De-orbiting Asteroids to earth's moon

[Widdekind]

Please ponder, a presumed metal-rich asteroid, in our sun's Main Asteroid Belt (MAB), circling our central star, in an assumedly circular orbit ([math]e \approx 0[/math]). And now, please ponder detonating (appropriate, shaped) nuclear charges, on its spin-ward side, to de-orbit the body, and 'drop' it down, onto an earth-crossing trajectory. Now, the largest such earth-crossing orbit, requiring the minimum [math]\Delta v[/math] de-orbit burn, out in the MAB, is the 'earth-touching', tangential, Hohmann transfer orbit:

 

minimal transfer orbit 'drops' body from [math]3 \rightarrow 2[/math]

Now, ignoring reduced mass effects, for a rough calculation, w.h.t.:

 

[math]\frac{L}{m} = \sqrt{G M_{\odot} a (1 - e^2)}[/math]

Before the de-orbit burn, the body begins, on an assumedly circular MAB orbit ([math]a_{ast} \approx 3 AU[/math]). After the burn, the body ends, on an elliptical earth-touching orbit ([math]2 a = a_{ast} + a_{\oplus}[/math]). Thus, [math]a \approx 2 AU[/math].

 

And, since our sun (essentially) resides at the perihelion focus, of that ellipse, w.h.t.:

 

[math]a_{\oplus} = a (1 - e)[/math]

 

[math]\therefore e = 1 - \frac{a_{\oplus}}{a} = 1 - \frac{2 a_{\oplus}}{a_{ast} + a_{\oplus}}[/math]

 

[math]\therefore e = \frac{a_{ast} - a_{\oplus}}{a_{ast} + a_{\oplus}}[/math]

Thus, [math]e \approx 1/2[/math].

 

And, so, the specific angular momentum, of the metallic meteoroid, must be muted, by the amount:

 

[math]\frac{L}{m} = \sqrt{G M_{\odot} \times 3 AU} \rightarrow \sqrt{G M_{\odot} \times 2 AU \times \frac{3}{4}}[/math]

Thus,

 

[math]\Delta \left( \frac{L}{m} \right) = \sqrt{G M_{\odot} \times 3 AU} \times \left( 1 - \sqrt{\frac{1}{2}}\right)[/math]

And, that specific angular momentum reduction is realized, by a burn, out at [math]3 AU[/math]:

 

[math]\Delta \left( \frac{L}{m} \right) \approx \Delta v \times 3 AU[/math]

Thus,

 

[math]\Delta v \approx \sqrt{G M_{\odot} / 3 AU} \times \left( 1 - \sqrt{\frac{1}{2}}\right) \approx 5 \, km \, s^{-1}[/math]

Now, 1 kiloton TNT equivalent yields roughly 4 TJ of energy. So, if, for the nuclear charges, [math]E \approx c \Delta p = c m_{ast} \Delta v[/math]; then, the specific de-orbital energy required, is approximately 400 kilotons per payload ton. Allowing for inefficiencies, then, to de-orbit asteroids, from the MAB, onto earth-crossing, moon-crashing orbits, would require 0.5 - 1 megaton per ton.

 

 

 

Proposal -- de-orbit asteroids, from MAB, to moon surface; and, transfer ore to earth, via Lunar space elevator

 

[swansont]

!

Moderator Note

Asteroid mining specifics split to new thread http://www.scienceforums.net/topic/56759-asteroid-mining/

[Widdekind]

If you 'moon crashed' a meteoroid, all the way from the MAB, onto our moon; and if it was composed of volatiles; then wouldn't it 'explode', and all the volatiles boil off, into space, on moon impact ?

 

What would be the relative 'over-take velocity', of the infalling ore body, relative to the earth-moon system? Please ponder, that infalling asteroid's 'space drop' transfer-to-intercept orbit. By conservation of (specific) angular momentum, w.h.t. (perihelion vs. aphelion):

 

[math]L_p = L_a[/math]

 

[math]\therefore r_p v_p = r_a v_a[/math]

 

[math]\therefore \frac{v_p}{v_a} = \frac{r_a}{r_p} \approx 3[/math]

Thus, the inbound body will be speeding through space ~3x faster, at intercept, then immediately after the -5 km s^-1 deceleration burn, back in the MAB. But, how fast was it then moving?

