<p>It's too bright to look at. Your astronaut has permanent retinal damage</p>

<p>The apparent magnitude of the sun from Earth is -26.7. From the asteroid belt it is -24.7 (<a href="http://www.1728.org/magntudj.htm" rel="nofollow noreferrer">http://www.1728.org/magntudj.htm</a>). So it is dimmer, but it is still far too bright for her to look at without eye protection.</p>

<p>Since she is standing in the vacuum of space, there is no atmosphere to protect her eyes from ultraviolet. Her eyes would be damaged by the intensity of the light. Within a few days her vision would become fuzzy, from the production of free radicals from ultraviolet exposure.</p>

<p>If she shaded her eyes from the sun, then the lack of atmosphere would help. With no atmosphere to scatter the sunlight, the sky around the sun would be dark, and the planets would be pretty bright. However, you specify "no light blockers" which means that our astronaut's eyes are basically toast.</p>

CC BY-SA 3.0
2017-03-21T21:05:28.183Z
I bookmarked the link; very handy.
catsteevens
CC BY-SA 3.0
2017-03-21T20:18:23.370Z
2017-03-21T20:18:23.370Z
James K
<blockquote>
  <p>What is the apparent magnitude of the Sun at that distance?</p>
</blockquote>

<p>Fortunately, there's a formula known as the <a href="https://en.wikipedia.org/wiki/Distance_modulus" rel="noreferrer">Distance Modulus</a> which is defined to calculate precisely this information. The equation is given by</p>

<p>$$m-M=5\log_{10}(d_{pc}) - 5$$</p>

<p>where $m$ is the <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="noreferrer">apparent magnitude</a> of your star, $M$ is the <a href="https://en.wikipedia.org/wiki/Absolute_magnitude" rel="noreferrer">absolute magnitude</a>, and $d$ is the distance from the star, in units of $\mathrm{parsecs}$. The absolute magnitude of the Sun is easily looked up and <a href="https://www.google.com/search?q=absolute%20magnitude%20sun&amp;ie=utf-8&amp;oe=utf-8" rel="noreferrer">found to be</a> $M_\odot = 4.83$. The distance, you provide as $2.5\:\mathrm{AU} = 1.2120342\times10^{-5}\:\mathrm{pc}$. Now its simple to solve for $m$.</p>

<p>$$m = 5\log_{10}(1.2120342\times10^{-5}) - 5 + 4.83$$</p>

<p>$$m = -24.75$$</p>

<p>As far as apparent magnitudes go, this is still pretty bright. You can see a table of typical apparent magnitudes for comparison <a href="https://en.wikipedia.org/wiki/Apparent_magnitude#Table_of_notable_celestial_objects" rel="noreferrer">right here</a>. Interestingly, this table lists $-25$ as the "minimum brightness that causes the typical eye slight pain to look at". At $m=-24.75$ (remember higher value = less bright) you'd be just above this threshold and so the Sun would be safe to look at without causing eye strain or pain.</p>

<blockquote>
  <p>Is that bright enough to blot out any of the inner planets?</p>
</blockquote>

<p>In order to determine the the brightness of the inner planets, we have to do some more math. Unfortunately, people don't really calculate the absolute magnitudes of planets so the form of the distance modulus listed above isn't entirely useful. Instead, we'll use a different form.</p>

<p>$$m_1 - m_2 = -5\log_{10}\left(\frac{d_2}{d_1}\right)$$</p>

<p>This equation relates an object with an apparent magnitude of $m_1$ at a distance $d_1$ to its apparent magnitude $m_2$ at a distance $d_2$. If we look up the apparent magnitude of the various inner systems from Earth, and figure out their distance from Earth at those times, we can figure out the apparent magnitude at the distance of your asteroid.</p>

<p>I'll work this out for Venus since Venus will likely be one of the brightest of the inner planets (given its high <a href="https://en.wikipedia.org/wiki/Albedo" rel="noreferrer">albedo</a>). Venus as viewed from Earth, at its brightest, has an apparent magnitude of $m_1 = −4.89$. This occurs when it is an the point of greatest elongation. As determined from this <a href="https://www.fourmilab.ch/cgi-bin/Solar" rel="noreferrer">solar system orrery</a>, the distance is approximately $d_1 = 0.75\:\mathrm{AU} = 0.36\times10^{-5}\:\mathrm{pc}$. On your asteroid, you'll be about $d_2 = 2.17\:\mathrm{AU} = 1.05\times10^{-5}\:\mathrm{pc}$ from Venus at the same point of maximum brightness (based on a little trig). From this we can calculate $m_2$ to be</p>

