Planetary Defense Experiment

[Shadowy Man]

Observatories like the Sloan Digital Sky Survey and the upcoming Vera Rubin observatory are not limited to the ecliptic.

That's a small percentage of all observatories, and the ecliptic is but a small slice of the sky from which a killer object might come.

I don’t understand your response. These observatories doing sky surveys cover the whole night sky. If things are moving anywhere they can be detected, given the detection limits of the instruments. They can’t however look too close to the Sun.

I doubt that any Oort Cloud object would come at us "out of the sun" - unless it was on its return trip, in which case it would have been spotted on its way into the solar system.

An object coming "straight in" from the Oort cloud would not initially show up as a moving object in one of those broad sky search surveys, or so I am told.

Who told you that?

Anything within the solar system will have noticeable parallax at least.

Aren’t long-period comets with high eccentricities essentially coming from the Oort Cloud in the way you describe? Isn’t that what instigated the hypothesis of the Oort Cloud in the first place?

Parallax measurements require multiple observatories, or they require observation over a long period.

Yes. And objects coming in from the Oort Cloud will take some time to get here.

 

[Shadowy Man]

In any case, 10 km asteroids on short notice are astronomically rare.
The one which killed dinosaurs had tens of millions of years notice. Dinosaurs did not have NASA.

What is our level of confidence that there are no "earth crossers" of that size that have yet to be cataloged?
If a new one were discovered today, how much notice would be required to do anything about it?

Seems to me that an object from the asteroid belt would be detected a lot earlier than say, a Kuyper Belt object whose orbit was somehow perturbed by collision or whatever, and took up an earth-crossing orbit in an orientation 60-90 degrees off the ecliptic. Such an object would probably blindside us.

Why would it blindside us? I think it’d be easier to see those objects coming from outside our orbit than those that would be coming from the sunward direction.

That's assuming someone is looking at the right place and the right time and looking for unknown objects. Most telescopes are going to be looking at known objects, one more point of light in the sky would go unnoticed. Only those looking for new objects are likely to notice a new dot in the sky and until they find it three separate times it won't mean anything.

That’s why I mentioned the sky surveys and cited a paper that shows that these sky surveys are identifying many many objects. The Rubin observatory will take Terabytes of data every night all over the sky. Will be able to find even more objects than the SDSS has.

 

[bilby]

You can doubt all you want, but there's a reason why mining companies don't save money and just put all the explosives into a single bore-hole when blasting.

20 meters fragments are considered small

Sure.

And a single large explosion won't reduce the asteroid to rubble that doesn't contain larger - probably much larger - fragments. Most likely (particularly if it's a highly non-homogeneous 'rubble pile'), a single large blast will fracture it along existing weaker fault lines, leaving some small stuff, plus a handful of pieces still large enough to be a serious problem.

From a practical perspective, the deflection option is far superior to the disintegration option. It's not possible to place an array of small charges, for a number of reasons; And a small number of large explosions won't do the job.

Your best bet is still to accelerate the thing into a safer orbit. Though you could do this in a number of clever ways - strapping a chemical rocket to it would require bringing a lot of fuel along, but I am sure you could MacGuyver up a probe with a nuclear power supply, that could hurl bits of the asteroid away in a controlled direction, using the existing material for reaction mass rather than lifting it out of Earth's gravity well.

Just scooping up shovelfuls of dirt and throwing them into space would suffice, if done for a long time and in such a way as to ensure that each scoop is flung in roughly the same direction as the last.

If the thing contains reasonable amounts of iron, perhaps you could build a railgun.

 

[barbos]

Disintegration option is when there is not enough time. It's better than doing nothing.

 

[lpetrich]

Here is how to find approximate results:

•  Back-of-the-envelope calculation --  Fermi problem
•  Dimensional analysis --  Scale analysis (mathematics) --  Buckingham π theorem --  Order of magnitude
•  Guesstimate --  Scientific wild-ass guess --  Hand-waving
•  Heuristic --  Rule of thumb
•  Sanity check

One can often find passable approximations much faster than high-quality results. I mention that because that's what we could be doing in this thread instead of going round and round in circles.

Here is a rather hand-waving summary of the math that's necessary.

(Momentum) = (mass) * (velocity)
(Velocity change) = (momentum change) / (mass)

(relative velocity change) = (velocity change) / (orbital velocity) = (momentum change) / (momentum in orbit)

One can see why Dimorphos was targeted. It is small and it moves slowly in its orbit around Didymos, thus having very little Didymos-orbit momentum, very little by asteroid standards.

