6 ms·

How to figure out the size of the moon yourself

86 points by cwillu 5y ago

mnw21cam 5y ago
It should be noted that we actually use this method to work out the size/shape of small asteroids that are too distant to clearly make out in a telescope. If you line up a few astronomers with telescopes over an area approximately the size of the object, and wait for it to pass in front of a distant star, then you can gather the measurements of when the star disappears from all those observers and get an outline of the object. https://sky-lights.org/2018/10/01/qa-determining-asteroid-shape-from-occultation-timings/ https://sky-lights.org/2018/10/01/qa-determining-asteroid-sh... https://occultations.org/ https://occultations.org/
epakai 5y ago
I searched duckduckgo for the surface area of the moon and it reports the magnitude for square miles, but gives the units as "m2".
Thomashuet 5y ago
The "yourself" part of the title is actually a lie, you need a bunch of trusted people spread over a large geographical area (as large as the moon itself) to make observations for you. When reading the title I understood the "yourself" as without having to trust anyone else and I was very much disappointed by the suggested method.
- metacritic12 5y ago
  Aristarchus's method [1] is actually something you can do yourself, and indeed was the historical way the size of the moon was discovered. I personally found that far more satisfying as you're relying on far less complexity to prove the point. [1] https://www.eg.bucknell.edu/physics/astronomy/astr101/specials/aristarchus.html#:~:text=The%20jump%20Aristarchus%20made%20from,measure%20both%20for%20the%20Moon https://www.eg.bucknell.edu/physics/astronomy/astr101/specia....
- scatters 5y ago
  Right, and I think that explains why the Greeks didn't use it; the world accessible to their civilisation extended maybe 2000 km north to south, from the shores of the Black Sea to Nubia. Maybe the Romans could have managed it, but only just (you try making astronomical observations on demand in northern England). Indeed, were there any premodern civilisations with a sufficient span of latitudes to perform this method? If there were, they'd have also been able to measure the distance to the Sun using the method of observing transits of Venus.
  - daniel-cussen 5y ago
    Olmecs and Mayans. They did all kinds of things with nothing, like just nothing. They tried to publish their works but the a Spanish bishop burned every codex. He tracked every codex down, too, presumably by interrogations. Four survived. We're currently in the Mayan Dark Ages in that sense, there's stuff they had we still can't do. They were a millennium ahead of Eurasian math, finding zero in 200BC versus 800AD.
    - jacobr1 5y ago
      > there's stuff they had we still can't do I'm skeptical, how would we know that? Are there vague inscriptions of things like Fermat's margin notes that imply missing knowledge?
      - labster 5y ago
        We still don’t have vertical basketball hoops. https://en.m.wikipedia.org/wiki/Maya_Ballgame https://en.m.wikipedia.org/wiki/Maya_Ballgame
        daniel-cussen 5y ago
        Those were the real ballers of the new world. That would have made great television!
      - daniel-cussen 5y ago
metacritic12 5y ago
Certainly an interesting concept -- use simultaneous measurements to find the size of the moon. Yet I can't help but wonder from a history of science point of view, how much the above exercise is "presuming the conclusion". Organizing an international star party to see where the moon is at the same time requires knowing what the concept of same time on very different places on earth. But historically, the concept of same time was heavily dependent on astronomical measurements -- both astronomical knowledge and advanced instruments. Now if you start using modern techs like GPS to figure out what the concept of same time is, you're already relying on satellites and can probably use that tech to make a direct measurement of the size of the moon.
- mannykannot 5y ago
  You do not need accurate time measurements for this method; the question to be answered by each observer is "was the given star occulted on the night in question?", and the ancients could track time to the precision of a day (as we are concerned with at most a hemisphere, there's no problems with date lines.) It does depend on accurate surveying over distances exceeding the diameter of the moon, and a good deal of communication, which I suspect is enough of a problem to answer the author's question of why it was not done, as far as we know. The method is not, however, begging the question.
- Smaug123 5y ago
