<p>According to the page on <a href="https://theskylive.com/1998or2-info" rel="nofollow noreferrer">The Sky Live</a> 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.</p>

<p><a href="https://i.sstatic.net/kWa7z.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/kWa7z.png" alt="enter image description here"></a></p>

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

CC BY-SA 4.0
2020-04-28T09:38:07.347Z
2020-04-28T09:38:07.347Z
MCG
<p>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.</p>

<p>This <a href="https://skyandtelescope.org/astronomy-blogs/explore-night-bob-king/chase-potentially-hazardous-asteroid-earth-flyby/" rel="nofollow noreferrer">Sky &amp; Telescope article</a> 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).</p>

<p>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!</p>

CC BY-SA 4.0
2020-04-29T03:55:49.463Z
2020-04-29T03:55:49.463Z
Mike G
<p>Based on this answer and sources therein, the absolute magnitude of an asteroid is given by</p>

<p><span class="math-container">$$M_{Abs} = 5 \left(\log_{10}(1329) -\frac{1}{2}\log_{10}(\text{albedo}) -\log_{10}(D_{km})\right)$$</span></p>

<p>where <span class="math-container">$D_{km}$</span> 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.</p>

<p>According to Wikipedia's <a href="https://en.wikipedia.org/wiki/(52768)_1998_OR2" rel="nofollow noreferrer">(52768) 1998 OR2</a> 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.</p>

<p>The linked answer also gives some information on the apparent magnitude:</p>

<blockquote>
  <p>The apparent magnitude <a href="https://space.stackexchange.com/a/25121/12102">from this answer</a>:</p>
  
  <blockquote>
    <p>Knowing the absolute magnitude of an object, you calculate the apparent magnitude <span class="math-container">$m$</span> using:</p>
    
    <p><span class="math-container">$$ m = M_{Abs} + 5 \log_{10}\left(\frac{d_{SR} \ d_{RE}}{1 \ \text{AU}^2 O(1)}\right), $$</span></p>
    
    <p>where <span class="math-container">$d_{SR}$</span> and <span class="math-container">$d_{RE}$</span> are the <s>Sun-Roadster and Roadster-Earth</s> Sun-satellite and satellite-Earth distances, each normalized by 1 AU, and the factor <span class="math-container">$O(1)$</span> is the <a href="https://en.wikipedia.org/wiki/Bond_albedo#Phase_integral" rel="nofollow noreferrer">phase integral</a>, 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 <a href="https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H)" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Absolute_magnitude#Solar_System_bodies_(H)</a>.</p>
  </blockquote>
</blockquote>

<p>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. </p>

<p>However, Wikipedia's <a href="https://en.wikipedia.org/wiki/(52768)_1998_OR2#2020_approach" rel="nofollow noreferrer">(52768) 1998 OR2; 2020 approach</a> says:</p>

<blockquote>
  <p>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.)</p>
</blockquote>

<p>So at 0.042 AU it will be 1/(0.042)<sup>2</sup> 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.</p>

CC BY-SA 4.0
2020-04-28T11:20:47.750Z
2020-04-28T11:20:47.750Z
uhoh
<p>Will I be able to see the <a href="https://en.wikipedia.org/wiki/(52768)_1998_OR2" rel="nofollow noreferrer">asteroid 1998 OR2</a> with the naked eye on April 29? </p>

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

CC BY-SA 4.0
2020-04-28T03:36:11.990Z
This is an interesting question! I've edited your question to add the name of the asteroid that will pass by on April 29.
uhoh
<amateur-observing><asteroids><naked-eye>
Can we see asteroid 1998 OR2 with unaided eye?
rbhat
# Can we see asteroid 1998 OR2 with unaided eye?

(Asked by: username_0 on 2020-04-28)

Will I be able to see the [asteroid 1998 OR2](https://en.wikipedia.org/wiki/(52768)_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?

**Tags:** amateur-observing, asteroids, naked-eye
> **Comments:**
> **username_1**: This is an interesting question! I've edited your question to add the name of the asteroid that will pass by on April 29.  

---

## 3 answers

### Answer by username_2 on 2020-04-28. Score: 2

According to the page on [The Sky Live](https://theskylive.com/1998or2-info) 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](https://i.sstatic.net/kWa7z.png)](https://i.sstatic.net/kWa7z.png)

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


---

### Answer by username_3 on 2020-04-29. Score: 1

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](https://skyandtelescope.org/astronomy-blogs/explore-night-bob-king/chase-potentially-hazardous-asteroid-earth-flyby/) 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!


---

### Answer by username_1 on 2020-04-28. Score: 0

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](https://en.wikipedia.org/wiki/(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](https://space.stackexchange.com/a/25121/12102):
>
> > 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](https://en.wikipedia.org/wiki/Bond_albedo#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](https://en.wikipedia.org/wiki/(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.


---

