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A common question asked at Star Parties, workshops, or other gatherings where Astronomy is discussed is:

Why can't we see the flag that the Apollo Astronauts left on the moon?

The answer is some basic Wakefield High School mathematics* as follows:

A.    Lets assume that:

Triangle Values

  • The height of the flag (a) pole in ft. = 6
  • The average distance to the Moon (approx) in miles = 221,000
  • Which means that the distance to the Moon (approx) in feet (b) = 1,166,880,000

B.    Given the above:

  • The tangent (t) of the above in radians [tan(a/b)] = 0.0000000051, in radians
  • Converted to Degrees [180*(t/pi)] = 0.000000295 degrees
  • This means that a telescope would have to be able to resolve an angle of 2.95 x 10-07 degrees to see the flag left on the moon.

C.    That means:

Since one Arc Degree (1/360 of a circle) =1.000000000 Deg.
And one Arc Minute (1/60 of a degree) =0.016666667 Deg.
And one Arc Second (1/60 of an arc-minute) =0.000277778 Deg.
We can calculate that: 
The Angle to be resolved is:0.000000295 Deg.
And the Angle to be resolved in Arc Seconds is:0.001060596

So, to see the flag, a telescope must be able to resolve an angle of about 1/1000th of an arc second.

*The formulas in bold above can be used in most spreadsheets

"So how big of a telescope would that require?", you ask.

The Dawes limit calculates how close two objects can be and still be resolved by a telescope.

The formula is:  Resolution (in arc seconds) = 4.56/Diameter.

So, to determine the required diameter, the formula is:  D = 4.56/Resolution(arc seconds)

Doing the math, that is 4.56/.001060596 arc seconds, to see the flag a telescope would require a mirror 4,299.47 inches wide! That is 358.29 Feet!

A super-telescope THAT big would have to be in orbit, to eliminate the turbulence of the earth's atmosphere.

So, what distances can we see on the moon?

Actual distances to the moon based on Rukl's Atlas of the Moon:
Closest (perogee)356,400221,457
Farthest (apogee)>406,700252,712
Lunar Diameter34762159.89

The Moon's Angular Diameter at perogee (when it is closest to the earth) =33' 28.8" (33 min. 28.8 sec.)
Lunar Diameter in feet =11,404,199.48
Total Angular diameter in Arc Seconds =2008.80
Lunar Surface per Arc Second =5677.12 (ft.)
For a 10" telescope, Dawes Limit is 4.56/10 or0.456000 Arc-Seconds
So, the resolvable Distance with a 10" telescope is (5677.12)/(.456), or:2588.77 ft.
Our super-telescope distance resolved at .00106 Arc Seconds is:6.02 ft.

Note: The Hubble Space Telescope can resolve .005 Arc Seconds, or about 280 Feet
(See Sky & Telescope 10/2004 pg 130)

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