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

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

**A.** Lets assume that:

- 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**to see the flag left on the moon.^{-07}degrees

**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

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.

Actual distances to the moon based on Rukl's Atlas of the Moon: | ||
---|---|---|

Kilometers | Miles | |

Closest (perogee) | 356,400 | 221,457 |

Average | 384,401 | 238,856 |

Farthest (apogee) | 406,700 | 252,712 |

Lunar Diameter | 3476 | 2159.89 |

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 or | 0.456000 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)*