  Once you've got a reliable portable clock, you can synchronise lots of them in the same place and then move them. Harrison had done this by 1750 (with clocks that supposedly drifted by only a few seconds per day, even after sea travel).
ezconnect 5y ago
I was thinking he is using ancient method or a novel idea incorporating ancient ones. He used data curated by scientific observers around the world.
lordnacho 5y ago
Doesn't the orbital period tell you how far away it must be? And from knowing that, you could measure how big an angle it takes up, giving you the actual diameter? It seems harder to recruit a load of people to look at the sky.
- dr_orpheus 5y ago
  Yes, if the orbit is circular and not eccentric, which the moon is very close to (eccentricity of 0.05). And you should be able to tell that it is close to circular by observing the relative size of the moon throughout the orbit. It will change slightly because of the eccentricity and your viewing distance to the moon but since it is far away enough these are pretty small contributing factors. I think part of the assumption is that this is still trying to use a method theoretically available to ancient Greeks and pre Brahe and Kepler before the laws of planetary motion were discovered. https://en.wikipedia.org/wiki/Orbit_of_the_Moon https://en.wikipedia.org/wiki/Orbit_of_the_Moon
blippage 5y ago
Can you get the size of the moon using similar triangles? You have to know the distance to the moon. But presumably you can calculate it because you know its angular velocity (it takes about 28 days to revolve around the earth). When you know that, you can calculate a distance based on the fact that the moon has a stable orbit. I can't derive a formula, but presumably it's possible.
- greeneggs 5y ago
  Sure. From Kepler's third law, assuming that the Moon's mass is negligible (it is actually ~1.2% of Earth's mass), the semi-major axis of the moon's orbit is given by (mu T^2 / 4 pi^2)^{1/3} [1]. Substituting T = 27.32 days and mu = G M_Earth = 3.986004418×10^14 m^3/s^2 [2], gives a distance of 3.83 x 10^8 m. Estimating the Moon's apparent angle at 31 arcminutes, its diameter is estimated to be 2 * (distance) * ArcSin[31/2 arcminutes], or 3.46 x 10^3 km. According to Google, the right answer is about 3.47 * 10^3 km. [1] https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbi... [2] https://en.wikipedia.org/wiki/Standard_gravitational_parameter https://en.wikipedia.org/wiki/Standard_gravitational_paramet... [3] https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2%29%2F%284+%5C%5BPi%5D%5E2%29%29%5E%28++1%2F3%29+%2F.+%7B%5C%5BMu%5D+-%3E+3.986004418+10%5E14%2C+T+-%3E+27.32+86400%7D https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2...
- marcodiego 5y ago
  I also thought that, but it needs some previous knowledge about the Earth, simplifications and approximations. Let's get to calculations: First we'll consider the mass of the Moon substantially smaller than that of the Earth and that its orbit is circular. In this case, in its orbit around the Earth, the weight of the Moon is the centripetal force: W = F_cp mg = mw^2R (R being orbit Radius) g = w^2R GM/R^2 = w^2R Assuming you have the gravity g_e on the surface of the Earth and its radius R_e, we can derive GM (M being the mass of the Earth): GM/R_e^2 = g_e GM = g_e * R_e^2 Replacing GM on the previous equation, we get: g_e * R_e^2/R^2 = w^2R R = (g_e * R_e^2 / w^2)^(1/3) Note that w (angular velocity of the Moon) is easy to calculate: 28 days per rotation and gravity on the surface of the Earth is around 9.8 meters per second. Radius of the Earth can be calculated by other means. Nevertheless, we now have an approximation for the orbital radius of the Moon. Considering the Radius of the Earth, distance to the Moon from someone at the surface of the Earth is given by: D = R - R_e Now, to calculate the radius of the visible Moon disk using triangle similarity, use a small disk (with radius r) and hold it at a distance d from your eye until it becomes the same apparent size of the Moon disk. Applying similarity of triangles, the radius r_m of the Moon is given by: r_m/D = r/d r_m = D*r/d Note that we are no calculating the radius of the Moon, but the radius of the visible Moon disk. Considering D is much bigger than d, it is a good approximation. I wonder how precise this will turn out though. Personal anecdote: a gifted uncle of mine once calculated the duration of total Moon eclipse using similar methods. Distance to the Moon was approximated to 1.5 light-seconds. He got the estimated time wrong by around 5 minutes. He did it using mental calculations only.
andrewflnr 5y ago
I don't understand the definition of occultation in the article, if there are only "several" occultations of stars per month. The moon is blocking several dozen to hundreds of naked eye stars at a given time, right? Are we only talking about really bright stars?
