Atomic/Nuclear Bombs in Space: Oxygen Required?

[Morbius]

So, your statement is to explode nuke warhead carried on interceptor missile after hit?.

Joe,

Not at all. You don't have to hit the asteroid. You want to explode it a few miles above the surface, so you arrange for the missile's trajectory to pass just in front of, or just behind ( depending on what new orbit is desired ) and you detonate the warhead at the proper time.

There's no "hitting" the asteroid involved.

Dr. Gregory Greenman

 

[Drakkith]

So, your statement is to explode nuke warhead carried on interceptor missile after hit?

If so:
• What deceleration that warhead will experience after hit before explosion?
• And admissible deceleration for reliable operation of mechanisms?

And I doubt that existing MCBMs can be used. I am sure that new interceptor should be developed. That would not be a problem on base of just today's technology. But that will be a new missile much more agile than MCBM.

As Morbius said there is no need to hit the asteroid. Current warheads are already equipped with the ability to detonate at different altitudes either based on time or proximity/impact. Actually, the cruise missiles I work on have impact fuzes in the nose for a ground burst detonation option. It is trivial to time the detonation right down to milliseconds or less.

 

[Joseph Chikva]

Joe,

Not at all. You don't have to hit the asteroid. You want to explode it a few miles above the surface, so you arrange for the missile's trajectory to pass just in front of, or just behind ( depending on what new orbit is desired ) and you detonate the warhead at the proper time.

There's no "hitting" the asteroid involved.

Dr. Gregory Greenman

Ok, Greg,
I understood that from your second another post. Thanks.
Very little amount of energy can be delivered in that case. Was that calculated?

 

[Joseph Chikva]

As Morbius said there is no need to hit the asteroid. Current warheads are already equipped with the ability to detonate at different altitudes either based on time or proximity/impact. Actually, the cruise missiles I work on have impact fuzes in the nose for a ground burst detonation option. It is trivial to time the detonation right down to milliseconds or less.

I know something about warheads, fuses, as well as proximity, point detonating, delay modes. Thanks.
As some times ago proposed to my country’s Government to produce modern fuses here in Georgia http://www.fuchs.co.za/technology/

 

[Drakkith]

Ok, Greg,
I understood that from your second another post. Thanks.
Very little amount of energy can be delivered in that case. Was that calculated?

Why would very little energy be delivered? I'd venture a guess and say that if you got really close you could get 40%+ of the energy of the nuke transferred to the asteroid.

 

[Joseph Chikva]

Why would very little energy be delivered? I'd venture a guess and say that if you got really close you could get 40%+ of the energy of the nuke transferred to the asteroid.

From what do you get numbers?
Not only in this case.
Radiation propagated to all directions (4pi steradian).
Distance to asteroid a few miles.
Bob mentioned asteroid’s mass M = 8.6 x 10^7 tonnes (metric tons).
It corresponds ~1.1 x 10^7 m3 of volume if asteroid is from iron and about 10^8 if from ice.
If asteroid spherical the diameter should has an order of a few hundred meters.
And if even if 100% radiation energy absorption much less than 40%. But some energy will be reflected.

 

[Drakkith]

Hrmm, I guess I should have been more specific. I meant that if a nuke is detonated above the surface of an asteroid, at least 50% of the blast will simply go into space above. The other 50% should interact with the asteroid somehow. I took 10% away simply because I don't think exactly 50% will hit the asteroid due to the altitude of the blast. Some should get radiated close to the asteroid but barely miss it.

 

[Joseph Chikva]

Hrmm, I guess I should have been more specific. I meant that if a nuke is detonated above the surface of an asteroid, at least 50% of the blast will simply go into space above. The other 50% should interact with the asteroid somehow. I took 10% away simply because I don't think exactly 50% will hit the asteroid due to the altitude of the blast. Some should get radiated close to the asteroid but barely miss it.

If you have an interest you can estimate that on base of provided by me data.
That is an elementary geometry exercise.
I am too lazy for that.
But think that on a few orders lower than you guess.

 

[Drakkith]

I'm not talking about the actual amount of energy absorbed or whatever by the asteroid. I'm just referring to the amount of the blast that should impact it. Please tell me someone understands what I'm saying.

 

[DaveC426913]

I'm not talking about the actual amount of energy absorbed or whatever by the asteroid. I'm just referring to the amount of the blast that should impact it. Please tell me someone understands what I'm saying.

Certainly. Of the expanding sphere that is the blast, a cone-shaped portion will intersect the asteroid, imparting energy.

The percentage of the sphere that is the cone is determined by the distance from cone apex to cone base (i.e. nuke's point of detonation to asteroid surface).

At distance zero, the cone is effectively a plane, making it 50% of the sphere.

 

[Drakkith]

That is exactly what I mean. Thank you Dave.

 

[Joseph Chikva]

That is exactly what I mean. Thank you Dave.

That is not a matter what you meant.
As by Morbius Dr. Dearborn who really was engaged with this problem told about explosion not near surface but at a few miles away.
So, much less than you guess.

 

[Drakkith]

Sure, depending on the size of it.

 

[DaveC426913]

That is not a matter what you meant.
As by Morbius Dr. Dearborn who really was engaged with this problem told about explosion not near surface but at a few miles away.
So, much less than you guess.

If the explosion is a few miles away from an asteroid that is a few miles in diameter, then the cone that intersects the asteroid has an apex angle on the order of 60 degrees.

One would then be able to calculate the area of the base of 60 degree cone as a percent of the area of the whole sphere.

 

[Morbius]

That is not a matter what you meant.
As by Morbius Dr. Dearborn who really was engaged with this problem told about explosion not near surface but at a few miles away.
So, much less than you guess.

