I'm sure this has come up before, but whenever I talk to someone about hoax people, we always wonder what keeps Hubble from snapping a picture of the landing site(s)?

I imagine that cost of tasking it is one reason that it hasn't been done. I can accept that they don't want to lend credit to the hoax believers. I also realize that it's like finding a needle in a large haystack.

But my question is, if those other problems were solved, is it technologically possible? I heard once that if Hubble were in San Francisco, it could see a quarter in Washington DC.

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"Lisa, in this house we obey the laws of thermodynamics!"

One of the reasons that hubble rarely takes pictures of the moon is because it is very hard to track it. The time it did take a pic (and I will use a quote from the Bad Astronomy book *bows*, becuase it sounds awesome) it was in "Ambush Mode". (That sounds so awesome 8) ) It aimed where the mood would be and snapped a pic at the time it arrived.

I agree that the hubble could take a pic of a quarter in WDC, but only if there wasn't an atmosphere... (this is just a guess, I have nothing to back that up.)

But, I would bet that if it could track the moon and stuff, it prolly could. (Again, this is just speculation on my part.)

[Edited for spelling and clarity]

The answer is no. Though someone will give you a more detailed explanation, the basics are a) the moon is too bright when there is sunlight on it, and b) when there isn't, the resolution of the Hubble could only detect things that are about 100m wide on the moon and the LM's landing structure is only about 10m which is too small to be seen.


There is a satilite photo floating about the shows a black dot at one of the landing sites though. Can't recall where off hand, and it's not too great again due to resolution sizes.

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Howling from the Shadows

It must be fun to lead a life completely unburdened by reality. --- JayUtah

You can't reason an irrational person out of an irrational belief. --- Noclevername

Apollo: The History and the Hoax
Enter the World of Athran

^^ He knows more than me, I'd put more money on his answer than mine . Ambush Mode... wow... that is the coolest phrase... Ok I'm done...

try reading this post, it will give you a better understanding as to why hubble can't be used for such things.

http://www.badastronomy.com/phpBB/viewtopic.php?t=12236

Other telescopes in the future will not even solve the problem. The woo-woos will claim that robotic missions put stuff up there to make it appear that the landings were real.

Heck you could even take them up to the moon personally and let them see the site for themselves. Then when they got back they would just claim that you fed them drugs to make them think they were there and saw it.

Welcome to the board. This thread may help with your question.
http://www.badastronomy.com/phpBB/viewtopic.php?t=10414

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This question comes up frequently enought that maybe the BA should add it to the website. I looked around but didn't see any mention.

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At night the stars put on a show for free (Carole King)

One Earth, One Sky - IYA 2009

This question has been pretty much answered already, but let me put in my 2 cents anyway. The following is a quote from my Web page:

Try here.

That looks like the one.

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Howling from the Shadows

It must be fun to lead a life completely unburdened by reality. --- JayUtah

You can't reason an irrational person out of an irrational belief. --- Noclevername

Apollo: The History and the Hoax
Enter the World of Athran

It's a case of physics. Given an aperture of the optical assembly and the wavelength being observed, there is only so small an angular distance that can be resolved.

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Freedom For Fission A breath of fresh Iodine-131

I'm an Aussie, so I don't really know the width of a quarter or the distance between San Francisco and Washington. But I imagine you do!

So as an exercise, you might like to calculate the angle subtended by the coin in the situation you describe above. It may well be that the coin would be visible, but it also may well be that the angle is greater than the angle subtended by a Lunar Module descent stage on the Moon.

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"If a tree is cut down in the rainforest, and is used to make paper to print a book, and the book is really bad, and there's nobody that will read it, do you still hear a sucking sound?"
Charlie in Dayton, A.AsC.

I myself have used 90 meters as the smallest object Hubble can resolve on the Moon, which is about equivalent to the 300 ft cited in an earlier post. Simple calculations using Dawes' Limit yield these numbers. But Dawes' Limit really only tells you how close both components of a double star can be to yet be detected as double, assuming both components are of the same magnitude. I've also seen Dawes' Limit used in telescope catalogues to calculate the smallest crater that can be seen on the Moon, which is similar to what we're doing here. Can Dawes' Limit really derive useful numbers for the detection of extended objects of low contrast?

Here's my BOE calculation.

The distance from SF to D.C. is 2,850 miles, or 1.806 e8 in. A quarter is just under an inch in diameter. For this I'll just round up.

Then I set up my right triangle to have an "opposite" leg, "o," of 0.5 inch and an "adjacent" leg, "a," of 1.806 e8 inches. So theta, the angle I'm looking for, has a tangent of o/a, or 2.769e-9

Take the arctan, and get 1.587e-7 degrees. Mulitply by 2 to get the full inch, so 3.174e-7 degrees. To get seconds, multiply by 3600.

That's roughly 0.001 seconds or arc.

I don't think Hubble could resolve that, even in the absence of an atmosphere.

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Jeff Schwarz
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Argh!! They booby-trapped their sun!!****--Invader ZIM

Dawes' Limit is the maximum Theoretical resolving power. Rarely are you actually able to resolve to that limit. The seeing and clarity of the atmosphere would have a lot to do with what you would actually be able to resolve. In the case of the Hubble, where there is no atmosphere to affect the resolving power, you should be able to approach the Limit (But there may be other things that may prevent the Hubble from reaching the limit. Phil, anyone else know?) . As far as an extended object, if it is larger than the limit, you should be able to resolve it, at least into dark and light areas, even if it is low contrast.

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Some try to tell me, thoughts they cannot defend,... - Moody Blues.

Since NASA is currently planning to take Hubble out by sending up a booster to de-orbit it, how much more difficult would it be to de-orbit into the moon, and take pictures on the way in? De-orbit as in make the orbit more and more elliptical until it intercepts the moon or use an ion engine like the probe that is taking 18 months to get to the moon. Unfortunately, if they wait until it is failing, and it fails before getting near the moon, it is more evidence of the conspiracy.

According to the Dawes' Limit formula, the theoretical resolving power of the HST is equivalent to resolving a one-meter object for every 4,250 km of distance. This relationship is linear.

It's lots more difficult. Deorbiting the HST can be accomplished with as little as a few hundred feet per second delta-v. Putting the HST on a translunar trajectory is a matter of about 10,000 feet per second.

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"Facts are stubborn things." --John Adams
Clavius Moon Base

And at the end of that day, NASA isn't going to bother. Why waste valuable time and money that would be better spent on actual science?

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Freedom For Fission A breath of fresh Iodine-131