- dmurray 5y ago
  No. The moon takes up about 1/100,000 of the night sky [0]. In excellent conditions you can see maybe 5,000 stars [1]. So on average the moon is blocking 0.05 of them. It's surprising how small the moon is in the sky, given its prominence, and how few stars there are even on the blackest night. Now correct for the fact that you won't have perfect viewing conditions everywhere in your survey area, and that you can make out far fewer stars when the moon is full, and a few opportunities a month starts to look reasonable. [0] https://www.quora.com/How-much-of-the-full-night-sky-does-the-full-moon-occupy https://www.quora.com/How-much-of-the-full-night-sky-does-th... [1] https://skyandtelescope.org/astronomy-blogs/how-many-stars-night-sky-09172014/ https://skyandtelescope.org/astronomy-blogs/how-many-stars-n...
typhonic 5y ago
In 1973, a high school student from another state told me that he became interested in astronomy because of a newsletter and activity he participated in. It seems there was a grad student somewhere trying to crowdsource astronomical observations. Periodically, he would identify a time and place where it was expected that one could observe a particular star as it was blocked from view and then reappeared from behind the moon. Participants would go to those locations in groups of three, one on the spot, one some distance to the East and another the same distance to the West. The teams of three would then send in the timing of their observations so the project leader could make corrections to astronomical data. The grad student, if I remember correctly, used the mailing list to continue the newsletter for some time after the project end date. I always liked that story and I imagined it was not the only effort of its kind.
- okamiueru 5y ago
  The distance to the sun was first calculated in very much the same way. When Venus passes in front of the sun, it crosses over at different places on the sun disk. Using Kepler's laws for orbits, we knew the distance to Venus. Together with the distance between the observations and angle of crossing, the only unknown was the distance to the sun. They were off by 0.8% in 1769. The story itself is pretty cool. The idea was presented in 1663, and it wasn't until 1761 and 1769 that the planets would align. And, this also doesn't happen all that often, although it comes in pairs. 1761 and 1769, 1874 and 1882, 2004 and 2012. And it won't happen again until 2117. So, this was considered so important, and the transit of Venus had to be observed from distances far away on earth. In spite of wars going on, these expeditions were given safe passage.
alberth 5y ago
Completely random: I love the site design. I guess it shouldn't surprise me that a site dedicated to scientific information would be dense in well, information (not trendy with flashy graphics, little content nor confusing color schemes where you can't figure out what is a link or not)
ineedasername 5y ago
1) Close 1 eye, hold thumb up to moon 2) Adjust distance of thumb from face until it just barely occludes vision of the moon. 3) Holding very still, use accurate calipers to measure thumb. Under many repetitions with many different experimenters, this should consistently yield a result showing that the width of the moon is one thumb wide.
- stavros 5y ago
  What does the calipers step achieve?
  - Doxin 5y ago
    It's fairly effective at briefly making you think it's a serious and effective method of measuring the size of the moon.
westcort 5y ago
When the moon is half full, the sun is just under 90 degrees away in the sky. This is how we know the moon is close by. You can use the law of sines to find out how many moon-earth distances are between Earth and then sun. Then, if you have a few friends who know what day it is in different countries, you can find out how large the moon's shadow is on the Earth. The rest is just math.
Tepix 5y ago
Two reasons why the ancient Greeks didn't use this method are: 1. They didn't know how far away the stars were. The stars might as well be illuminated dots on a sphere that is surrounding the earth not too far away. 2. Also you'd have to measure the time very precisely at several quite distant points on the Earth's surface. Not an easy feat.
Tepix 5y ago
The idea of crowdsourcing the occultation data is used by the eVscope telescope, a small (11cm mirror) telescope that replaces the optical viewfinder with a Sony IMX224 sensor. Despite being expensive, it had successful kickstarter campaigns and via its app you can join campaigns with your telescope and contribute to science. Kinda neat.