Joe,

Why the big concern about the percentage of the bomb's energy that is delivered?

What counts is that the amount that is delivered is enough to alter the orbit.
It's not like the bomb is stretched for delivering the energy.

Dr. Dearborn is calculating using a device with a yield in the kilotons. That hardly taxes the state of the art in weapons, as some devices go into the megatons.

Dr. Gregory Greenman

 

[Joseph Chikva]

Joe,

Why the big concern about the percentage of the bomb's energy that is delivered?

What counts is that the amount that is delivered is enough to alter the orbit.
It's not like the bomb is stretched for delivering the energy.

Dr. Dearborn is calculating using a device with a yield in the kilotons. That hardly taxes the state of the art in weapons, as some devices go into the megatons.

Dr. Gregory Greenman

I have not any concern. Simply interesting. Thanks.

 

[Joseph Chikva]

If the explosion is a few miles away from an asteroid that is a few miles in diameter, then the cone that intersects the asteroid has an apex angle on the order of 60 degrees.

One would then be able to calculate the area of the base of 60 degree cone as a percent of the area of the whole sphere.

Sorry, I missed your comment.
If we are interested the share of energy of nuke device that can be absorbed by asteroid, we have a task to calculate not area but ratio between apex angle of cone to whole apex angle of sphere (4pi).
And I think that typical asteroid's linear dimension not a few miles but on an order lower (few hundreds meters). On base of mass estimation provided by Bob.

 

[DaveC426913]

If we are interested the share of energy of nuke device that can be absorbed by asteroid, we have a task to calculate not area but ratio between apex angle of cone to whole apex angle of sphere (4pi).

That's what I said. To wit:

What fraction (or percent, or, if you wish, ratio) is the area of the cone compared to the area of the entire sphere.

 

[Joseph Chikva]

That's what I said. To wit:

What fraction (or percent, or, if you wish, ratio) is the area of the cone compared to the area of the entire sphere.

Oh sorry. Thanks.

 

[BobG]

Except for the '92-93 Observer mission, all the http://en.wikipedia.org/wiki/Exploration_of_Mars#Timeline_of_Mars_exploration" with a successful launch arrived at Mars. I doubt the landing issues that plagued a couple of the missions are relevant to the delivery of a nuclear weapon targeted at an astronomical body with no atmosphere and negligible gravity.

I think you're making a mistake simplifying it. We're pretty familiar with planetary probe procedures and yet we still have a high screw up rate. We have very littel experience landing on small tumbling bodies whose orbits are not nice, neat and low eccentricity and whose delta v is quite different from Earth's.

I think the ways it's more difficult outnumber the ways it's easier.

The orbit part isn't that tough. The tumbling part could be bad.

The rotation of a tumbling asteroid can look very ugly since the asteroid doesn't have enough gravity to smooth out the features, but there has to be two points on the surface that line up with the asteroid's angular momentum vector and that would be a relatively stable place to land if the spacecraft 's rotation rate matched the asteroid's rotation rate - at least if that were the only problem.

The asteroid's odd shape and low mass means the angular momentum vector could have a fairly high rate of precession, and, depending on the composition of the asteroid, you could have some nutation, as well (probably not much, since I think an asteroid of any size will be pretty solid and rigid).

However, you could situate the spacecraft very near the asteroid's surface without landing since the asteroid exerts practically no gravitational pull (if you're talking about aphophis, the mass is only 2.7x10^10kg instead of the mass Bob tossed out there, plus the velocity of apophis when it's close to Earth is about 28,500 m/sec). The spacecraft only has to be far enough away to make sure the 'highest' points of a 320 meter asteroid miss it.

So the 'landing' part isn't as big a deal as one might think.

 

[BobG]

Joe,

Why the big concern about the percentage of the bomb's energy that is delivered?

What counts is that the amount that is delivered is enough to alter the orbit.
It's not like the bomb is stretched for delivering the energy.

Dr. Dearborn is calculating using a device with a yield in the kilotons. That hardly taxes the state of the art in weapons, as some devices go into the megatons.

Dr. Gregory Greenman

Because you don't explain how much of that total energy is converted to kinetic energy (or how it's converted); how the energy from an electromagnetic wave can transfer momentum to the object it hits.

 

[Morbius]

Because you don't explain how much of that total energy is converted to kinetic energy (or how it's converted); how the energy from an electromagnetic wave can transfer momentum to the object it hits.

Bob,

That's elementary radiation hydrodynamics.

The radiation heats / vaporizes the material and it blows off.

The specifics of how much is complex and is not a "back of the envelope" type of calculation - it's something we do with large computer codes that account for a myriad of different physics that is going on.

Look up laser fusion, it works essentially the same way. In an indirect drive laser fusion scheme, the lasers hit the inside of the hohlraum and heat it to extreme temperatures. The hohlraum glows in X-rays, and those X-rays hit / vaporize the surface of the fusion pellet. The force created by the blow off of the surface of the pellet implodes it to extreme density, pressure, and temperature to give thermonuclear fusion:

https://lasers.llnl.gov/programs/nic/icf/

https://lasers.llnl.gov/multimedia/publications/photons_fusion/2009/november_december.php

Greg

 

[rollcast]

Would it not be more effective to hit the asteroid with serveral nukes that are synced to go off at the smae time so you can achieve a larger surface area being vaporized?

 

[Drakkith]

There's no need to hit it all at once. The same effect can be achieved by staggering them I believe.

 

[Morbius]

Would it not be more effective to hit the asteroid with serveral nukes that are synced to go off at the smae time so you can achieve a larger surface area being vaporized?

rc1102,

Because you don't need several nukes - a single nuke will do the job. In fact, it doesn't
require even one of our largest nukes - a fairly small nuke will do the job.

Greg