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  #1121  
Old December 23rd 07, 11:41 PM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1116/1120]
  #1122  
Old December 23rd 07, 11:41 PM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1117/1120]
  #1123  
Old December 23rd 07, 11:41 PM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1118/1120]
  #1124  
Old December 23rd 07, 11:41 PM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1119/1120]
  #1125  
Old December 23rd 07, 11:41 PM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1120/1120]
  #1126  
Old December 24th 07, 12:15 AM posted to alt.binaries.pictures.astro
Ralph Hertle
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Posts: 1,193
Default Moon MOON-SHADOW-5000-3.tif [1143/1143]
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  #1127  
Old December 24th 07, 10:16 AM posted to alt.binaries.pictures.astro
Geoff[_4_]
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Posts: 36
Default Moon MOON-SHADOW-5000-3.tif [0001/1120]
The line-of-sight distance (LOSD) formula is well known. The distance
to the horizon for any height, h, above a spherical body is d =
sqrt(2*R*h + h^2) where d is distance to horizon and R is mean radius
of the spherical body. Where h, d and R are all expressed in units of
meters, the Earth horizon for a 1.7 meter tall observer at sea level,
the horizon is at 4.65km. (R=6371km)

For Luna, using R=1737.1, the horizon is 2.43km away for the same
observer.

Of course the length of a shadow is infinite at any point along the
terminator but if the sun angle is known then h can be calculated for
any object casting a finite shadow of length d on that body. If A is
the sun angle with respect to a plane normal to the base of h, the
length of the shadow would be proportional to R cos(A).

Since the sun for all intents is infinitely far away from the moon
compared to an observer on a crater rim, the line of sight from the
rim to a point on the surface at the shadow edge can be considered to
be on the line from the sun to the shadow. (Equivalent to the horizon
for that observer looking away from the sun.)

So if we can measure the length of the shadow at maximum extent,
(minimum sun angle), before the shadow goes infinite or reaches the
opposite rim, we can get a pretty good idea how high above the crater
floor the rim would be.

For a 1000 meter tall object, the LOSD is 58.92km.

You can try it out here.
http://newton.ex.ac.uk/research/qsys...ysics/horizon/
 
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