<p>$$m_2 = −4.89 + 5\log_{10}\left(\frac{1.05\times10^{-5}\:\mathrm{pc}}{0.36\times10^{-5}\:\mathrm{pc}}\right) = -2.57$$</p>

<p>From the comparison table listed above, this would make Venus, at its brightest, about as dim as the "minimum brightness of the Moon". This is a rough calculation, but I don't expect the actual answer to vary too much from this value.</p>

<p>Keep in mind, that's the maximum brightness of one of the brightest inner planets. Any other time Venus will be dimmer and all other planets will never be brighter than this. So all in all, the inner planets will likely be outshined by the much brighter Sun. There's a chance, with the right conditions, if you managed to properly block the light from the Sun using some sort of occulting disk or <a href="https://exoplanets.nasa.gov/resources/1015/" rel="noreferrer">flower starshade</a>, that you could see the inner planets with the naked eye as they still would be pretty bright in their own right. You just need to get rid of the sunlight outshining them. </p>

CC BY-SA 3.0
2017-03-21T20:16:56.147Z
For a quicker calculation, each +1 magnitude is 2.5x dimmer, so if you're 2.5 times as far away, that is +2 magnitude (square the distance), or -27 to -25 for the Sun.
Trip Space-Parasite
CC BY-SA 3.0
2017-03-21T21:10:56.720Z
I was told that, at its greatest extent, Earth would be 22 deg away from the sun. Do you think it would visible by the unaided naked eye at that point?
catsteevens
CC BY-SA 3.0
2017-03-21T21:29:20.113Z
@catsteevens like I said, possibly if you could block out the Sun with a star shade. It's certainly gonna be bright, the problem is that the Sun is much brighter.
zephyr
CC BY-SA 3.0
2017-03-21T20:12:26.107Z
2017-03-21T20:12:26.107Z
zephyr
<p>Imagine an astronaut standing on an asteroid 2.5 A.U. distant from the Sun. He is gazing, naked-eye, towards the Sun.
<a href="https://i.sstatic.net/Bt7TX.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/Bt7TX.png" alt="enter image description here"></a></p>

<p>What is the apparent magnitude of the Sun at that distance, and is that bright enough to blot out any of the inner planets? No light blockers and solely viewing with the Eyeball Mark I?</p>

<the-sun><apparent-magnitude>
Sun's apparent magnitude at 2.5 AU?
catsteevens
# Sun's apparent magnitude at 2.5 AU?

(Asked by: username_0 on 2017-03-21)

Imagine an astronaut standing on an asteroid 2.5 A.U. distant from the Sun. He is gazing, naked-eye, towards the Sun.
[![enter image description here](https://i.sstatic.net/Bt7TX.png)](https://i.sstatic.net/Bt7TX.png)

What is the apparent magnitude of the Sun at that distance, and is that bright enough to blot out any of the inner planets? No light blockers and solely viewing with the Eyeball Mark I?

**Tags:** the-sun, apparent-magnitude


---

## 2 answers

### Answer by username_2 on 2017-03-21. Score: 7 | Accepted answer

> What is the apparent magnitude of the Sun at that distance?

Fortunately, there's a formula known as the [Distance Modulus](https://en.wikipedia.org/wiki/Distance_modulus) which is defined to calculate precisely this information. The equation is given by

$$m-M=5\log\_{10}(d\_{pc}) - 5$$

where $m$ is the [apparent magnitude](https://en.wikipedia.org/wiki/Apparent_magnitude) of your star, $M$ is the [absolute magnitude](https://en.wikipedia.org/wiki/Absolute_magnitude), and $d$ is the distance from the star, in units of $\mathrm{parsecs}$. The absolute magnitude of the Sun is easily looked up and [found to be](https://www.google.com/search?q=absolute%20magnitude%20sun&ie=utf-8&oe=utf-8) $M\_\odot = 4.83$. The distance, you provide as $2.5\:\mathrm{AU} = 1.2120342\times10^{-5}\:\mathrm{pc}$. Now its simple to solve for $m$.

$$m = 5\log\_{10}(1.2120342\times10^{-5}) - 5 + 4.83$$

$$m = -24.75$$

As far as apparent magnitudes go, this is still pretty bright. You can see a table of typical apparent magnitudes for comparison [right here](https://en.wikipedia.org/wiki/Apparent_magnitude#Table_of_notable_celestial_objects). Interestingly, this table lists $-25$ as the "minimum brightness that causes the typical eye slight pain to look at". At $m=-24.75$ (remember higher value = less bright) you'd be just above this threshold and so the Sun would be safe to look at without causing eye strain or pain.

