I found this list, but it doesn't say how close it is to the ground https://echo.jpl.nasa.gov/~lance/delta_v/delta_v.rendezvous.html
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asked Mar 19, 2020 at 16:13
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1$\begingroup$ Right now, and for the near future, none. In the far future, maybe. This is not to say that it is not worthwhile to retrieve a sample, which is a very reasonable goal. $\endgroup$– David HammenMar 19, 2020 at 16:19
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$\begingroup$ Re, "how close it is to the ground." The Earth orbits the Sun. The asteroids also orbit the Sun, but much further out. en.wikipedia.org/wiki/Asteroid#/media/File:Asteroid_Belt.svg $\endgroup$– Solomon SlowMar 19, 2020 at 22:04
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2$\begingroup$ Most of 'em, anyway. upload.wikimedia.org/wikipedia/commons/f/f3/… $\endgroup$– Solomon SlowMar 19, 2020 at 22:07
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3 Answers
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The list doesn't specify a distance to the Earth (the "ground") because this is always changing as the asteroid travels around its orbit. You can get an idea of the asteroid's orbit using the JPL Small Body Browser and typing the name into the search box. (The first entry (2018 AV2) won't work as it has been removed as likely to be part of the Apollo 10 mission). Clicking 'Orbit Diagram' will give you a customizable plot of the orbit.
Clicking on the 'Ephemeris' link in the menu will take you to the JPL HORIZONS system which will let you compute an ephemeris. The delta column will give you the distance from the Earth to the object in au. One other thing to bear in mind is the size of the object. This is not normally well measured by you can get idea from the $H$ magnitude listed in the table. There is a formula for converting $H$ to diameter $D$ (in km), if you assume an albedo ($a$; can assume $a=0.15$ in absence of other information):
$$
D= 10^{3.1236 - 0.5\log10(a) - 0.2H}
$$
(This comes from this Center for NEO Studies page which also has a calculator to plug values in and a table of common values). Many of the objects in the table are very small (few meters in size) and there orbits are not well known. This is because they were only seen for a short fraction of their orbit when they close to the Earth and bright enough to be measured (the $H$ is the magnitude the asteroid would have at 1 au from the Earth and the Sun; magnitude ~27 is the faintest NEO that has been observed with a 8-meter telescope and those are very rarely used for observing NEOs).
There is a bias in that table towards small objects as they are more common but can only be detected by the NEO surveys if they come very close to the Earth. If they come very close to the Earth, and assuming they don't have a high inclination, that will tend to make the delta-$v$ for a rendezvous smaller.
answered Mar 19, 2020 at 16:51
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2$\begingroup$ Not seeing an answer here to the question in the title. $\endgroup$ Mar 19, 2020 at 17:02
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$\begingroup$ Without some idea of a merit function for "worth mining" means, it seems difficult/impossible to answer the question except from only a delta-v requirement which says nothing about travel time to the asteroid or whether there is anything there to mine. I answered the text in the question which asks how "close to the ground" it is, but raised some of the other issues with size, position uncertainty and distance/transit time $\endgroup$ Mar 19, 2020 at 17:14
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$\begingroup$ Sorry if I misunderstand, I mean that the total mineral extracted is capable of paying the mission and obtaining profits, however minimal they may be $\endgroup$ Mar 19, 2020 at 18:12
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1$\begingroup$ @ValentinoZaffrani - Minerals? None. Helium 3? None. The only items that might be worth extracting in the foreseeable future are volatiles such as water and methane that might be put into use in space, but even that has become suspect given how SpaceX has reduced and is continuing to reduce the cost of going into space. $\endgroup$ Mar 19, 2020 at 19:22
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$\begingroup$ Your question is a valid one but subject to many unknowns. As discussed at the bottom of the CNEOS page I linked, a quite likely uncertainty of 0.1 in the albedo (0.05 vs 0.15 say) means the diameter is uncertain by a factor of 2 and the volume by a factor of 8. So for many of the NEOs on that list, how much rock is actually there to mine is uncertain by a factor of 8 also (this assumes we know the orbit well enough to get there). There is a lot of uncertainty on the actual mining side as well for yield etc but I know much less about that area than planetary science so can't comment. $\endgroup$ Mar 19, 2020 at 19:24
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Worth mining, as in making money from mining an asteroid, the answer is we don't know.
