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# Planetary Persceptive?

8 replies to this topic

### #1 cpsTN

cpsTN

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Posted 30 September 2013 - 09:13 PM

Since it is Ice Giant season, I was wondering what I could say to give people perspective on what they are seeing, in terms they could (and I) could understand. Something like, viewing Neptune from Earth is like seeing a pea in Seattle from Newark or something to that effect. Does anyone know where I could get information like this to put things into terrestrial perspective? For all the planets and stars, for that matter, not just U and N.

### #2 GlennLeDrew

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Posted 30 September 2013 - 09:43 PM

You already know the angular sizes of the planets at any time. The matter at hand is to determine at what distance a common objects subtends the same angle. We know that an object subtends an angular diameter of 1 arcsecond when it lies at a distance equal to 206,265 times its diameter.

For example, a 1/4" ball bearing would appear 1 arcsecond across when 206,265 * 0.25" = 51,566", or 4297', or 1,432 yd away.

To appear 10" across, the 1/4" ball bearing would have to be 1,432 / 10 = 143.2 yd.

And so your formula for distance at which an object of measured size subtends a given angular diameter (arcseconds):

206,265 * obj. diam. / arcseconds

### #3 Asbytec

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Posted 30 September 2013 - 10:45 PM

So, Uranus (4" arc) would be the size of a dime (0.7 inch) at ~1000 yards.

Another interesting site...
http://www.explorato...h/solar_system/

### #4 cpsTN

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Posted 01 October 2013 - 05:05 AM

Oops. Yes, gas giant. lol Thanks guys.

### #5 cpsTN

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Posted 01 October 2013 - 07:52 PM

If something is one arcsecond when it is 206,265x its diameter, does that mean if it is only 1/4 the distance, it will subtend 4 arcseconds or is it not a linear measure?

### #6 Asbytec

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Posted 01 October 2013 - 08:27 PM

Glenn's equation above is linear, so without doing the math the result would have to be linear.

### #7 GlennLeDrew

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Posted 01 October 2013 - 10:10 PM

Over these small angles, a linear approach is perfectly valid. Only when going past about 10 degrees or so would non-linearity become worth considering (depending on requirements for accuracy.)

### #8 Asbytec

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Posted 01 October 2013 - 10:29 PM

So, working this out, the Earth (7,926 miles) would appear 4" arc (same as Uranus's apparent angular size) at ~409 million miles. Uranus (31,250 miles) appears that same size at 1,610 million miles - 4 times the distance.

"The diameter of Uranus is 51,118 km. Just for comparison, this about 4 times bigger than the diameter of the Earth, at 12,742 km across."

That's it? Uranus is only 4 times the diameter of Earth? I though it was much larger.

### #9 azure1961p

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Posted 01 October 2013 - 11:10 PM

It's actually 4.007 Earths wide.

Pete

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