 

Now, for the assumedly circular orbits, of the MAB body, and the earth, [math]v^2 = G M_{\odot} / a[/math]. This gives velocity values, of roughly 30 km s^-1 (earth), and 17 km s^-1 (MAB). After a -5 km s^-1 de-orbit burn, then, the now-infalling ore-load was moving at roughly 17 - 5 = 12 km s^-1; and, it accelerates, in-bound, upto ~3x that, or roughly 36 km s^-1. Thus, the over-take velocity, is ~6 km s^-1. The moon orbits around the earth, at roughly 1 km s^-1. Thus, by timing the space-drops, on an appropriate monthly basis, the infalling payloads would overtake the moon, as it was 'moving away' from them. Then, their relative velocities would be 6 - 1 = 5 km s^-1.

 

The point being, that's more-than-cannon-shot-fast ! Anything that slammed into the moon, at ~5 km s^-1, would take a profound pounding. What other than metal, could survive such speeds?

[pwagen]

I'm curious how we would send the asteroid our way. I won't even pretend to understand the math at this point, but wouldn't a nuclear blast mostly dissipate into space, wasting most of the "push"? While probably taking a wee bit longer, how would a "gravity tractor" hold up against that method?

 

http://www.msnbc.msn.com/id/16788616/ns/technology_and_science-space/

 

 

Also, an asteroid is quite massive. What if we miss the moon and send the asteroid tumbling down here? It's not like the thing has breaks!

[Widdekind]

Using our moon's ~1 km s^-1 orbital velocity, to 'cushion the landing', of lunar-delivered MAB asteroids, would reduce their relative overtake velocities, to ~5 km s^-1. Thus, 'space dropping' ore bodies, from the MAB, to our moon, would require deceleration [math]\Delta Vs[/math], of roughly - 5 km s^-1 'at both ends', both 'up' in the MAB, and again immediately prior to moon impact.

 

So, even with optimal nuclear-energy-to-momentum-transfer efficiencies, it would require 2 x 0.5 = 1 megaton per payload ton, to 'space drop' MAB bodies, to our moon. 'Space drops' would occur every New Moon, when our moon was on the 'outside' of earth's orbit, moving 'forward', +1 km s^-1 faster around than the earth (to cushion the delivery landings).

 

And if so, then such a scheme, would amount to 'nuclear mining', for each ton delivered... to the moon, not even as far as the earth. Could that possibly be economical ??

 

 

Economics:

 

Since the 1940s AD, the U.S.A. has spent roughly 400 billion dollars, building nearly 70 thousand nuclear warheads. For sake of sci-fi, then, one could conceive of costs near $5 million per device, or $10 million dollars per pair of devices (both the 'drop' and 'deliver' [math]\Delta Vs[/math], at the 'top' and 'bottom' of the MAB-to-moon transfer trajectory).

 

What, then, is worth >$10 million per payload ton ? At current earth market prices, most metals sell for $3 - $30 thousand per ton -- of order 1% minimum space-mining costs. Only precious metals (palladium, gold, platinum) sell for $20-60 million per ton. Thus, this space-mining proposal could, conceivably, work well, for precious metal extraction, from space deposits.

Edited April 28, 2011 by Widdekind

[Widdekind]

Space Mining amounts to 'hurling whole hills into orbit', and so requires 'nuclear-arsenal Planet-Killer power'

 

On earth, mining requires moving mountains worth of material. And, space mining, to economically compete, with terrestrial mining, must move similar amounts of materials. Thus, space mining requires 'land-scaping our solar system' -- to wit, 'space scaping' -- moving mountain-sized meteoroids, from MAB orbits (out at 3 AU), onto terrestrial-intercepting trajectories (1 AU), and then 'soft-landing' them on world (with, maybe, the moon in some step in between).

