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Old 11-February-2008, 04:56 PM
toothdust toothdust is offline
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Default Moon Landing leftovers?

My question is, can't you just look up with a powerful enough telescope and see the remains of the moon landings? I mean I know it would have to be REALLY powerful, but can it be done, like with a large ground based scope, or Hubble?
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Old 11-February-2008, 05:01 PM
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No. You would need a telescope an order of magnitude (pardon the pun) larger than any existing Earthbound telescope. Search on "Hubble Apollo" to see various discussions in this forum.
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Old 11-February-2008, 05:42 PM
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270m for the machinery, 2 miles for the footprints...
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Old 11-February-2008, 07:14 PM
HypothesisTesting HypothesisTesting is offline
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To resolve footprints, about 0.1 meters in size, you'd need a resolving angle about:

resolving angle = 0.1 m/ 4 X 10(8) m = 3 X 10(-10) radians = 5/ 100,000 arc-seconds.

This seems way above the resolving ability of any telescope yet built. I would estimate best resolving power with adaptive/active optics would be 1/100 arc-seconds. This would be the diffraction limit for a single 10 m circular aperature, like Keck. (I would estimate with adaptive optics we should be able to reach somewhere in the ball-park of the theoretical limit).

The only solution I see is an optical interferometer . Maybe one of those upcoming planet searching telescope arrays will have this feature and will be able to see the footprints? The Terrestrial Planet Finder looks like somewhat of an interferometer array (but that might be IR which I don't think is useful for footprints, maybe for the lunar dune buggies). I doubt Buzz would have gotten so worked up and punched that guy if he really didn't land on moon.
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Last edited by HypothesisTesting : 11-February-2008 at 08:30 PM.
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Old 11-February-2008, 09:08 PM
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Which would be easier, seen another earth sized exo-planet directly, or seen foortprints on the moon? I seriously want to know.
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Old 11-February-2008, 09:16 PM
korjik korjik is offline
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Quote:
Originally Posted by ravens_cry View Post
Which would be easier, seen another earth sized exo-planet directly, or seen foortprints on the moon? I seriously want to know.
Using Hypo' Testings numbers:

resolving angle= width/distance

goes to

Distance= width/ resolving angle

D= 1.2x10^6 m / 3x10^-10
D= .4x10^16 m= about .4 light years

so an earth sized exoplanet would be harder to resolve if it were further away then .4 light-years.
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Old 11-February-2008, 09:41 PM
JustAFriend JustAFriend is offline
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Telescopes, like computers, aren't magic.

Unfortunately people see movies and TV that make them think they are....
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Old 11-February-2008, 11:11 PM
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going by what i've seen on CSI:Miami, they should be able to take any run of the mill digital pic of the moon and enhance it a little bit to make out every individual grain of sand on the moon.
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Old 11-February-2008, 11:26 PM
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^ Only a security camera can do that. So if the moon uses a gas station or takes cash from an ATM maybe then we can image Apollo hardware.
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Old 12-February-2008, 12:03 AM
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Some more answers on this page.

http://www.tass-survey.org/richmond/...ar_lander.html

This is the closest we've ever come to revisiting an Apollo landing site. (Scroll to bottom.)
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Old 12-February-2008, 12:07 AM
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Well, When I was on a train ride through the Canadian Praries, I saw it so low on the horizen, it looked liek it was on the street of the town we passed by, maybe something happned then. lol. On the plus side, if a telescope to resolve exo-planets directly gets built (a more likely goal for the billions of dollers needed) then we WILL be able to take pics of the site.
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Old 12-February-2008, 03:09 PM
HypothesisTesting HypothesisTesting is offline
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Quote:
Originally Posted by korjik View Post
Using Hypo' Testings numbers:

resolving angle= width/distance

goes to

Distance= width/ resolving angle

D= 1.2x10^6 m / 3x10^-10
D= .4x10^16 m= about .4 light years

so an earth sized exoplanet would be harder to resolve if it were further away then .4 light-years.

great.

The closest earth-like planet would be further than 4 LY. So to resolve this planet would take

angle = 1.2X10(6) / 4 X 10(16) = 3 X 10^-11 radians = 5 X 10^-6 arc-sec.

Wow! Also, to resolve continents on these planets would take even more resolution.

To estimate the size of aperature required:

diffraction limit angle = (wavelength)/ diameter

diameter = wavelength / angle = 5 X 10^-7 / 3 X 10^-11 = 20 kilometer (aprox.).

So I would conclude we need an optical interferometer at least 20 km in effective diameter to resolve the closest earth-sized planets. To really get medium distance exoplanets and resolve surface features would probably take about a 1,000 km optical interferometer array (with great computing and navigation/pointing ability). Is NASA/ESA planning such an array (or something in that ball-park to resolve earth-like planets). They'd probably use coronographs to reduce starlight also.

When I was young, some texts taught that imaging exoplanets was sci fi. But I'm not so sure any more. We could quibble about what the major advances since then are, but I think computers and electronics were the biggest advance . I think NASA/ESA will someday (maybe 20 years?) show images of exoplanets.
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Last edited by HypothesisTesting : 12-February-2008 at 04:38 PM.
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Old 19-February-2008, 04:06 PM
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I just got an excellent answer to the same question

Last edited by clint : 19-February-2008 at 04:07 PM. Reason: edited wrong link
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