Let's look at orbit parameters:

(Semimajor axis) -- relative -- amount of change ~ (original value) * RVC
(Eccentricity, orientation angles, position angle in orbit at collision time) -- absolute -- amount of change ~ RVC

One can condense these two results into
(Position of the orbit) ~ (orbit size) * RVC

One also finds
(Position change in orbit after n orbits) -- (2*pi)*n * (orbit size) * RVC

It is cumulative from a change in the semimajor axis. That change is most efficiently made by imparting some momentum either in the target's direction of motion or in the opposite direction. Either will work, and what's to avoid is imparting that momentum in the perpendicular direction.

So if one hits an asteroid far enough in advance, it may end up either too early or too late for hitting the Earth.

 

[steve_bank]

It is like calculating that the energy of a nuclear bomb is not capable of causing the detraction that they do.

Which takes more energy, deflection or accelerating in the direction of travel?

An engine attached to the object is the best solution. Nuclear powered.

 

[Jimmy Higgins]

It is like calculating that the energy of a nuclear bomb is not capable of causing the detraction that they do.

Which takes more energy, deflection or accelerating in the direction of travel?

An engine attached to the object is the best solution. Nuclear powered.

Why not slap an engine on a much bigger trojan and use that to deflect?

 

[lpetrich]

What one needs is momentum, and one can estimate how much momentum that one can get for some energy input.

Let us consider kinetic energy, since that is the sort of energy most directly related to momentum. The math:

Reaction mass: m
Its velocity: v
Its momentum: p = m * v
Its kinetic energy: T = (1/2) * m * v^2

Let us see what momentum one can get for some amount of kinetic energy:

v = sqrt(2*T/m)
p = sqrt(2*m*T)

So if one has some amount of energy available, one needs to increase the reaction mass m of one's asteroid kicker. This mass can come from the asteroid itself if necessary.

So if one wants to use a nuclear bomb, one ought to dig a pit in the asteroid, place the bomb at the base of the pit, then bury that bomb with the removed material from digging that pit. Exploding the bomb will accelerate this rubble, thus producing a much bigger kick than what one can get from the bomb by itself.

 W80 (nuclear warhead) has some numbers.

• Mass: 290 lb (130 kg)
• Length: 31.4 inches (80 cm)
• Diameter: 11.8 in (30 cm)
• Blast yield: 5 or 150 kilotonnes of TNT (21 or 628 TJ)

So it should be easy to carry aboard a spacecraft.

The best case, conversion of its explosion energy into a single direction of motion, gives a momentum kick of 4*10⁸ kg * (m/s) at 3.1*10⁶ m/s. If it was buried under 1,300 metric tons of rubble, one gets 4*10¹⁰ kg * (m/s) at 3.1*10⁴ m/s. That means a pit some 8 - 10 meters deep.

In actual fact, the efficiency will be much lower, from imperfect conversion of thermal into kinetic energy, and from imperfect directionality. Using a factor of 2 for each factor gives a momentum about half as large: 2*10⁸ kg * (m/s) by itself and 2*10¹⁰ kg * (m/s) for 1,300 tons of rubble.

Using my earlier estimates for the masses of Dimorphos and Didymos, 5.2*10⁹ kg and 5.40*10¹¹ kg, I find 0.038 m/s and 0.00037 m/s for a nuclear bomb by itself and 3.8 m/s and 0.037 m/s for a buried nuclear bomb.

Relative to their orbital velocity around the Sun, an average of 23 km/s, that is a relative velocity change of 1.6*10^-6 and 1.6*10^-8 for a bomb by itself and 1.6*10^-4 and 1.6*10^-6 for a buried bomb.

Using 1 AU of distance, 1.5*10⁸ km, I find 250 km and 2.4 km for a bomb by itself, and 25,000 km and 240 km for a buried bomb.

So a nuclear bomb will work, at least if one uses it far enough in advance.

 

[Loren Pechtel]

Parallax measurements require multiple observatories, or they require observation over a long period.

That's how asteroids are being detected

Because there's no rush. And they're not looking much for stuff out of the plane of the ecliptic.

 

[Loren Pechtel]

Check your math. You're looking at the ballpark of 1 MJ/kg.

I checked, still 10MJ per ton of atmosphere.

10km is the dinosaur killer. You're way low on it's impact energy.

How do you know I am way low? Have you calculated or you use gut feelings?

It's called Google. I looked it up.

 

[Loren Pechtel]

You can doubt all you want, but there's a reason why mining companies don't save money and just put all the explosives into a single bore-hole when blasting.

20 meters fragments are considered small

Sure.

And a single large explosion won't reduce the asteroid to rubble that doesn't contain larger - probably much larger - fragments. Most likely (particularly if it's a highly non-homogeneous 'rubble pile'), a single large blast will fracture it along existing weaker fault lines, leaving some small stuff, plus a handful of pieces still large enough to be a serious problem.

