Super Telephoto versus a Telescope for Lunar/Planetary Photography

Started Nov 6, 2012 | Discussions thread
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RustierOne
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Super Telephoto versus a Telescope for Lunar/Planetary Photography
Nov 6, 2012

Would a super telephoto or a telescope work better? For a super telephoto (say a 500 mm mirror-lens) the answer depends on what you intend to do. For full disc Moon photography, the 500 mm telephoto would be OK. For planets, the answer is largely no. What follows is the reason why. When I use the term “apparent size”, I’m referring to the angular diameter of the celestial object, not its physical diameter in miles or kilometers.

The Moon has a relatively large apparent size – about ½°. The planets have much smaller apparent sizes. Take for example Jupiter, with the largest apparent diameter of the planets. Currently its angular diameter is about 47 arc-seconds. This equates to an apparent size for Jupiter of 0.0131° (see * below). Putting this in other terms, the Moon’s angular size is over 38 times that of Jupiter’s (1/2° divided by 0.0131°). So compared to the Moon, Jupiter appears about as big as an average size (say 50 or 60-mile) Moon crater. And that’s for the largest planet. This is something that few aspiring astrophotographers appreciate – just how tiny the planets appear.

Let’s draw some conclusions. For a 500 mm focal length telephoto, the Moon forms an image on the sensor of 4.4 mm, while Jupiter’s image is only about 0.1 mm. See ** below. Since an APS-C sensor is about 15 mm high, the Moon would encompass quite a few pixels, while Jupiter would not. So the 500 mm telephoto would work OK for the Moon, but not for Jupiter. What about a telescope? Let’s take two examples.

First is a Celestron-8 (w/ f/6.3 focal reducer) – about 1260 mm focal length. For the Moon, the image size is 11 mm – a real nice match for the APS-C’s 15 mm sensor height. What about Jupiter? Its image size would be 0.3 mm – not a good match-up.

The next example is a Celestron-11 telescope with 2800 mm focal length. The Moon’s image size is about 24 mm – not good for full-disc shots, but OK for more zoomed-in views. For Jupiter, the image size is only about 0.6 mm! So even with this long focal length, something else is needed. And that is “eyepiece projection”, which greatly increases the focal length of the telescope for planetary imaging. That is a subject for later discussion.

To sum up:

  • Super-telephoto, on Moon (full-disc) – OK
  • Super-telephoto, on planets – not recommended ***
  • Telescope, on Moon (full disc or zoomed) – OK depending on focal length
  • Telescope, on planets – best with eyepiece projection

* There are 60 arc-seconds per arc-minute and 60 arc-minutes per degree.
So one arc-second is 1/3600 of a degree. So Jupiter’s 47 arc-second size is 47/3600of a degree or 0.0131°.

** Image size = tangent (angular size) X Focal Length

*** For super-telephoto users, you may be able get some useable results on planets, with tele-converters to increase focal length.

The above discussion is for your consideration. Comments and critique are welcomed.

Best Regards,
Russ

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