Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As you may have seen, a small asteroid had a near miss with the earth a few days ago.

As a physics teacher teaching momentum, I think this could be an excellent problem for my students, however, I don't know how to determine the mass or velocity of this object. I could get a decent estimate of the mass simply by assuming its a spherical rock, but I still don't see how to determine the velocity. I can get the orbital elements: Orbital Elements of 2011 CQ1.

Is the 'V' term the magnitude of the velocity, and if so, what are those units? km/s? How could I get a heading for the asteroid?

share|cite|improve this question
up vote 0 down vote accepted

I guess it's better to read articles optimized for human readers, e.g.

The relative velocity in the Earth's frame was 10 km/s when it was closest, 5471 km. Your webpage says that MOID (closest distance) was 0.0001 AU but it turned out to be 3 times closer at the end. A picture of the direction is included on the page I linked, too. The radius 1 meter wouldn't pose a problem if it hit us.

In the table, you have the ephemeris:

Here, YYYY MM DD.## is the date (fourth of February 2011) followed by numbers 00-06 which are probably just labels (in the table I see above). It's separated by a space from the HH MM.MM time, changing from 8:20 to 8:21 or so (decimal point in minutes in between - about a minute was listed by you). What follows is R.A. or right ascension

(the entries are 14) which is the latitude on the Earth above which it occurs (where we see it), with the rotation of the Earth subtracted. Note that it is not moving much because the "horizontal" orbits of both bodies around the Sun are relatively synchronized. Then you get the declination

which is the ordinary Earth's latitude above which the object is located. It's changing between 37 and 43 (plus means North Hemisphere): the planes of orbits differ. From the changes of these two coordinates, you may deduce the angular velocity. The declination changed from 37.5 to 42.5 i.e. by 5 degrees = $pi/36$ radian. At distance of 5471 km I announced above, it's 477 kilometers. It took some time - the difference between 8:21.60 and 8:20.73 which is 0.87 minute i.e. 52 seconds, so you can see that the speed was about 10 km/s in your table

The table also includes Delta which I am not sure about now; distance from the Sun in AU called $r$ (which is obviously not too far from one), and elongation

which is the angle between the object and the Sun, as seen from the Earth. The distance from the Earth is not directly listed because it's not directly measured. One only knows the angle and the redshift. $V$ is not the velocity you want - which I calculated above.

share|cite|improve this answer
I think mass can only be guessed from estimating both the diameter from visual and infrared brightness of the object, and density. Both of which are highly uncertain. It is way too small for its graviational effect on other objects to be measured. If it had had some interaction with the atmosphere, decelleration due to drag might also be used to constrain the mass. Something this small probably can't be detected unless it is near the earth. Velocity is messy, as you have the rotation and revolution of the earth to take account of. – Omega Centauri Feb 7 '11 at 16:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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