86 points by cwillu 5y ago

It should be noted that we actually use this method to work out the size/shape of small asteroids that are too distant to clearly make out in a telescope. If you line up a few astronomers with telescopes over an area approximately the size of the object, and wait for it to pass in front of a distant star, then you can gather the measurements of when the star disappears from all those observers and get an outline of the object. https://sky-lights.org/2018/10/01/qa-determining-asteroid-shape-from-occultation-timings/ https://sky-lights.org/2018/10/01/qa-determining-asteroid-sh... https://occultations.org/ https://occultations.org/

epakai 5y ago

I searched duckduckgo for the surface area of the moon and it reports the magnitude for square miles, but gives the units as "m2".

Thomashuet 5y ago

The "yourself" part of the title is actually a lie, you need a bunch of trusted people spread over a large geographical area (as large as the moon itself) to make observations for you. When reading the title I understood the "yourself" as without having to trust anyone else and I was very much disappointed by the suggested method.

metacritic12 5y ago
Aristarchus's method [1] is actually something you can do yourself, and indeed was the historical way the size of the moon was discovered. I personally found that far more satisfying as you're relying on far less complexity to prove the point. [1] https://www.eg.bucknell.edu/physics/astronomy/astr101/specials/aristarchus.html#:~:text=The%20jump%20Aristarchus%20made%20from,measure%20both%20for%20the%20Moon https://www.eg.bucknell.edu/physics/astronomy/astr101/specia....
scatters 5y ago
Right, and I think that explains why the Greeks didn't use it; the world accessible to their civilisation extended maybe 2000 km north to south, from the shores of the Black Sea to Nubia. Maybe the Romans could have managed it, but only just (you try making astronomical observations on demand in northern England). Indeed, were there any premodern civilisations with a sufficient span of latitudes to perform this method? If there were, they'd have also been able to measure the distance to the Sun using the method of observing transits of Venus.
- daniel-cussen 5y ago
  Olmecs and Mayans. They did all kinds of things with nothing, like just nothing. They tried to publish their works but the a Spanish bishop burned every codex. He tracked every codex down, too, presumably by interrogations. Four survived. We're currently in the Mayan Dark Ages in that sense, there's stuff they had we still can't do. They were a millennium ahead of Eurasian math, finding zero in 200BC versus 800AD.
  - jacobr1 5y ago
    > there's stuff they had we still can't do I'm skeptical, how would we know that? Are there vague inscriptions of things like Fermat's margin notes that imply missing knowledge?
    - labster 5y ago
      We still don’t have vertical basketball hoops. https://en.m.wikipedia.org/wiki/Maya_Ballgame https://en.m.wikipedia.org/wiki/Maya_Ballgame
      - daniel-cussen 5y ago
        Those were the real ballers of the new world. That would have made great television!
    - daniel-cussen 5y ago

metacritic12 5y ago

Certainly an interesting concept -- use simultaneous measurements to find the size of the moon. Yet I can't help but wonder from a history of science point of view, how much the above exercise is "presuming the conclusion". Organizing an international star party to see where the moon is at the same time requires knowing what the concept of same time on very different places on earth. But historically, the concept of same time was heavily dependent on astronomical measurements -- both astronomical knowledge and advanced instruments. Now if you start using modern techs like GPS to figure out what the concept of same time is, you're already relying on satellites and can probably use that tech to make a direct measurement of the size of the moon.

mannykannot 5y ago
You do not need accurate time measurements for this method; the question to be answered by each observer is "was the given star occulted on the night in question?", and the ancients could track time to the precision of a day (as we are concerned with at most a hemisphere, there's no problems with date lines.) It does depend on accurate surveying over distances exceeding the diameter of the moon, and a good deal of communication, which I suspect is enough of a problem to answer the author's question of why it was not done, as far as we know. The method is not, however, begging the question.
Smaug123 5y ago
Once you've got a reliable portable clock, you can synchronise lots of them in the same place and then move them. Harrison had done this by 1750 (with clocks that supposedly drifted by only a few seconds per day, even after sea travel).

ezconnect 5y ago

I was thinking he is using ancient method or a novel idea incorporating ancient ones. He used data curated by scientific observers around the world.

lordnacho 5y ago

Doesn't the orbital period tell you how far away it must be? And from knowing that, you could measure how big an angle it takes up, giving you the actual diameter? It seems harder to recruit a load of people to look at the sky.

