<p>Yes, brightness is inversely proportional to the square of the distance, and the square of 632 is about 400,000. It's probably appropriate to round the value to 1s.f. to give "600AU".  Of course, at that distance the sun appears point-like.</p>

<p>The distance to the Oort cloud is rather hard to define. It's not like you reach 1000 AU and see a <em>Star Wars</em> like "Asteroid field" of comet nuclei. For example Wikipedia gives a distance of 2000AU. </p>

CC BY-SA 4.0
2019-01-03T13:00:01.453Z
Thanks.  I did not think of the Oort Cloud as a wall starting a 1000AU (I was using NASA's estimate).  I had pictured it as a few comet nuclei spread out thinly at that distance.  I also assumed getting deeper into the Cloud it was more like the asteroid belt between Mars and Jupiter, than Star Wars weaving and twisting between objects.
Bob516
CC BY-SA 4.0
2019-01-03T13:00:52.667Z
What does 1s.f. mean?
Bob516
CC BY-SA 4.0
2019-01-03T17:04:25.360Z
1 significant figure.  We can round the distance figure the to the same degree that the brightness figure has been rounded.
James K
CC BY-SA 4.0
2020-02-15T16:43:50.033Z
The Oort cloud actually begins at about 10,000 au.
user30007
CC BY-SA 4.0
2019-01-03T08:32:54.173Z
2019-01-03T08:32:54.173Z
James K
<p>Is this correct?  <a href="http://blogs.discovermagazine.com/badastronomy/2012/03/15/bafact-math-how-bright-is-the-sun-from-pluto/#.XC2clS-ZNTY" rel="noreferrer">If the Sun is 400,000 times brighter than the full moon</a> am I correct in figuring that one would have to be about 632AU from the Sun for it to appear as bright as the full moon.  <a href="https://solarsystem.nasa.gov/solar-system/oort-cloud/overview/" rel="noreferrer">If the Oort Cloud begins about 1000AU from the Sun</a>, one would have to be more than 60% the way to the Oort Cloud for the Sun to appear as bright as the full moon.</p>

<the-sun><the-moon><apparent-magnitude>
Would one have to be more than halfway to the Oort Cloud for the Sun to appear as bright as the full moon?
Bob516
# Would one have to be more than halfway to the Oort Cloud for the Sun to appear as bright as the full moon?

(Asked by: username_0 on 2019-01-03)

Is this correct? [If the Sun is 400,000 times brighter than the full moon](http://blogs.discovermagazine.com/badastronomy/2012/03/15/bafact-math-how-bright-is-the-sun-from-pluto/#.XC2clS-ZNTY) am I correct in figuring that one would have to be about 632AU from the Sun for it to appear as bright as the full moon. [If the Oort Cloud begins about 1000AU from the Sun](https://solarsystem.nasa.gov/solar-system/oort-cloud/overview/), one would have to be more than 60% the way to the Oort Cloud for the Sun to appear as bright as the full moon.

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


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## 1 answers

### Answer by username_1 on 2019-01-03. Score: 5 | Accepted answer

Yes, brightness is inversely proportional to the square of the distance, and the square of 632 is about 400,000. It's probably appropriate to round the value to 1s.f. to give "600AU". Of course, at that distance the sun appears point-like.

The distance to the Oort cloud is rather hard to define. It's not like you reach 1000 AU and see a *Star Wars* like "Asteroid field" of comet nuclei. For example Wikipedia gives a distance of 2000AU.

> **Comments:**
> **username_0**: Thanks.  I did not think of the Oort Cloud as a wall starting a 1000AU (I was using NASA's estimate).  I had pictured it as a few comet nuclei spread out thinly at that distance.  I also assumed getting deeper into the Cloud it was more like the asteroid belt between Mars and Jupiter, than Star Wars weaving and twisting between objects.  
> **username_0**: What does 1s.f. mean?  
> **username_1**: 1 significant figure.  We can round the distance figure the to the same degree that the brightness figure has been rounded.  
> **username_2**: The Oort cloud actually begins at about 10,000 au.  

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