 

Now, to do so, requires accelerating the 'space mountains' ('space-bergs ?') with [math]\Delta_Vs[/math] of 2 x 5 = 10 km/s. And, that's earth escape velocity (~11 km/s) ! And again, that's ~10x faster, than a speeding tank-cannon artillery shell !

 

So, please ponder, that space mining, which requires 'space-scaping' our solar system, energetically amounts to 'hurling whole hills', off-world, from earth's surface, into orbit, and out into deep space beyond. And so, now, please ponder, that when the reverse occurs -- to wit, when 'whole hills' and 'massive mountainous meteoroids' impinge & impact upon this world, they can cause major Mass Extinction events (e.g., 'Dinosaur-killer' 65 Mya). Thus, the power required, to mine space resources -- which requires energies equivalent to 'super-cannon-shotting' whole 'hills', or even more massive 'mountains', off-world -- is equally equivalent, to dropping 'Planet Killer' asteroids, onto earth, from the MAB. To wit, it requires Planet-Killer power, which is energetically equivalent to 'Dinosaur-killer-class' impacts, per payload:

 

visual equivalents, of energy requirements, of actual Space Mining

 

 

 

 

[csmyth3025]

Space Mining amounts to 'hurling whole hills into orbit', and so requires 'nuclear-arsenal Planet-Killer power'

 

On earth, mining requires moving mountains worth of material. And, space mining, to economically compete, with terrestrial mining, must move similar amounts of materials. Thus, space mining requires 'land-scaping our solar system' -- to wit, 'space scaping' -- moving mountain-sized meteoroids, from MAB orbits (out at 3 AU), onto terrestrial-intercepting trajectories (1 AU), and then 'soft-landing' them on world (with, maybe, the moon in some step in between).

I believe that the current approach to "space mining" is the capture and extraction of usable materials from near Earth asteroids (NEA's). The intent is to use this material to further the exploration and exploitation of space as a living and working environment. I don't think that those proposing space mining intend for this activity to supply the Earth with a significant alternate source of mineral resources. You might want to take a look at the Wiki article on this subject: http://en.wikipedia....Asteroid_mining

 

Chris

[Widdekind]

I believe that the current approach to "space mining" is the capture and extraction of usable materials from near Earth asteroids (NEA's). The intent is to use this material to further the exploration and exploitation of space as a living and working environment. I don't think that those proposing space mining intend for this activity to supply the Earth with a significant alternate source of mineral resources. You might want to take a look at the Wiki article on this subject: http://en.wikipedia....Asteroid_mining

 

Chris

 

 

In 2004 AD, earth-world steel production exceeded 1 million metric tons, derived from over 1 billion metric tons of iron ore (98% of which is devoted to steel production). Thus, the mass-ratio, of desired output (steel), to raw input (iron ore), is roughly 1:1000. And thus, excepting other (extreme) compensating factors; and due to the extreme energy costs, of 'moving mountains of mass', in-and-out, hundreds of millions of kilometers, through inter-planetary space; so, seemingly, steel 'should' be refined in situ, and the finished products only shipped back to earth (rather than the raw space ore).

 

Note, that on earth, the cost cargo transportation, by train, is roughly 0.2 MJ per ton-km = 30e6 MJ per ton-AU = 30 TJ per ton-AU = 8 kiloton-TNT per payload-ton-AU. Thus, order-of-magnitude, terrestrial train transport costs, applied to MAB transportation distances (~2 AU), would be of-order 0.01 megatons per ton. Wheres, nuclear decelerations, required for 'space dropping' MAB asteroids onto our moon, amount to roughly 1 megaton per ton, or, two orders of magnitude more. Thus, if steel could be refined in situ, out in space, in the MAB, reducing needed transportation masses, by three orders of magnitude, then refined steel could be 'space dropped' onto our moon, at one-tenth the transportation cost, per 'product-ton'. And, the use of nuclear power (nuclear deceleration charges, for space mining), over chemical power (diesel trains, on earth), might amount to additional monetary savings.

