Using an Isp of 465 which is about what the space shuttle main engine gets and is pretty good for a chemical rocket I get a mass ratio of 1.00011 for the 0.5 m/s dV above.

So you just need 0.00011 times the mass of the asteroid in fuel - that's just shy of 1 billlion Kg Any plans on how to get that much gas up there?

Other alternatives include solar sails, solar panels with ion engines, etc. But these are very low thrust, so you beter get there early!

 

[everman]

http://neo.jpl.nasa.gov/cgi-bin/ip?2.2e-02 :Q

 

[Soccerman06]

Originally posted by: Armitage

Originally posted by: Soccerman06

Originally posted by: Armitage

Originally posted by: Soccerman06

Originally posted by: AbsolutDealage
Quit feeding this guy. Seriously.

If there were ever an argument for being able to hide posts from a specific user, it would be him. Just read around his HT posts and you will see what I mean.

I'll shutup now, b/c this forum is not for flaming.

-----

Back on topic, everyone else is right piddle. You cannot use E=MC² in this case, because you are not going to be turning the asteroid into energy. What you are essentially calculating is the energy of the asteroid if it were travelling at c... which means nothing. What you really want is the kinetic energy of the asteroid, which any high school physics book will tell you, is K=mv^2/2.

I hope you meen Ke=mv^2.

The thing is with asteroids, meteors, comets and whatnot is if one was going to hit Earth, and if we detected and launched some device as early as say 10 milion miles away

10 million miles isn't that far on this scale - only about 3.8 degrees of the earths orbital path. If that's the soonest we can intercept, I suspect we're toast. If you have less then the orbital period of the asteroid to work with before impact you want to do your manuever about 180 degrees away from the impact point for in-plane manuevers, or 90 degrees away for out of plane (inclination) manuevers. Out of plane takes alot more energy, so forget about it for the moment. In this in-plane case case you are hoping to raise or lower the orbit of the impactor by at least half the radius of the earth.

Let's pick a very simple case - an asteroid with a aphelion @ 300,000,000 Km and perihelion @ the earth - 150,000,000 Km. So let's say we want to raise the perihelion so the object no longer crosses earth orbit - make it an even 13,000 Km.

The energy of the asteroid orbit is: E = -u/2a = 1.3271544e11/(2*225000000) = -294.9232 km^2/s^2

Orbital velocity is given by E = V^2/2 - u/r

For the initial orbit:
Vaphelion = 17.173328157 Km/s
Vperihelion = 34.34665631 Km/s

for the raised orbit:
Vaphelion = 17.173824254 Km/s
Vperihelion = 34.34467197 Km/s

So the dV needed at aphelion is 0.4961 meters/second

Let's assume a 1Km diameter spherical stone asteroid. Given a density of 2 g/cm^3 its mass will be approximately 8.3776e12 Kg

So the energy needed to give it that dV will be 1/2 M V^2 = 1.0309e12 kg m^2/s^2

This works out to 0.2464 kilotons

Quite a bit less then I was expecting - I better check my math . However, when you consider bombs, I suspect that setting off a bomb next to an asteroid will be a very inefficient & unpredictable way to change the velocity of a rock in space. If the asteroid has a lower perihelion, the energy is higher, so anything you do with it will take more energy as well.

Now if you have several revs to deal with it, you can get away with less dV. You don't have to raise the orbit by that amount, but simply alter the period enough so that even though it crosses earths orbit, the timing is changed and the earth isn't there at the same time as the asteroid. This is left as an exercise for the reader

, we would easily be able to deflect it out of the way. Even if you nudge it less than 1 degree, that is still way more than enough deflect it out of Earths path.

One degree of what?

1 degree if you think about the earth and the meteor/asteroid on an xy axis.

But I ment 10 billion sorry bout that.

Sure, what's a few orders of magnitude between friends! But how do you plan to spot an earth impacting asteroid or comet at a distance more then double the radius of Neptunes orbit?

