Asteroid 2012 DA14 will pass by close to Earth. Very close. So close, in fact, that it's inside the orbit of the moon and even inside the orbit of geostationary satellites, as shown by this illustration:

2012 DA14 approach to Earth Source: NASA.

What the figure doesn't show is the angle of its path relative to the orbital plane of the moon and the orbital plane of geostationary satellites. What is this angle? What is the risk that it hits either the moon, or a geostationary satellite?

I guess it's quite small, to quote Douglas Adams:

"Space," it says, "is big. Really big. You just won't believe how vastly, hugely, mindbogglingly big it is. I mean, you may think it's a long way down the road to the chemist's, but that's just peanuts to space, listen..."

but I'd be interested in a slightly more quantitative analysis.

(Googling this event is quite amusing, with one source godlikepredictions claiming that Asteroid 2012 DA14 will pass earth within .09 AU, missing Earth, but what is located at 1 AU from Earth???? Our moon!!!!!! (SIC).

  • $\begingroup$ I doubt you can get more specific than Douglas Adams, unless you have extremely accurate data on satellite orbits for all the myriad satellites up there (and you'd probably want data for space junk as well, as that could deflect the thing or be deflected by it, putting it on a collision course with something else in a game of snooker, potting a rock in a satellite eventually :) $\endgroup$ – jwenting Feb 6 '13 at 9:46
  • 1
    $\begingroup$ Suppose it's in the same orbital plane as the geostationary ring, then one could get a 1st order estimate which would simply be twice the fraction of the ring occupied by satellites. The same for the moon. Exact orbits and higher-order gravitational interactions would be needed for a more accurate estimate, but not for the order of magnitude of the risk. $\endgroup$ – gerrit Feb 6 '13 at 9:54

See the JPL Java applet to see how the orbit is tilted wrt the Earth. From playing about with the applet I'd say it's orbit is tilted by something between 10 and 15 degrees relative to us. That means it isn't in the same plane as the Moon or geostationary satellites and therefore has zero chance of hitting either.

It could hit any satellites orbiting at the distance of closest approach, 27,700 km. However I don't think there are many satellites at that distance. The GPS satellites get closest, but I think the highest of those are around 26,000 km, so the chance of hitting a satellite is zero as well.

Boring really :-)

Response to comment:

Because the orbit of 2012-DA14 is tilted wrt to the Earth's orbit it only intersects the plane of the Earth's orbit at two points. Only one of these is relevant to this discussion, and it's the one at the closest approach of 27,700 km. So it can only intersect the orbit of a satellite if the altitude of that satellite is 27,700 km. This makes it impossible to hit the moon or and geostationary satellites i.e. the chance of collision is zero not "very small".

The chance of hitting a GPS satellite is also zero because the highest of them are at about 26,000 km.

I don't know if there are any satellites at an altitude of 27,700 km (though I don't think so), and if there are satellites at this altitude the chance would be just "virtually zero".

  • $\begingroup$ "Zero" meaning "virtually zero", as the planes still intersect? $\endgroup$ – gerrit Feb 6 '13 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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