Earth Angle Above Horizon
"NiteCatty" has brought to my attention three Apollo images that show the earth at angles above the moon's alleged horizon:
NiteCatty indicated that in image A, the earth is about 15 degrees above the horizon, in image B the earth is about 40 degrees above the horizon, and in image C the earth is about 30 degrees above the horizon.
" We know that the moon always faces the earth with the same side. So when looking at the earth from the lunar surface, the earth should always show up at the same position in the sky, and at a fixed angle above the lunar horizen. Different photoes should show the earth at the same height above the lunar horizen.
" Calculating from the coordinate of the A17 lunar landing site (20.16 North and 30.77 East), the earth should be at 54 degree above the lunar horizen. That's pretty high in the sky. None of the three photoes is even close to this figure.
" How do you estimate the angle of the earth above the horizen from
photoes? We know that the diameter of the earth should have a visual
1.9 degree viewed from lunar distance. So you can use the visual size
earth in the photo as a good reference to estimate the angle."
On my screen, the earth diameter is 12 mm, and the distance to the "local horizon" is 100 mm.1.9 degrees x 100 mm / 12 mm = 16 degrees above the "local horizon", with "local horizon" defined as "the actual lower boundary of the observed sky".
On my screen, the earth diameter is 13 mm, and the distance to the horizon is 190 mm. There is a hill at the horizon, so I would estimate an ideal "flat" horizon distance of 220 mm.1.9 degrees x 220 mm / 13 mm = 32 degrees above the estimated ideal "flat" horizon.
On my screen, the earth diameter is 11 mm, and the distance to the estimated "flat" horizon is 195 mm.1.9 degrees x 195 mm / 11 mm = 34 degrees above the estimated ideal "flat" horizon.
NiteCatty's statement, "Calculating from the coordinate of the A17
lunar landing site (20.16 North and 30.77 East), the earth should be at
degree above the lunar horizen", is also calculated to be correct when
referring to the ideal "flat" horizon:
the landing coordinates for Apollo 17 are 20.2N, 30.8E.
Letting the radius of the moon = r,
the circumferential distance on the surface of the moon from 0 N/S, 0 E/W to 0 N/S, 30.8 E is
C(E) = Angle(E) x r
letting Angle(E) = [ ( 2 x PI ) / 360 ] x 30.8 = 0.54 radians.
giving C(E) = 0.54 r.
The circumferential distance on the surface of the moon from 0 N/S, 0 E/W to 20.2 N, 0 E/W is
C(N) = Angle(N) x r
letting Angle(N) = [ (2 x PI) / 360 ] x 20.2 = 0.35 radians
giving C(N) = 0.35 r.
The circumferential distance on the surface of the moon from 0 N/S, 0 E/W to 20.2 N, 30.8 E is
C(NE) = Angle (NE) x r
Using the Pythagorean theorem as a close approximation,
C(NE) ~ SQRT [ (C(N))^2 +
C (NE) ~ 0.64 r
Angle (NE) ~ C (NE) / r ~ 0.64
radians = 36.6 degrees
Finally, Nitecatty's satement "We know that the diameter of the
should have a visual angle of 1.9 degree viewed from lunar distance" is
confirmed as well:
Visual angle of the earth from the moon = visual angle of the moon
earth x [ D(earth) / D(moon)]
"Visual angle of the sun or moon = 0.5 deg "
" ... the sun and the moon each subtend about 1/2 degree"
Visual angle of the earth from the moon = 0.5 degree x [ 12,753 km /
(the higher-resolution version is available at
http://www.hq.nasa.gov/office/pao/History/alsj/a17/as17-134-20387-cropped.jpg), what would you estimate for the:
1. earth to "massif" (background hill) ridge angle (i.e. earth to "massif" local horizon angle)?
2. earth to "flat" (no "massif", all flat terrain) ridge angle (i.e. earth to estimated ideal "flat" horizon angle)?
considering also http://www.hq.nasa.gov/alsj/a17/20117337.jpg (the higher-resolution version is available at http://www.hq.nasa.gov/office/pao/History/alsj/a17/AS17-134-20386.jpg).
I believe these images were taken with a 70mm Hasselblad with lens
length 60mm and field of view 50 degrees horizontal, 66 degrees
Two more Apollo 17 EVA 2 images show the earth at a low angle above the local horizon.
On my screen, for both of these images, the earth diameter is 15 mm, and the distance to the local horizon is 125 mm.
1.9 degrees x 125 mm / 15 mm = 16 degrees above the local horizon.
A view of the Apollo 16 Command/Service Module over the moon, NASA Photo ID AS16-113-18282, at http://science.ksc.nasa.gov/mirrors/images/images/pao/AS16/10075829.jpg, shows a range finding crosshair missing where the C/SM is located.
During the Apollo era, the technology existed to produce fake images involving one image superimposed over another. Evidence for this comes from http://www.hq.nasa.gov/office/pao/History/alsj/Doble11.JPG. We know that the earth should be about 65 degrees above the lunar true horizon when viewed by the Apollo 11 astronauts. In the reflection in the Apollo 11 astronauts' visors, the earth is shown less than five degrees above the horizon.
Would superimposition of astronauts and equipment over range finding crosshairs indicate that the images were made by pasting a foreground image over a background image, probably by using some sort of darkroom or computer photographic processing technique?
The following image, from the 1969 World Book Science Year, page 22, could very well answer this question.
This is a close-up of the LM dish at the green ledge of the C/SM, from which the image is allegedly taken from.
Cover the light grey moon portion with a straight edge, such as a
paper, so that as much green, but only green, is showing. You will
a portion of the dish is still visible.
Email concerning this web page may be sent to David Wozney at email@example.com.
© 1996-2015 David P. Wozney