Slashdot Mirror
Easily-Captured Asteroids Identified
Hugh Pickens DOT Com writes "Long overlooked as mere rocky chunks leftover from the formation of the solar system, asteroids have recently gotten a lot more scrutiny as NASA moves forward with plans to capture, tow, and place a small asteroid somewhere near our planet. Two different private space companies, Planetary Resources and Deep Space Industries, plan to seek out and mine precious metals and water from near-Earth asteroids. Now Adam Mann reports that astronomers have identified 12 candidate Easily Retrievable Objects (EROs) ranging in size from approximately 2 meters to 60 meters in diameter that already come (cosmically) close enough to our planet — close enough that it would take a relatively small push to put them into orbits at Lagrange points near Earth using existing rocket technology. For example, 2006 RH120 could be sent into orbit at L2 by changing its velocity by just 58 meters per second with a single burn on 1 February 2021. Moving one of these EROs would be a 'logical stepping stone towards more ambitious scenarios of asteroid exploration and exploitation, and possibly the easiest feasible attempt for humans to modify the Solar System environment outside of Earth (PDF),' write the authors in Celestial Mechanics and Dynamical Astronomy. None of the 12 ERO asteroids are new to astronomers; in fact, one of them became briefly famous when it was found to be temporarily orbiting the Earth until 2007. But until now nobody had realized just how easily these bodies could be captured."
Looks like it's time to build a foundry in space so we can begin the construction of satellites, space stations and long range spacecraft with materials readily available in space, so we don't have to keep carting it up there. Between that and robots and assembly machines, we should be able to build out stuff in the next couple decades.
Watch for Penguins, they eat Apples and throw rocks at Windows.
It doesn't have to hit Earth to affect it. Consider the tides. Our global eco system has evolved to expect tides. It would be difficult if not impossible to predict the full extent of the harm that could result if tidal patterns are altered. All sorts of life could flourish or die under such changes.
I'm not exactly a tree-hugger, but I certainly appreciate the factors and influences over life on this planet. This would affect the oceans in all sorts of ways. That which affects the oceans and the life within them will affect us and possibly even global weather patterns.
Because a 2~60m diameter stone in space can significantly alter tides.
It doesn't have to hit Earth to affect it. Consider the tides.
Why not consider the Lily? Look, the largest of these objects is sixty meters in diameter. I'm math-challenged, but a quick back-of-the-napkin calculation reveals marinara sauce and a little olive.
"You're right," Fisheye says. "I should have set it on 'whip' or 'chop.'"
There's stuff whizzing past us all the time with the gravitational attractive force that these rocks will have. It's not going to impact tidal patterns until we start capturing relatively large objects... like relative to the moon kind of size.
You know you only have to stand about 6 feet away from somebody to have the same gravitational pull on them as Mars has on you when it's closest to earth?
Mars already impacts our tidal patterns more than these rocks.
Well, let's consider the damage from the impact of a rocky asteroid, 60m in diameter. Plug this into the excellent Earth Impact Effects program at http://impact.ese.ic.ac.uk/ImpactEffects/. Assume a velocity of 17 km/s, which they say is "typical for asteroids," and an impact angle of 45 degrees.
The calculator says:
The projectile begins to breakup at an altitude of 54000 meters = 177000 ft
The projectile bursts into a cloud of fragments at an altitude of 4700 meters = 15400 ft
The residual velocity of the projectile fragments after the burst is 4.77 km/s = 2.96 miles/s
The energy of the airburst is 4.52 x 1016 Joules = 1.08 MegaTons.
No crater is formed, although large fragments may strike the surface.
Clearly you wouldn't want to be right underneath it, but even as close as 20 km, the air blast effects seem rather anticlimactic:
Peak Overpressure: 18900 Pa = 0.189 bars = 2.69 psi
Max wind velocity: 41.4 m/s = 92.6 mph
Sound Intensity: 86 dB (Loud as heavy traffic)
Damage Description:
Glass windows will shatter.
About 30 percent of trees blown down; remainder have some branches and leaves blown off.
So it'd be like BOOM! But not like KA-FOOOM!
For comparison, the Chelyabinsk meteor was estimated at 17-20m, with an airburst energy of 0.4 MegaTons.