 

According to the Wikipedia article that you cited,

 

all the gold, cobalt, iron, manganese, molybdenum, nickel, osmium, palladium, platinum, rhenium, rhodium and ruthenium that we now mine from the Earth's crust, and that are essential for economic and technological progress, came originally from the rain of asteroids that hit the Earth after the crust cooled

And, scientists seemingly suggest, that our moon has, over the aeons, absorbed many of the meteoroids that might have impacted our planet. And if so, then the crust of our moon, may be economically minable, as ore-bearing source-rock, for shipment back to earth.

 

 

 

 

use of latent GPE, stored within 'moon-dropped' meteoroids, and released upon moon impact, could create a 'lunar lava lake', which would melt & 'self-smelt' the moon-delivered raw ore (?!)

 

The (relative) KE, of a 'moon-dropped' meteoroid, impacting upon the moon, with [math]\Delta Vs \approx 5-7 km/s[/math], would be 10-20 GJ per ton. And, melting one ton of raw ore, currently consumes only 2 GJ per ton, roughly one order-of-magnitude less. Thus, 'moon-dropped' meteoroids store sufficient (relative) GPE, to completely melt themselves, on moon impact. So, with sufficient delivery rate, a lunar 'lava lake' could be created and maintained, into which meteoroids could be deposited, as a 'world-scale smelter system'.

 

The lunar lava lake could be allowed, to gravitationally chemically differentiate, so that the slag rose to the surface. Then, that surfaced slag could be 'super-sleuced', through artificial canals, into the lunar maria.

 

Then, after the deposited molten iron cooled sufficiently, it could be lunar-surface strip mined:

 

Earth-impinging meteor-showers, thrown off by the deposit impacts, could cause collateral damages, on earth, which would have to be covered by insurance:

 

Cost-of-Doing-Business --

"Sorry about the wind-shield, keep the rock"

Edited May 2, 2011 by Widdekind

[Widdekind]

'Obliteration Mass' of given nuclear charge

 

For sake of simplicity, let us characterize a given asteroid (or comet) with a mass, [math]M[/math], and size-scale, [math]L[/math], such that [math]M = \rho L^3[/math]. Then, the 'obliteration mass', of a given nuclear charge:

 

[math]E \equiv \frac{G M^2}{L} \approx \frac{G M^2}{\left( \frac{M}{\rho}\right)^{1/3}}[/math]

 

[math]= G \rho^{1/3} M^{5/3}[/math]

so that:

 

[math]\therefore M_{obliteration} \equiv \left( \frac{E}{G \rho^{1/3}} \right)^{3/5}[/math]

 

[math]\approx 600 \; gigatons \times \left( \frac{\frac{E}{1 \; megaton-TNT}}{G \left( \frac{\rho}{3000 \; kg/m^3} \right)^{1/3}} \right)^{3/5}[/math]

That seems way too high -- could that calculation be correct?? That would seemingly say, that a 1 megaton-TNT nuke, could obliterate, a ~6 km asteroid ([math]\rho \approx 3000 \; kg/m^3[/math]), or a ~7 km comet ([math]\rho \approx 1700 \; kg/m^3[/math]).

 

Indeed, re-formulated, in terms of characteristic size-scale, w.h.t.:

 

[math]E \equiv G \rho^2 L^5[/math]

 

[math]\therefore L \approx \left( \frac{E}{G \rho^2}\right)^{1/5}[/math]

 

[math]\approx 6 \; km \times \left( \frac{\frac{E}{1 \; megaton-TNT}}{G \left(\frac{\rho}{3000 \; kg/m^3}\right)^2}\right)^{1/5}[/math]

If so, then 1 gigaton-TNT charges, could obliterate 20-30 km asteroids / comets. Even 1 kiloton-TNT charges, could obliterate bodies of ~2 km.

 

How would obliteration, affect impact (and ore delivery)?? It would reduce the 'impact pressure', spreading out the 'spray of sand' across a wide swath of lunar surface. Note, that this seems to say -- since the same-sized nuclear charges can 'barely budge' those bodies' overall solar-orbital velocities -- that MAB asteroids / comets have much more (solar-orbital) KE, than GPE. To wit, the 'rubble piles' are barely held together (low GPE), but are 'galloping around' rather rapidly (high KE).

