<p>I've been presented with this problem:    </p>

<blockquote>
  <p>Say that Jupiter, with its diameter of 142,000 km, was located where Mars now orbits.  What would be the angular size of (the newly-relocated) Jupiter during a close approach, when its distance would be 79,300,000 km?  Would we be able to see Jupiter as a round object with our unaided eye, or only as a point of light?      </p>
</blockquote>

<p>The angular size is easy enough to calculate given $ \frac{Angular \space Size}{206,000} = \frac{Linear \space Size}{Distance} $.<br>
So $\frac{Angular \space Size}{206,000} = \frac{142000}{79300000}$ meaning Angular Size = 369 seconds of arc.     </p>

<p>However, would the unaided eye be able to see this? Knowing that the eye can detect 1 minute of arc across, surely  it can detect 6.15? Any help appreciated, thanks!</p>

CC BY-SA 3.0
2016-01-26T20:12:52.930Z
One way to look at it is, moon is about 30, so about 1/5 the diameter of the moon. Yes, certainly visible as a "round" object.
Jeff Y
<naked-eye><angular-resolution>
Seconds of Arc and the Unaided Eye
Jester Jeffrey
# Seconds of Arc and the Unaided Eye

(Asked by: username_0 on 2016-01-26)

I've been presented with this problem:

> Say that Jupiter, with its diameter of 142,000 km, was located where Mars now orbits. What would be the angular size of (the newly-relocated) Jupiter during a close approach, when its distance would be 79,300,000 km? Would we be able to see Jupiter as a round object with our unaided eye, or only as a point of light?

The angular size is easy enough to calculate given $ \frac{Angular \space Size}{206,000} = \frac{Linear \space Size}{Distance} $.  
So $\frac{Angular \space Size}{206,000} = \frac{142000}{79300000}$ meaning Angular Size = 369 seconds of arc.

However, would the unaided eye be able to see this? Knowing that the eye can detect 1 minute of arc across, surely it can detect 6.15? Any help appreciated, thanks!

**Tags:** naked-eye, angular-resolution
> **Comments:**
> **username_1**: One way to look at it is, moon is about 30, so about 1/5 the diameter of the moon. Yes, certainly visible as a "round" object.  

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## 0 answers