From a practical perspective, the deflection option is far superior to the disintegration option. It's not possible to place an array of small charges, for a number of reasons; And a small number of large explosions won't do the job.

Your best bet is still to accelerate the thing into a safer orbit. Though you could do this in a number of clever ways - strapping a chemical rocket to it would require bringing a lot of fuel along, but I am sure you could MacGuyver up a probe with a nuclear power supply, that could hurl bits of the asteroid away in a controlled direction, using the existing material for reaction mass rather than lifting it out of Earth's gravity well.

Just scooping up shovelfuls of dirt and throwing them into space would suffice, if done for a long time and in such a way as to ensure that each scoop is flung in roughly the same direction as the last.

If the thing contains reasonable amounts of iron, perhaps you could build a railgun.

It doesn't need to contain iron. You bring your launcher, you throw bucketfuls of crap in a safe direction, don't discard the bucket. Not only is it expensive but you can recover most of the energy that was used accelerating it.

You still have to land on it for this approach and it's slow. Fine if the problem is a NEO detected years out, not so fine if it's a rogue object.

 

[Loren Pechtel]

What one needs is momentum, and one can estimate how much momentum that one can get for some energy input.

Let us consider kinetic energy, since that is the sort of energy most directly related to momentum. The math:

Reaction mass: m
Its velocity: v
Its momentum: p = m * v
Its kinetic energy: T = (1/2) * m * v^2

Let us see what momentum one can get for some amount of kinetic energy:

v = sqrt(2*T/m)
p = sqrt(2*m*T)

So if one has some amount of energy available, one needs to increase the reaction mass m of one's asteroid kicker. This mass can come from the asteroid itself if necessary.

So if one wants to use a nuclear bomb, one ought to dig a pit in the asteroid, place the bomb at the base of the pit, then bury that bomb with the removed material from digging that pit. Exploding the bomb will accelerate this rubble, thus producing a much bigger kick than what one can get from the bomb by itself.

 W80 (nuclear warhead) has some numbers.

• Mass: 290 lb (130 kg)
• Length: 31.4 inches (80 cm)
• Diameter: 11.8 in (30 cm)
• Blast yield: 5 or 150 kilotonnes of TNT (21 or 628 TJ)

So it should be easy to carry aboard a spacecraft.

The best case, conversion of its explosion energy into a single direction of motion, gives a momentum kick of 4*10⁸ kg * (m/s) at 3.1*10⁶ m/s. If it was buried under 1,300 metric tons of rubble, one gets 4*10¹⁰ kg * (m/s) at 3.1*10⁴ m/s. That means a pit some 8 - 10 meters deep.

In actual fact, the efficiency will be much lower, from imperfect conversion of thermal into kinetic energy, and from imperfect directionality. Using a factor of 2 for each factor gives a momentum about half as large: 2*10⁸ kg * (m/s) by itself and 2*10¹⁰ kg * (m/s) for 1,300 tons of rubble.

Using my earlier estimates for the masses of Dimorphos and Didymos, 5.2*10⁹ kg and 5.40*10¹¹ kg, I find 0.038 m/s and 0.00037 m/s for a nuclear bomb by itself and 3.8 m/s and 0.037 m/s for a buried nuclear bomb.

Relative to their orbital velocity around the Sun, an average of 23 km/s, that is a relative velocity change of 1.6*10^-6 and 1.6*10^-8 for a bomb by itself and 1.6*10^-4 and 1.6*10^-6 for a buried bomb.

Using 1 AU of distance, 1.5*10⁸ km, I find 250 km and 2.4 km for a bomb by itself, and 25,000 km and 240 km for a buried bomb.

So a nuclear bomb will work, at least if one uses it far enough in advance.

Digging a pit is by far the most efficient means of employing the bomb's energy. However, let's look at the DART mission--it's going to be hauling ass at 6,600m/s when it hits. If you want to bury your bomb that means a 6,600m/s burn and bringing along digging equipment. The limiting factor is almost certainly going to be launch capability--you can get a lot more boom on target if it's standoff detonation (and the standoff distance can be very low, you just have to ensure the bomb isn't smashed before it fires.)

Note that NASA doesn't agree with your 1% efficiency for standoff--they think more like 10% for an off-the-shelf bomb.

 

[steve_bank]

It is like calculating that the energy of a nuclear bomb is not capable of causing the detraction that they do.

Which takes more energy, deflection or accelerating in the direction of travel?

An engine attached to the object is the best solution. Nuclear powered.

Why not slap an engine on a much bigger trojan and use that to deflect?

I was thinking about that. How fast does say a 10kg slug need to be going to defect the ateroid?

