40 ms·
The apparent size of the Andromeda galaxy (M31) is 3.167 square degrees [1], and the angular diameter of Moon ranges from 29.3 to 34.1 arcminutes [2]. Using are
by bumbledraven 4y ago
The apparent size of the Andromeda galaxy (M31) is 3.167 square degrees [1], and the angular diameter of Moon ranges from 29.3 to 34.1 arcminutes [2]. Using area = πr², the apparent size of Moon ranges from 0.1873 to 0.2537 square degrees. Hence M31 should appear to be roughly 12.5 to 16.9 times the size of Moon.
[1] https://en.wikipedia.org/wiki/Andromeda_Galaxy https://en.wikipedia.org/wiki/Andromeda_Galaxy
[2] https://en.wikipedia.org/wiki/Moon https://en.wikipedia.org/wiki/Moon
- deleted 4y ago[deleted]
- rendall 4y ago> Hence M31 should appear... I don't think square degrees should be used here. People don't visually approximate that way when fitting round objects into a space. Andromeda 3.167° across its longest length. Moon 0.518°. That's about 6 moons across the longest length. Andromeda is 1° across the shorter axis. Approximately 2 moons. So, appearing about 10-12 moons in area maximum, shaving some off at the corners.
- bumbledraven 4y agoI should have mentioned that area = π r² is only an approximation for the moon’s apparent size in square degrees. What we really want to calculate is the surface area of the spherical cap (on a unit sphere) with the same angular diameter as the moon. A formula for that is 2π(1 – cos r) [1], where r again is the radius of the moon in degrees. In this case, the two formulas give the same result to four significant digits (e.g. [2]). [1] https://en.m.wikipedia.org/wiki/Spherical_cap https://en.m.wikipedia.org/wiki/Spherical_cap [2] http://www.wolframalpha.com/input/?i=2pi%281-cos%5B%2834.1%2F60%2F2%29%2Api%2F180%5D%29%20steradians%20to%20square%20degrees http://www.wolframalpha.com/input/?i=2pi%281-cos%5B%2834.1%2...