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Given the size of sphere 1, the dimensions of one reference point, the visible curve on sphere 1. What size is sphere 2 at 238,900 mi away?

[Picture from Apollo 11]

https://images-assets.nasa.gov/image/6900994/6900994~orig.jpg

[Moon Radius] Mean radius 1737.1 km

[Moon Distance] The Moon's average orbital distance is 384,402 km (238,856 mi) https://en.wikipedia.org/wiki/Moon

[Lander Dimensions] Dimensions 23 feet 1 inch (7.04 m) high 31 feet (9.4 m) wide 31 feet (9.4 m) deep overall, landing gear deployed https://en.wikipedia.org/wiki/Apollo_Lunar_Module

[What has sparked my curiosity]

https://epic.gsfc.nasa.gov/galleries/2015/lunar_transit/video?fbclid=IwAR23jlQ2wpIVWP5b4c21o3uT5qcLlbvAUDiapsMq4uM35_l0btWRvdwkNRc

How can I calculate this? And also prove my equations and calculations are correct? As of now, any hints or suggestions are appreciated.

[Clarification] What I am interested in, is using the visible curve of the moon in link 1 some how in an equation involving the dimensions of the lander to calculate the size of the earth in photograph from link 1. And even perhaps testing and verifying with two balls, a camera and a lego block.

[Alternative photo] Maybe it would be easier or better to use this photo? Going to have to see more about exact dimensions or how much their variance would change the results. https://www.hq.nasa.gov/alsj/a11/AS11-40-5924.jpg [This photo can be used as reference] https://www.hq.nasa.gov/alsj/a11/AS11-40-5929.jpg [Found huge trove of pictures, not many look useful] https://www.hq.nasa.gov/alsj/a11/AS11-40-5930HR.jpg [Odd lighting and shadows angels] https://www.hq.nasa.gov/alsj/a11/images11.html I still think my best bet is with what I listed within [Clarification]

Anthony Meadows
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1 Answers1

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Just based on the numbers you have given, the answer is not yet determined. We don't know how far the camera is from the moon or from the lander. Knowing the actual sizes of the moon and lander we could estimate the ratio of distances, that is, a number $k$ such that $$ d_L = k d_M, $$ where $d_L$ is the distance to the lander and $d_M$ is the distance to the moon.

But as far as we can say from the given numbers, the camera could be $239,000$ miles from the Moon, equipped with a telescopic lens powerful enough to fill the viewfield with just this small part of the Moon, and the other round object could be a beach ball $100$ miles away from the camera (so $238,900$ miles from the Moon) that someone has painted to look like the Earth.

Knowing the size of the Earth would put everything into scale. Knowing a few more distances would let us find the size of the Earth.

We might be able to use perspective to figure the distance to the lander if we had more detailed information (including the sizes a near part of the lander and a farther part of the lander, and the distance between them), though the camera seems to be far enough from the lander that this kind of "rangefinding" would be difficult.

In theory it should be possible to map the visible features on the Moon's surface to a three-dimensional map of the Moon, and use these to estimate the distance from the camera to the Moon.

The lunar transit video was taken from a location much farther from the Earth than the Moon is.

David K
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  • On July 6, 2015, DSCOVR returned its first publicly released view of the entire sunlit side of Earth from 1,475,207 km (916,651 mi) away, taken by the EPIC instrument. https://en.wikipedia.org/wiki/Deep_Space_Climate_Observatory#Operation Based off these observations(Containing a picture of both the moon and the earth), is it not safe to assume that the camera is no greater than or no less than 916,651 miles away? – Anthony Meadows Aug 17 '19 at 23:30
  • For the lunar transit video, the camera is far enough from both the Moon and the Earth that you can get a reasonably good estimate of the Earth's size simply by multiplying the known size of the Moon by the ratio of the apparent sizes of the Moon and Earth in the video. The lunar lander photo is a very different situation. – David K Aug 17 '19 at 23:33
  • any zoom in on the pictures from wikipedia ( or https://epic.gsfc.nasa.gov/galleries/2015/lunar_transit/video?fbclid=IwAR23jlQ2wpIVWP5b4c21o3uT5qcLlbvAUDiapsMq4uM35_l0btWRvdwkNRc) cannot shrink the size of the earth while increasing the size of the moon.. What I am interested in, is using the visible curve of the moon in link 1 some how in an equation involving the dimensions of the lander to calculate the size of the earth in photograph from link 1(https://images-assets.nasa.gov/image/6900994/6900994~orig.jpg). – Anthony Meadows Aug 17 '19 at 23:38
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    You can't change the relative sizes by zooming, but what we cannot tell from the picture is: is the Earth closer to us than the Moon, or the other way around? In the lunar transit it is obviously "the other way around". – David K Aug 17 '19 at 23:40
  • And even perhaps testing and verifying with two balls, a camera and a lego block. – Anthony Meadows Aug 17 '19 at 23:46
  • pussies, florida gypt... me SOS evil. I am the REAL " IS REAL "... math too fucking difficult huh?! quran 2:275 – Anthony Meadows Jan 09 '20 at 01:15
  • Looks like someone's account got hacked. – David K Jan 09 '20 at 02:15