As technology progresses could something like a Super Hubble see satellite the Earth to have a time delay to have any use?

Super Hubble A (or a huge mirror) is 30 light minutes away looking at Earth and Super Hubble B. Super Hubble B is near Earth looking at Super Hubble A.

Could you back track a plane, satellite, or car that disappeared on a live 30-min delayed feed of Earth without having to store data?

  • 1
    $\begingroup$ sounds like a very high tech version of a mercury delay tube. $\endgroup$ – JCRM Apr 9 '18 at 9:15
  • 2
    $\begingroup$ I mean you could also take a photo with your phone every second and email them to yourself. Then when you open your email, you can see photos taken in the past! $\endgroup$ – Dragongeek Apr 9 '18 at 11:14
  • $\begingroup$ voting to close as not related to space exploration. $\endgroup$ – Organic Marble Apr 9 '18 at 13:05
  • $\begingroup$ @OrganicMarble is this better? $\endgroup$ – Muze the good Troll. Apr 16 '19 at 5:44
  • 1
    $\begingroup$ This is a physics question. Patience is a challenge to creative people ;-) $\endgroup$ – uhoh Apr 16 '19 at 10:21

Yes, but this is overly complicated and have absolutely no application.

  • You'll be limited by A resolution, so you can remove B entirely.
  • It's the same as having a recorder and looking 30min back in time.
  • At 30 light minutes away, you are too far to have any good resolution on earth. You can achieve better by being closer to earth. You just need to record the data anyway
  • Your laser projector serve no purpose, it's just a data link.
| improve this answer | |
  • $\begingroup$ @Muze I don't think you understand how telescopes works... If a telescope is 30lm away it will be too far to see anything. If you use an other telescope to relay informations, you will be limited to its resolution in the first place, so this won't help you. $\endgroup$ – Antzi Apr 10 '18 at 0:30
  • $\begingroup$ Yes, but you are only going to be able to see at the mirror resolution. Which at 30lm away will be incredibly small. $\endgroup$ – Antzi Apr 10 '18 at 1:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.