> Is that bright enough to blot out any of the inner planets?

In order to determine the the brightness of the inner planets, we have to do some more math. Unfortunately, people don't really calculate the absolute magnitudes of planets so the form of the distance modulus listed above isn't entirely useful. Instead, we'll use a different form.

$$m\_1 - m\_2 = -5\log\_{10}\left(\frac{d\_2}{d\_1}\right)$$

This equation relates an object with an apparent magnitude of $m\_1$ at a distance $d\_1$ to its apparent magnitude $m\_2$ at a distance $d\_2$. If we look up the apparent magnitude of the various inner systems from Earth, and figure out their distance from Earth at those times, we can figure out the apparent magnitude at the distance of your asteroid.

I'll work this out for Venus since Venus will likely be one of the brightest of the inner planets (given its high [albedo](https://en.wikipedia.org/wiki/Albedo)). Venus as viewed from Earth, at its brightest, has an apparent magnitude of $m\_1 = −4.89$. This occurs when it is an the point of greatest elongation. As determined from this [solar system orrery](https://www.fourmilab.ch/cgi-bin/Solar), the distance is approximately $d\_1 = 0.75\:\mathrm{AU} = 0.36\times10^{-5}\:\mathrm{pc}$. On your asteroid, you'll be about $d\_2 = 2.17\:\mathrm{AU} = 1.05\times10^{-5}\:\mathrm{pc}$ from Venus at the same point of maximum brightness (based on a little trig). From this we can calculate $m\_2$ to be

$$m\_2 = −4.89 + 5\log\_{10}\left(\frac{1.05\times10^{-5}\:\mathrm{pc}}{0.36\times10^{-5}\:\mathrm{pc}}\right) = -2.57$$

From the comparison table listed above, this would make Venus, at its brightest, about as dim as the "minimum brightness of the Moon". This is a rough calculation, but I don't expect the actual answer to vary too much from this value.

Keep in mind, that's the maximum brightness of one of the brightest inner planets. Any other time Venus will be dimmer and all other planets will never be brighter than this. So all in all, the inner planets will likely be outshined by the much brighter Sun. There's a chance, with the right conditions, if you managed to properly block the light from the Sun using some sort of occulting disk or [flower starshade](https://exoplanets.nasa.gov/resources/1015/), that you could see the inner planets with the naked eye as they still would be pretty bright in their own right. You just need to get rid of the sunlight outshining them.

> **Comments:**
> **username_3**: For a quicker calculation, each +1 magnitude is 2.5x dimmer, so if you're 2.5 times as far away, that is +2 magnitude (square the distance), or -27 to -25 for the Sun.  
> **username_0**: I was told that, at its greatest extent, Earth would be 22 deg away from the sun. Do you think it would visible by the unaided naked eye at that point?  
> **username_2**: @username_0 like I said, possibly if you could block out the Sun with a star shade. It's certainly gonna be bright, the problem is that the Sun is much brighter.  

---

### Answer by username_1 on 2017-03-21. Score: 4

It's too bright to look at. Your astronaut has permanent retinal damage

The apparent magnitude of the sun from Earth is -26.7. From the asteroid belt it is -24.7 (<http://www.1728.org/magntudj.htm>). So it is dimmer, but it is still far too bright for her to look at without eye protection.

Since she is standing in the vacuum of space, there is no atmosphere to protect her eyes from ultraviolet. Her eyes would be damaged by the intensity of the light. Within a few days her vision would become fuzzy, from the production of free radicals from ultraviolet exposure.

If she shaded her eyes from the sun, then the lack of atmosphere would help. With no atmosphere to scatter the sunlight, the sky around the sun would be dark, and the planets would be pretty bright. However, you specify "no light blockers" which means that our astronaut's eyes are basically toast.

> **Comments:**
> **username_0**: I bookmarked the link; very handy.  

---

