Stargazing Spatial and Angular Resolution of a Earth observing Telescope

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The discussion focuses on determining the required diameter of a primary mirror for an Earth observing telescope, with participants sharing equations for angular and spatial resolution. The key equations mentioned are Sinθ = 1.220 λ/D for angular resolution and Δl = 1.220 dλ/D for spatial resolution, where D is the diameter, d is the distance, and λ is the wavelength. Confusion arises regarding the correct angle to use, with clarification that the angle should be the object size divided by altitude, and it's recommended to use radians for small angles. A participant highlights that while the formulas provide a rough estimate, factors like optical aberrations also significantly impact resolution, suggesting that the simple diffraction limit may not be sufficient for accurate design. Overall, the discussion emphasizes the complexities involved in designing an effective Earth observing telescope.
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Does anybody know of a good reference for determining the required diameter of an Earth observing telescope's primary mirror? I am trying to find determine a rough estimate for a design I am working on. So far I have found the equation

Sin\theta = 1.220 \lambda/D

for angular resolution and

\Deltal = 1.220 d\lambda/D

for spatial resolution, where D is the Diameter, d is the distance, lambda is the wavelength. I've never really dealt with optics like this (minus like a class in high school long ago) but I do know my results are ridiculous at a 12 cm primary for a 500Km orbit

My confusion comes from what the angle I use is. The satellite would be orbiting between 200 and 500 Km and be taking 25 x 25 km images in full colour.

If anyone can help me with this it would be greatly appreciated.
 
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The angle is the object size (I assume you're talking about an orbiting telescope looking down on Earth?) divided by the altitude. You can forget the sines and tangents, the angles are tiny so just use the angle in radians. Note the angle is the half-angle, so the diffraction limit to "resolve" an object of diameter L requires a telescope diameter D given by D = 2.44*d*lamda/L. I use quotes on "resolve" because it's actually a lot more complicated than this simple formula.
 
Thank you. I just had just found something saying just this and this confirms that I have been using every angle but the right one! This formula should do for a first approximation. I just need to give the guys in the structures and mechanisms group an idea of how big this is going to be (seeing as they need to get it into the launch vehicle fairing). This has been a great help.
 
Hmmm... the data still seems wonky. According to this for 1m resolution at 600km I only need a mirror of about 1.1 m, which doesn't seem right given that every commercial sat I know has a mirror around 1.6 m and has a working resolution in colour of around 4m.
 
It's not just limited by diffraction, in fact that's probably the least of your worries.
 
Optical aberrations being what I need to reduce.
 
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UC Berkely, December 16, 2025 https://news.berkeley.edu/2025/12/16/whats-powering-these-mysterious-bright-blue-cosmic-flashes-astronomers-find-a-clue/ AT 2024wpp, a luminous fast blue optical transient, or LFBOT, is the bright blue spot at the upper right edge of its host galaxy, which is 1.1 billion light-years from Earth in (or near) a galaxy far, far away. Such objects are very bright (obiously) and very energetic. The article indicates that AT 2024wpp had a peak luminosity of 2-4 x...

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