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


3 Answers 3


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.

  • 2
    $\begingroup$ Not seeing an answer here to the question in the title. $\endgroup$ Mar 19, 2020 at 17:02
  • $\begingroup$ 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
  • $\begingroup$ 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
  • 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
  • $\begingroup$ 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

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.

  • 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 :) $\endgroup$
    – Samuel Low
    Mar 20, 2020 at 14:45
  • 2
    $\begingroup$ We do know that Psyche has valuable metals/materials, only it's not so close. en.wikipedia.org/wiki/16_Psyche $\endgroup$
    – Cornelis
    Mar 20, 2020 at 15:21
  • 1
    $\begingroup$ @SamLow: I totally agree with you. Ceres could be a good source of waster or carbon. $\endgroup$
    – Fred
    Mar 20, 2020 at 16:31

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 


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:




Delta v, km/s

Delta v, compared with going to the Moon

Delta v, compared with going to Mars


Semimajor axis, AU


Inclination, degrees

Radar detection flag


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.