I tried looking around but couldn't find any sources attempting to answer this particular question. It's obviously impossible to know the exact answer, but all I'm looking for is an estimate. My best guess would be that it's about a tenth (or less) of the mass of the asteroid belt, with a composition roughly 10-15% metallic and 85-90% carbonaceous or silicate. Is anyone aware of any sources that attempt to give an estimate? If not, what would you say is a reasonable estimate?

Thanks in advance.


1 Answer 1


Short answer: The total mass of the known NEOS is about $8 \times 10^{17}\rm\,kg$, if we assume average density of $2\rm\,g/cm^3$.

Long Answer: I used the NASA JPL data query website. Instead of mass, they have diameter for equivalent spherical shape in kilometers for NEOs.

Here are the top 10 results from the query:

1036 Ganymed (A924 UB)      37.675
109P/Swift-Tuttle           26
3552 Don Quixote (1983 SA)  19.0
433 Eros (A898 PA)          16.84
C/2001 OG108 (LONEOS)       13.6
C/1991 L3 (Levy)            11.6
1P/Halley                   11.0
4954 Eric (1990 SQ)         10.8
1627 Ivar (1929 SH)         9.12
1866 Sisyphus (1972 XA)     8.48

Out of 24877 entries for NEOs, there are only 1268 NEOs with an entry for the diameter. If we convert these to volumes using $V= \frac{4}{3}\pi r^3$ and sum, we get $4.21 \times 10^{5}\rm\, km^3$ of total NEO volume.

If we assume an average asteroid density of $2\rm\, g/cm^3$ = $2000\rm\, kg/m^3$ = $2 \times 10^{12}\rm\,kg/km^3$ , then the total mass is $8.42 \times 10^{17}\rm\, kg$.

As a comparison, the total mass of the asteroid belt is estimated at a whopping $2.39 \times 10^{21}\rm\, kg$. So, even with comets, the NEOs are much less massive than the asteroid belt.

This estimate of total mass may be somewhat conservative, because it leaves out the mass of most of the known NEOs. However, I don't think it is too conservative since about 90% of the mass estimate is accounted for in the largest 10 asteroids/comets listed above. The smaller stuff doesn't contribute much to the overall estimate.


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