The problem I see is accuracy, how accurately do we know the trajectory of the asteroid and where to hit it. An engine that can automatically vector thrust seems to me like the best solution.

I thought NASA tried that remotely and they had problems.

Someone is going to have to go there and install whatever the solution is. I remember something about Obama retasting the new moon vehicle fr asteroid intercept.

Calling Buck Rogers....

 

[barbos]

Check your math. You're looking at the ballpark of 1 MJ/kg.

I checked, still 10MJ per ton of atmosphere.

10km is the dinosaur killer. You're way low on it's impact energy.

How do you know I am way low? Have you calculated or you use gut feelings?

It's called Google. I looked it up.

Well. you are still wrong.

 

[Bomb#20]

What one needs is momentum, and one can estimate how much momentum that one can get for some energy input.
...
So if one has some amount of energy available, one needs to increase the reaction mass m of one's asteroid kicker. This mass can come from the asteroid itself if necessary.

So if one wants to use a nuclear bomb, one ought to dig a pit in the asteroid, place the bomb at the base of the pit, then bury that bomb with the removed material from digging that pit. Exploding the bomb will accelerate this rubble, thus producing a much bigger kick than what one can get from the bomb by itself.
...

Digging a pit is by far the most efficient means of employing the bomb's energy. However, let's look at the DART mission--it's going to be hauling ass at 6,600m/s when it hits. If you want to bury your bomb that means a 6,600m/s burn and bringing along digging equipment. The limiting factor is almost certainly going to be launch capability--you can get a lot more boom on target if it's standoff detonation (and the standoff distance can be very low, you just have to ensure the bomb isn't smashed before it fires.)

Note that NASA doesn't agree with your 1% efficiency for standoff--they think more like 10% for an off-the-shelf bomb.

^^^^ This ^^^^
lpetrich appears to be assuming that if you don't bury the bomb then the bomb is your reaction mass. That's not how it works. A standoff detonation will vaporize tons of the asteroid and blast the gaseous rock into space.

But even if it were true that burying the bomb is a hundred times more efficient, the great virtue of a standoff detonation is you can set it off while you're coming up on the asteroid at 7-odd km per second, which makes the fuel requirements enormously less. You're not likely to be able to make the combined mass of the digging equipment and the fuel to match speed with the asteroid for a soft landing come out less than a hundred times the mass of the warhead.

(Also, digging is complicated. The more moving parts, the more likely something will go wrong. A standoff detonation is simple.)

 

[barbos]

100% efficiency means asteroid is split into 2 equal pieces and all bomb energy is then converted into giving them opposite momentum. There is no way standoff nuke can achieve 10% efficiency.

 

[Bomb#20]

100% efficiency means asteroid is split into 2 equal pieces and all bomb energy is then converted into giving them opposite momentum. There is no way standoff nuke can achieve 10% efficiency.

But there is no way burying a bomb will mean the asteroid is split into 2 equal pieces and all bomb energy is then converted into giving them opposite momentum. So we're normalizing, to get relative efficiency. The question is: what fraction of the delta-v burying a bomb delivers to the asteroid can a standoff detonation deliver? And if that's, say, 10%, then the follow-up question is: can we fly more than ten bombs to a standoff detonation point with the same amount of resources it would take to bury one bomb in a mine shaft we have to dig? IMHO, yes, we probably can.

 

[steve_bank]

One of the main reasons for early supercomputers at a national lab was to simulate nuclear bombs instead of physical testing.

I expect given data from actual tests the simulation of nuclear explosions are accurate. A detonation near am asteroid and the reaction forces on the asteroid are likely simulated. I take it on faith NASA knows what it is doing. They do not launch an experiment to an asteroid on a whim.

Here is an idea.

Imagine an iron sphere of metal in space 1km diameter. Assume the surface is a pefect black body radiator in W/m^2.

It is a vacuum so the only way to get rid of heat is by radiation. Put a nuclear reactor in the center generating more heat than the capacity to radiate.

How long for the temperature to reach the melting point?

 

[Loren Pechtel]

I was thinking about that. How fast does say a 10kg slug need to be going to defect the ateroid?

Asteroids have quality control?

 

[Loren Pechtel]

100% efficiency means asteroid is split into 2 equal pieces and all bomb energy is then converted into giving them opposite momentum. There is no way standoff nuke can achieve 10% efficiency.

They were measuring it as bomb energy converted to movement of the asteroid.

And your approach isn't the most efficient, anyway. Splitting it in two means expending some of your energy on pushing the asteroid apart and it ensures you need even more velocity unless the original course was for an absolutely dead center impact. Maximum efficiency is when you shove it aside sufficiently to generate a miss.