There is another thing you can do besides using nukes. Basically launch many rockets into space and put them on an intercept course with the meteor/asteroid. Once they have gotten into the proper position, they would shoot out a tow hook which would ancor into the asteroid. Then powerful chemical jets would propel the rock out of our way. There would have to be many rockets depending on which direction you would want the rock to go, but the farther away, the less required. This solution could only work if the rock was not rotating at a major speed.

Sure, that'll work.
Using an Isp of 465 which is about what the space shuttle main engine gets and is pretty good for a chemical rocket I get a mass ratio of 1.00011 for the 0.5 m/s dV above.

So you just need 0.00011 times the mass of the asteroid in fuel - that's just shy of 1 billlion Kg Any plans on how to get that much gas up there?

Other alternatives include solar sails, solar panels with ion engines, etc. But these are very low thrust, so you beter get there early!

I havent thought of getting it there, thats not my job, I think of things, others figure out the problems

Solar sails would work but the problem is it would have to be soo big to counteract the weight of the rock that it would be easier to just use rockets, and you cant put it at an angle because there would be too little oomph from other sources, so the only real direction you could use would be slowing the rock down, but that also creates a problem of it might hit the earth at a later point in time. Ion drives would be an intriging thought but they need a nuclear reactor to be of any use, because batteries would die way too soon and solar panels would barely put enough energy off at that great of distance.

Nasa has been thinking up ways to use nukes as a propulsion device, but that would require large amounts of testing before we could actually use one. But think about using many small nukes as a way to give a meteor a nudge in the right direction. They also are thinking about using plasma as a propulsion but how do you heat up gasses above millions of degrees without using a nuke and for a very long time, in space? :laugh: someone stop my babbling

 

[Gilby]

Originally posted by: piddlefoot
So for a generallisation that people can understand , like me some of em drink beer alot, and some are very young, and can understand or relate to E=MC² , K=mc^2-m0c^2 to me means squat if i saw it, so l used a known proven law , one we all know and my comments stand E=MC² tells me if its big enough and / or fast enough then yea bloody big bag. How is that technicaly wrong ?.

Hrm.

I don't think you really need to worry about scientific laws.

Rather, you need to study and apply an old proverb:

'Tis better to remain silent and be thought a fool that to speak and remove all doubt.

For the record, there really is no doubt at this point.

 

[Woodchuck2000]

Originally posted by: Gilby

Originally posted by: piddlefoot
So for a generallisation that people can understand , like me some of em drink beer alot, and some are very young, and can understand or relate to E=MC² , K=mc^2-m0c^2 to me means squat if i saw it, so l used a known proven law , one we all know and my comments stand E=MC² tells me if its big enough and / or fast enough then yea bloody big bag. How is that technicaly wrong ?.

Hrm.

I don't think you really need to worry about scientific laws.

Rather, you need to study and apply an old proverb:

'Tis better to remain silent and be thought a fool that to speak and remove all doubt.

For the record, there really is no doubt at this point.

 

[Gibsons]

Originally posted by: Soccerman06

Solar sails would work but the problem is it would have to be soo big to counteract the weight of the rock that it would be easier to just use rockets, and you cant put it at an angle because there would be too little oomph from other sources, so the only real direction you could use would be slowing the rock down, but that also creates a problem of it might hit the earth at a later point in time. Ion drives would be an intriging thought but they need a nuclear reactor to be of any use, because batteries would die way too soon and solar panels would barely put enough energy off at that great of distance.

Nasa has been thinking up ways to use nukes as a propulsion device, but that would require large amounts of testing before we could actually use one. But think about using many small nukes as a way to give a meteor a nudge in the right direction. They also are thinking about using plasma as a propulsion but how do you heat up gasses above millions of degrees without using a nuke and for a very long time, in space? :laugh: someone stop my babbling

A solar sail is easy, you just attach it and forget about it. They weigh a lot less and are easier to handle than large solid rocket boosters. But like Armitage pointed out, you do need to get there real early. Spin could be a problem. And the solar sail won't necessarily just slow it down, it depends on the direction the meteor is traveling relative to the sun. Speeding it up could be just as good as moving it aside.

The plasma engines just use onboard electricity to do the heating. They don't have tons of electricity, so they heat up very small amounts at a time. You get low impulse but very high efficiency.