Edited May 7, 2011 by Widdekind

[Widdekind]

Nasa simulates journey to an asteroid by rigging up giant rock underwater | Mail Online

 

Nasa simulates journey to an asteroid by rigging up giant rock underwater

 

 

COMMENT: I offer the following, for developing a 'physicist's feel', for the physical phenomena. For, an asteroid does not merely 'gallop' through space, at earth-escape-super-speeds; it also tumbles, like a gigantic boulder tumbling down the side of Mt. Everest, in a miles-long rock-fall. Thus, even if you match speeds, with an asteroid, it will, even still, generally be spinning & tumbling -- that is the 'hard-hit ground ball' that you have to catch, before you can even begin to manipulate the mountain. Asteroid mining amounts to "Wrangling Mountains". You cannot just 'dock a tug' up to the 'barge', when that boat is 'swirling around in a vortex'. Asteroid mining will require the ability to 'de-spin' payloads, before further (linear) accelerations are applied.

 

 

 

[lemur]

Can an asteroid rotate around more than one axis somehow? If not, wouldn't it be possible to shoot it with explosives at an angle that brakes the spin?

[csmyth3025]

Can an asteroid rotate around more than one axis somehow? If not, wouldn't it be possible to shoot it with explosives at an angle that brakes the spin?

As far as I know, an asteroid can rotate (tumble) around all three axes. (I looked it up -that really is the plural of axis)

 

Chris

[Ophiolite]

Most asteroids are thought to be rubble piles. Detonating a large explosive on, in, or next to one will most likely blow it apart. If the explosion is relatively small, most of the material will then gradually reassemble itself.

 

You need slow, gentle acceleration, best provided by a mass driver using the asteroids own materials both to eject and the volatiles to provide power generation.

[imatfaal]

During the panic about 20 years that there was a NEO that was a bit too near-earth for comfort - the leading runner to deflect it from earth was a huge array of mirrors focussing suns light on a small spot - this vapourised and expanded causing a reactive force

[Widdekind]

Rubble piles are probably primordial, unrefined, rock. They would not typically be worthwhile, for economic exploitation. Only metallic asteroids, which evidently come from fragments of larger, gravito-chemically differentiated bodies, would be worthwhile. And, they would be one big block of metal. As for rotation, I think that, generally, most such bodies would 'wobble' like an imbalanced top (?). To 'wrangle' such an asteroid might be difficult.

[Ophiolite]

Rubble piles are probably primordial, unrefined, rock.

This is incorrect. The clue is in the name: rubble pile. Rubble pile asteroids are the result of multiple collisions between planetesimals, both differentiated and undifferentiated. They have high porosities.

 

Primordial objects are coherent and dense. Eros is almost certainly an example of such. The last time I checked the consensus was that the majority of asteroids would be rubble piles.

[Widdekind]

This is incorrect. The clue is in the name: rubble pile. Rubble pile asteroids are the result of multiple collisions between planetesimals, both differentiated and undifferentiated. They have high porosities.

 

Primordial objects are coherent and dense. Eros is almost certainly an example of such. The last time I checked the consensus was that the majority of asteroids would be rubble piles.

 

True. But space mining is inherently expensive, and it will probably be a long time, before space-smelting an entire rubble pile, for a few flakes off of some differentiated body, would be worth while. Certainly, the "grade A" asteroids would be those that are mostly metal. Of course, some sort of "space survey", for magnetic anomalies, indicating a giant chunk of metal in the middle of a rubble pile, might make mining the same a sound economic proposition.

[Widdekind]

'Rampart Craters' on Mars show signs of mud-flows, over-land, w/o ejecta spray, through the air. They may occur when impactors fall on ice-saturated ground, which ice absorbs allot of energy in melting & liquifying. And, Mars has more than twice as strong a surface gravity as our moon. Thus, could one arrange for "sloppy landings", of de-orbited asteroids, onto prepared "ice-mud-flats", e.g., on the floors of large pre-existing craters ??