<p>Just divide the actual distance by the magnification.  In your example that would give 750 million /50, which is 15 million miles.  (Actually saturn is currently over 800 million miles from earth).</p>

<p>The reason this works is that a magnification of M means the object is M times its apparent linear size, which requires being M times closer.  This is a slight approximation, but is fine for distant objects.</p>

CC BY-SA 3.0
2017-08-04T13:44:29.780Z
Hi. Thanks Dr. Chuck. I'm glad it's a simple matter. So I would need to be 15 million miles from Saturn or 745 or 785 million miles from Earth. Amazing bit of distance jumping through a simple telescope! Thanks again
Kevin
CC BY-SA 3.0
2017-08-03T20:39:27.350Z
2017-08-03T20:39:27.350Z
Dr Chuck
<p>I'm looking at Saturn through a modest telescope at 50x. What point in space, expressed as the distance from Saturn, let's say it's 750 million miles from Earth, would I need to view Saturn with the naked eye and see it at the same apparent size as viewed from Earth at 50x. How do I estimate this in general for other objects as well. Thanks!</p>

<telescope><observational-astronomy><amateur-observing>
Telescope magnification compared to distance from an object
Kevin
# Telescope magnification compared to distance from an object

(Asked by: username_0 on 2017-08-03)

I'm looking at Saturn through a modest telescope at 50x. What point in space, expressed as the distance from Saturn, let's say it's 750 million miles from Earth, would I need to view Saturn with the naked eye and see it at the same apparent size as viewed from Earth at 50x. How do I estimate this in general for other objects as well. Thanks!

**Tags:** telescope, observational-astronomy, amateur-observing


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## 1 answers

### Answer by username_1 on 2017-08-03. Score: 5 | Accepted answer

Just divide the actual distance by the magnification. In your example that would give 750 million /50, which is 15 million miles. (Actually saturn is currently over 800 million miles from earth).

The reason this works is that a magnification of M means the object is M times its apparent linear size, which requires being M times closer. This is a slight approximation, but is fine for distant objects.

> **Comments:**
> **username_0**: Hi. Thanks Dr. Chuck. I'm glad it's a simple matter. So I would need to be 15 million miles from Saturn or 745 or 785 million miles from Earth. Amazing bit of distance jumping through a simple telescope! Thanks again  

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