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Lunar landers.

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16 years 7 months ago #60949 by pj30something
Lunar landers. was created by pj30something
We all know man went to the moon (least i hope we all know and believe it).

I'm just curious......................

With all the high powered telescopes around today has any of them actually photographed the area's of the moon where the lunar landers were left behind and can they be seen in those images?

Or can no telescope get that close?

Also it doesnt help that the lunar landers were white..........they kinda blend in with the background surface.

Just thought i'd ask.

Paul C
My next scope is going to be a Vixen VMC200L Catadioptric OTA

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16 years 7 months ago #60951 by philiplardner
Replied by philiplardner on topic Re: Lunar landers.
The Moon is currently (01.21am 25Jan08) 383892km away from the Earth. All that is left of the lunar landers on the surface of the moon is the descent module (about 2 or 3m in diameter.) You'd need an awfully big scope (and no atmosphere) to be able to resolve a 2m object at 383,892,000m. I'll work out just how big your scope would need to be tomorrow (just for gits & shiggles) but right now I'm gonna catch some Zs.

Phil.

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16 years 7 months ago #60954 by pj30something
Replied by pj30something on topic Re: Lunar landers.
Thanks Phil. I look forward to your findings.

Paul C
My next scope is going to be a Vixen VMC200L Catadioptric OTA

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16 years 7 months ago #60971 by dave_lillis
Replied by dave_lillis on topic Re: Lunar landers.
No ground based scope in existance has a hope of detecting it.

Dave L. on facebook , See my images in flickr
Chairman. Shannonside Astronomy Club (Limerick)

Carrying around my 20" obsession is going to kill me,
but what a way to go. :)
+ 12"LX200, MK67, Meade2045, 4"refractor

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16 years 7 months ago #61003 by philiplardner
Replied by philiplardner on topic Re: Lunar landers.
Ok, the resolving power of any telescope is defined as the angular resolution that the scope is capable of seeing. This angle is easy to calculate: R=lambda/D where R is the minimum angle resolvable (in radians), lambda is the wavelength of light you are observing at (in our case the human eye is most sensitive to light at about about 555nm) and D is the diameter of the telescope.

So, a 6" telescope should be capable of resolving detail as small as 0.00000555 / 152 = 0.000003651radians = 0.000209187degrees = 0.75arcseconds.

Hubble has a 2.4m mirror and can resolve detail as small as 0.048arcseconds

Now using a little Pythagoras theorum for right-angle triangles we can work out that a 2m diameter object on the surface of the moon at a distance of 383,892,000m subtends an angle of 0.000,000,298degrees = 0.0010728arcseconds. So no hope of seeing your moonlander with a 6" scope then... or even with the Hubble! You would need a mirror 44.7 times as big as Hubble (or 107.4m in diameter) to hit the mark. You're gonna need a big 4x4 to carry that monster in the boot of the car! Oh and I haven't allowed for the atmosphere... which will cause you a bit of grief!

Phil.

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16 years 7 months ago #61005 by dave_lillis
Replied by dave_lillis on topic Re: Lunar landers.

0.0010728arcseconds.
Phil.

thats actually bigger then I imagined !

You're gonna need a big 4x4 to carry that monster in the boot of the car! Oh and I haven't allowed for the atmosphere... which will cause you a bit of grief!
Phil.


I reckon the 2 of us could lug it around, are you up to the challenge Phil?

Dave L. on facebook , See my images in flickr
Chairman. Shannonside Astronomy Club (Limerick)

Carrying around my 20" obsession is going to kill me,
but what a way to go. :)
+ 12"LX200, MK67, Meade2045, 4"refractor

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