Will I be able to see the asteroid 1998 OR2 with the naked eye on April 29?

If so, where can I find more information on when and where I'll be able to see it?

  • $\begingroup$ This is an interesting question! I've edited your question to add the name of the asteroid that will pass by on April 29. $\endgroup$ – uhoh Apr 28 at 3:36
  • $\begingroup$ According to HORIZONS (ssd.jpl.nasa.gov/horizons.cgi) it will only brighten to 10.79 magnitude, so probably not. $\endgroup$ – user21 Apr 28 at 4:24

According to the page on The Sky Live the asteroid is currently at a visual magnitude of +11, which is also what Sky Safari says. It is not due to get much brighter so no, it will not be visible with the naked eye. According to the predicted light curve, there is a chance it will get to naked eye brightness (from a dark sky site) around April 2079.... So there is quite a wait! And these predictions are based on current models which can change with new information.

enter image description here

You can use theskylive to check on things like comets and asteroids, or if you prefer a mobile app, I always recommend Sky Safari

| improve this answer | |

I observed 1998 OR2 in a 30 cm telescope on 2020-04-28 around 02:00 UT. As predicted, its apparent magnitude was around 11, much too faint for the unaided eye.

This Sky & Telescope article has finder charts, which indicate the correct position for the time I observed it. I prefer to see more faint stars on the map for positive identification, so I used Stellarium with downloaded star catalogs to magnitude 13.5 (Configuration: Extras).

In Stellarium's solar system editor (Configuration: Plugins), I imported orbital elements for 1998 OR2 from list "MPCORB: near-Earth asteroids (NEAs)," and the predicted position agreed with my observation. Online search got slightly different elements which put the asteroid a few minutes ahead of its actual position. I do not recommend the MPC PHA list, whose elements put it ahead by more than an hour!

| improve this answer | |

Based on this answer and sources therein, the absolute magnitude of an asteroid is given by

$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$

where $D_{km}$ is the asteroid's diameter in kilometers, and the albedo (depending on the kind of albedo, is usually between zero and 1) is probably between 0.1 and 0.2 roughly.

According to Wikipedia's (52768) 1998 OR2 it's albedo is about 0.2 and its diameter is about 2 km, but the diameter is based on its *measured absolute magnitude of +15.7 since we usually can't independently measure most asteroid diameters, so let's go with that.

The linked answer also gives some information on the apparent magnitude:

The apparent magnitude from this answer:

Knowing the absolute magnitude of an object, you calculate the apparent magnitude $m$ using:

$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$

where $d_{SR}$ and $d_{RE}$ are the Sun-Roadster and Roadster-Earth Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor $O(1)$ is the phase integral, of order unity, taking into account the angular difference between the direction of illumination and the direction of viewing. In an order of magnitude calculation, this only becomes really significant when the body moves between the Sun and the viewer. See https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H).

So if the Sun were right behind us and we viewed the asteroid at 1 AU, it would have a apparent magnitude roughly equal to its absolute magnitude.

However, Wikipedia's (52768) 1998 OR2; 2020 approach says:

On 29 April 2020 at 09:56 UTC, the asteroid will safely pass 0.042 AU (6.3 million km; 16 LD) from Earth. With observations as recent as April 2020 and a 32-year observation arc, the 2020 close approach distance is known with an accuracy of roughly ± 7 km. (For comparison, Venus will be 0.29 AU (43 million km; 110 LD) from Earth on 3 June 2020.)

So at 0.042 AU it will be 1/(0.042)2 or 6.9 magnitudes brighter but only if it were fully sunlit. Chances are it won't be, but if it were it would be magnitude +8.8 or still not visible.

| improve this answer | |

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.