XPost: sci.physics.relativity   
   From: PointedEars@web.de   
      
   J. J. Lodder wrote:   
   > Paul.B.Andersen  wrote:   
   >> Den 17.07.2025 02:18, skrev Bertitaylor:   
   >>> Hubble and Webb can find so called black holes and planets light years   
   >>> away but cannot show a footprint on the Moon!   
   >>   
   >> Hubble photos of Moon.   
   >> []   
   >>   
   >> The best Hubble can do:   
   >> [> moon/>]   
   >>   
   >> "Hubble can resolve features as small as 280 feet across."   
   >   
   > Just what you expect, given that Hubble is diffraction-limited,   
      
   [sic!]   
      
   (AFAIK) *every* optical system (i.e., one based on light) has this   
   limitation.  That is merely a consequence of light being an electromagnetic   
   wave.  The best theoretical resolution is   
      
     θ ≈ 1.22 λ/D,   
      
   where λ is the wavelength and D is the aperture:   
      
      
      
   This is also the case for radio telescopes, for example (radio waves are   
   just long-wavelength electromagnetic waves), which is why one uses several   
   (sometimes many small) radio telescopes observing the same object, and   
   computers to combine their observations, to increase the resolution:   
      
      
      
   The Hubble Space Telescope (HST) is currently sensitive in the following   
   ranges:   
      
   Advanced Camera for Surveys (ACS), 350–1100 nm (visible to near-infrared);   
   Cosmic Origins Spectrograph (COS),  90– 320 nm (UV);   
   Near Infrared Camera and Multi-Object Spectrometer (NICMOS):   
                                      800–2500 nm (near-infrared);   
   Space Telescope Imaging Spectrograph (STIS),   
                                      115–1030 nm (UV to near-infrared);   
   Wide Field Camera 3 (WFC3),        200–1700 nm (UV to near-infrared).   
      
      
      
   A photo of a footprint on the Moon would have to be taken in the visible   
   range, 380–780 nm (definitions vary).   
      
   The diameter of the HST's reflector, and thus approximately its effective   
   aperture (typically not the entire reflector surface can be used), is 2.4 m:   
      
      
   (Infobox)   
      
      
      
   Thus in the visible range it has a maximum theoretical resolution of   
      
     θ ≈ 1.22 λ/D = (1.93166667 × 10^-7 to 3.96500 × 10^-7).   
      
   The resolution corresponds to the minimum diameter ∆ of the imaged object   
   according to tan(θ/2) = (∆/2)/d = ∆/(2d):   
      
                                       __   
                           ___...---''' ^   
               ___...---'''             |   
     _...---''' θ _ _ _ _ _ _ _ _ _ _ _ | ∆   
      '''---...___                      |   
                  '''---...___          |   
                              '''---..._v   
      
     <-------------------------------->   
                     d   
      
   Luna is d ≈ 384 000 km away on average, so for an object on the lunar   
   surface to be resolvable by the HST in the visible range, it has to have a   
   diameter ∆ of at least   
      
     ∆ ≈ 2d tan(θ/2) ≈ 2d θ/2 = d θ ≈ (74 to 152) m.   
      
   A footprint is /a lot/ smaller than that.   
      
   --   
   PointedEars   
      
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