Never attribute to malice that which can be explained by mere idiocy.
It doesn't have to hit Earth to affect it. Consider the tides. Our global eco system has evolved to expect tides. It would be difficult if not impossible to predict the full extent of the harm that could result if tidal patterns are altered. All sorts of life could flourish or die under such changes.
I'm not exactly a tree-hugger, but I certainly appreciate the factors and influences over life on this planet. This would affect the oceans in all sorts of ways. That which affects the oceans and the life within them will affect us and possibly even global weather patterns.
Because a 2~60m diameter stone in space can significantly alter tides.
The level of numerical illiteracy* of the general public (i.e. the GP) is appalling, and combined with the boatloads of self-esteem fed to them during school years, it resulted in people worse than being totally ignorant.
A totally ignorant person would either ask the above question without assumption, e.g. "Is it possible for the captured asteroid to affect the Earth in any meaningful way?", or just assume the experts have already thought about it. Only those who knew just enough to be dangerous would both assume their imagination (considerations that is not based on hard facts nor experience is no different than imagining things) is correct, AND the experts have not considered it already.
* - by that, I mean the lack of sense in numerical scales and numbers. The radius of the Moon is in the order of ~1000km, so a 60m asteroid (round to 100m) is 4 orders of magnitude in linear dimension and thus 12 orders of magnitude in volume. How lack of numerical sense do you need to be to think that something 12 orders of magnitude smaller can have any impact?
It's probably the most precious commodity in space.
Patents Drive Free Software as Hurricanes Drive Construction Industry
i think you should be taking your own advice.
one thing you forgot to consider was distance, and the moon si really really far out there.
also, 12 orders of magnitude is 10^12. given that the moon is ~3474km in diameter (1700km radius), compared to a 60m object the moon is only 5.79e+4 times larger....which is no where near "12 orders of magnitude". but the size that we really need to consider isnt dimensional anyway, but mass.
so let's explore:
Remember the formula is F=G*m1*m2 / d^2. The gravitational force is inversely proportional to the square of the separation distance between the two. So we can hold the factors other than d unitary to determine the relative strengths at the following distances (truncated for space):
~380k km (roughly the moon's average distance) = 6.925e-12
~36000 km (typical geosynchronous orbit, ie, GPS) = 7.716e-10
~2000 km (medium earth orbit) = 2.5e-7
So an object at MEO has 324x as much pull as the same object at typical geosynchronous distance, and >36000x as much pull as the same object at the moon's distance. So an object the size of the moon at the moons distance can have the same pull as an object 1/36000 the mass of the moon but in MEO*. Given the moon's mass is 7.3477e+22 kg, this gives us an equivalent mass of 2.041e18 kg at MEO, or 2.26e+20 kg at geosynchronous distance**. Then we can take the moons density of ~3346 kg/m^3. This gives us volumes of ~6.0998e+14 m^3 (MEO) and ~6.754e+16 m^3 (GS), which in turn give shperical diameters of 105.22 km (MEO) and 505.27 km (GS).
So we end up with objects only 0.0302 and 0.1454 the diameter of the moon at MEO and GS to have the same effect as the moon, assuming the same density as the moon. If we instead assume say an asteroid largely composed of Iron (density 7,870 km/m^3) we get diameters of ~79 and 380 km. An iridium asteroid is about the densest thing we might find out there, and even then our diameters calculate to ~56 and ~268 km.
So this is neat stuff, and now we get a real sense of what it would take to have an effect equivalent to the moon. But that's not to say there would no effect. while the distance relationship is an inverse square, the effect of mass is directly proportional, so something with half the mass will have half the effect. and while the poster mentioning hundreds of thousands of these things misses the logistical problems, having a sufficient number number of solid or metallic core examples of these things could have a measurable impact, particularly in terms of periodic reinforcement. and now im running out of time for thought experiment math (gotta get back to work).
*(force vector going to center of a theoretical main body, and thus ignoring for now the angles of distributed force vectors in the real situation being far different between an object in MEO and an object at the moons distance as they effect a fluid on the surface of said main body)
**(ignoring for now the orbital velocities or distances required for such objects to remain in stable orbit)
The guy who said the election was rigged won the presidency with the second-most votes.