To know what an asteroid is worth, we have to know what it's composed of and how much of any valuable metals/materials the asteroid contains. Currently we don't have that type of information.
Also, I not aware of any viable asteroid capturing or harvesting system that would allow asteroids to me mined. Once such systems are available the cost of using them can be established.
Without these two critical pieces of information all we have is wishful thinking an hope for what might be.
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2$\begingroup$ Might also be worth pointing out 'valuable' and 'worth' is very subjective. It could be valuable to Earth dwellers, with rare elements like platinum. But chances are, by the time we get access to asteroid mining technology, we'd also be doing quite a bit of space travelling, where we find other new materials more valuable. Who's to say water being mined from asteroids (read - Planetary Resources) wouldn't be of utmost value to space farers in future? I'm sure water will be worth more its weight than say, Au or Pt in space :) $\endgroup$ Mar 20, 2020 at 14:45
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2$\begingroup$ We do know that Psyche has valuable metals/materials, only it's not so close. en.wikipedia.org/wiki/16_Psyche $\endgroup$– CornelisMar 20, 2020 at 15:21
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1$\begingroup$ @SamLow: I totally agree with you. Ceres could be a good source of waster or carbon. $\endgroup$– FredMar 20, 2020 at 16:31
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Although "worth mining" is subjective and unanswerable at the moment, "close has an objective measure: the amount of $\Delta v$ required for a rocket to get from here to there. NASA has a list of near-Earth asteroids ranked in order of increasing $\Delta v$. The ten asteroids at the top of the list, requiring the least $\Delta v$, are given below. Due to alignment issues when I copiedxand pasted I mention that the $\Delta v$ value in km/s is the first figure after the asteroid name.
N = 17607
For comparison, delta-v for transferring from low-Earth orbit to rendezvous
with the Moon and Mars:
Moon: 6.0 km/s
Mars: 6.3 km/s
DELTA-V (ASTEROID)/
PROVISIONAL DELTA-V DELTA-V FOR
RANK PERCENTILE ASTEROID NAME DESIGNATION (KM/S)
THE MOON MARS H (mag) a (AU) e i (deg)
==== ========== ============= =========== ======= ======== ==== ======= ====== = =======
01 99.99 2018 AV2 3.758 0.626 0.596 28.8 1.045 0.041 0.1 RADAR
02 99.99 2006 RH120 3.820 0.637 0.606 29.5 1.033 0.024 0.6 RADAR
03 99.98 2007 UN12 3.823 0.637 0.607 28.7 1.054 0.060 0.2
04 99.98 2010 UE51 3.829 0.638 0.608 28.3 1.055 0.060 0.6
05 99.97 2012 TF79 3.867 0.644 0.614 27.4 1.050 0.038 1.0
06 99.97 2009 BD 3.870 0.645 0.614 28.1 1.062 0.052 1.3
07 99.96 2017 FJ3 3.880 0.647 0.616 29.9 1.133 0.118 1.0
08 99.95 2008 HU4 3.910 0.652 0.621 28.3 1.071 0.056 1.4
09 99.95 2010 VQ98 3.924 0.654 0.623 28.2 1.023 0.027 1.5
10 99.94 2014 UV210 3.931 0.655 0.624 26.9 1.159 0.134 0.6 RADAR
The complete list, involving more than 17,000 asteroids, is found at https://echo.jpl.nasa.gov/~lance/delta_v/delta_v.rendezvous.html.
Someone with better formatting skills can help me here, as I cannot get the column labels lined up. There are supposed to be, from left to right:
Rank
Percentile
Name
Delta v, km/s
Delta v, compared with going to the Moon
Delta v, compared with going to Mars
Magnitude
Semimajor axis, AU
eccentricity
Inclination, degrees
Radar detection flag
answered Mar 21, 2020 at 0:07