Last week's tweet by Ron Baalke (of JPL) mentions five near Earth asteroids that will come within five Lunar distances (5 LD) of the Earth in 2017.

One of them, 2012 TC4, will pass very close to Earth. Looking at the table, I see a Nominal Closest Approach (CA) Distance of 0.15 LD, but a Minimum CA Distance of only 0.03 LD!

Is nominal the most likely value, and minimum the smallest likely, or smallest possible, or 3-sigma closest? Is there any chance it could be even closer?

Will these numbers improve as the time of 2012 TC4's pass grows near?

partial tweet

above: Cropped from this tweet.

below: "Asteroid 2012 TC4 as seen by the Remanzacco Observatory team of Ernesto Guido, Giovanni Sostero, Nick Howes on Oct. 9, 2012" From Phys.org's 2015 article Will asteroid 2012 TC4 hit Earth in October 2017?

enter image description here


1 Answer 1


It looks like the tweet is screenshot of JPL's NEO close approach table (found here). JPL has definitions of many terms here.The 'nominal' miss distance is based off the latest/greatest prediction of the asteroid's trajectory, and the 'minimum' is the minimum of all the 3-sigma possibilities for it's range at close approach. This is the 3-sigma error ellipsoid projected onto the B-Plane for the approach, see the diagram below. The B-Plane is defined as the plane containing the Earth and target object at the time of closest approach, where the plane is perpendicular to the approach velocity of the target.

Diagram of miss distances on the B-Plane

The site below has more information on the close approach, as well as information that goes into the close approach prediction:

JPL Small Body Database Browser (2012 TC4)

It's maintained by the Solar System Dyanamics group at JPL.

If you look, you can see in the Orbit Determination Parameters table that the object was last tracked in 2012, so it's prediction is pretty poor for the 2017 close approach. I would expect as the object gets closer over the course of this year, there will be additional tracking, and the orbit solution will be re-computed, and get better.

As @chirlu pointed out, 2012 TC4 won't be visible until August, so the solution won't be updated until then.

The better solution (paired with shorter prediction time) will likely change the predicted close approach distance, but more importantly, it will reduce the uncertainty in the prediction.

However, it's impossible to say if the close approach will get closer or further; it will just have less uncertainty.

  • $\begingroup$ It looks like you are on the right track to finding an answer. These are the same values shown in the question, right? The key seems to be within the parenthetical phrase, but I don't understand just what it means: "minimum distance (the minimum distance between the 3-sigma Earth target-plane error ellipse and the Earth’s surface)." It would be best to try to find a diagram of that. $\endgroup$
    – uhoh
    Commented Jun 1, 2017 at 16:26
  • $\begingroup$ Also, it's a good idea to explain clearly within your answer itself, at least the key points in the link. That way if the link breaks (they usually do sooner or later) the key point of your answer will be here in your answer. Thanks! $\endgroup$
    – uhoh
    Commented Jun 1, 2017 at 16:27
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    $\begingroup$ @uhoh, I think their definition is basically the error ellipse projected onto the B-Plane.... but I'm not sure, so I'll dig a bit more to try and be sure. Once I have something I'll update the post. $\endgroup$ Commented Jun 2, 2017 at 13:46
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    $\begingroup$ According to newton.dm.unipi.it/neodys/index.php?pc=4.1&ots=r, the asteroid is not expected to be visible again before end of August, at the earliest. $\endgroup$
    – chirlu
    Commented Jun 2, 2017 at 18:06
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    $\begingroup$ The way JPL has it defined, the minimum miss distance can actually be less than zero. It's kind of a bad naming convention on their part; it's really the worst-case miss distance. If the distance is zero or less than zero, there is a statistical chance of an impact. Then you get into Probability of Collision calculations, which are a bit different. $\endgroup$ Commented Jun 7, 2017 at 18:45

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