dr_orpheus 5y ago
Yes, if the orbit is circular and not eccentric, which the moon is very close to (eccentricity of 0.05). And you should be able to tell that it is close to circular by observing the relative size of the moon throughout the orbit. It will change slightly because of the eccentricity and your viewing distance to the moon but since it is far away enough these are pretty small contributing factors. I think part of the assumption is that this is still trying to use a method theoretically available to ancient Greeks and pre Brahe and Kepler before the laws of planetary motion were discovered. https://en.wikipedia.org/wiki/Orbit_of_the_Moon https://en.wikipedia.org/wiki/Orbit_of_the_Moon

blippage 5y ago

Can you get the size of the moon using similar triangles? You have to know the distance to the moon. But presumably you can calculate it because you know its angular velocity (it takes about 28 days to revolve around the earth). When you know that, you can calculate a distance based on the fact that the moon has a stable orbit. I can't derive a formula, but presumably it's possible.

greeneggs 5y ago
Sure. From Kepler's third law, assuming that the Moon's mass is negligible (it is actually ~1.2% of Earth's mass), the semi-major axis of the moon's orbit is given by (mu T^2 / 4 pi^2)^{1/3} [1]. Substituting T = 27.32 days and mu = G M_Earth = 3.986004418×10^14 m^3/s^2 [2], gives a distance of 3.83 x 10^8 m. Estimating the Moon's apparent angle at 31 arcminutes, its diameter is estimated to be 2 * (distance) * ArcSin[31/2 arcminutes], or 3.46 x 10^3 km. According to Google, the right answer is about 3.47 * 10^3 km. [1] https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbi... [2] https://en.wikipedia.org/wiki/Standard_gravitational_parameter https://en.wikipedia.org/wiki/Standard_gravitational_paramet... [3] https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2%29%2F%284+%5C%5BPi%5D%5E2%29%29%5E%28++1%2F3%29+%2F.+%7B%5C%5BMu%5D+-%3E+3.986004418+10%5E14%2C+T+-%3E+27.32+86400%7D https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2...
marcodiego 5y ago
I also thought that, but it needs some previous knowledge about the Earth, simplifications and approximations. Let's get to calculations: First we'll consider the mass of the Moon substantially smaller than that of the Earth and that its orbit is circular. In this case, in its orbit around the Earth, the weight of the Moon is the centripetal force: W = F_cp mg = mw^2R (R being orbit Radius) g = w^2R GM/R^2 = w^2R Assuming you have the gravity g_e on the surface of the Earth and its radius R_e, we can derive GM (M being the mass of the Earth): GM/R_e^2 = g_e GM = g_e * R_e^2 Replacing GM on the previous equation, we get: g_e * R_e^2/R^2 = w^2R R = (g_e * R_e^2 / w^2)^(1/3) Note that w (angular velocity of the Moon) is easy to calculate: 28 days per rotation and gravity on the surface of the Earth is around 9.8 meters per second. Radius of the Earth can be calculated by other means. Nevertheless, we now have an approximation for the orbital radius of the Moon. Considering the Radius of the Earth, distance to the Moon from someone at the surface of the Earth is given by: D = R - R_e Now, to calculate the radius of the visible Moon disk using triangle similarity, use a small disk (with radius r) and hold it at a distance d from your eye until it becomes the same apparent size of the Moon disk. Applying similarity of triangles, the radius r_m of the Moon is given by: r_m/D = r/d r_m = D*r/d Note that we are no calculating the radius of the Moon, but the radius of the visible Moon disk. Considering D is much bigger than d, it is a good approximation. I wonder how precise this will turn out though. Personal anecdote: a gifted uncle of mine once calculated the duration of total Moon eclipse using similar methods. Distance to the Moon was approximated to 1.5 light-seconds. He got the estimated time wrong by around 5 minutes. He did it using mental calculations only.

andrewflnr 5y ago

I don't understand the definition of occultation in the article, if there are only "several" occultations of stars per month. The moon is blocking several dozen to hundreds of naked eye stars at a given time, right? Are we only talking about really bright stars?

dmurray 5y ago
No. The moon takes up about 1/100,000 of the night sky [0]. In excellent conditions you can see maybe 5,000 stars [1]. So on average the moon is blocking 0.05 of them. It's surprising how small the moon is in the sky, given its prominence, and how few stars there are even on the blackest night. Now correct for the fact that you won't have perfect viewing conditions everywhere in your survey area, and that you can make out far fewer stars when the moon is full, and a few opportunities a month starts to look reasonable. [0] https://www.quora.com/How-much-of-the-full-night-sky-does-the-full-moon-occupy https://www.quora.com/How-much-of-the-full-night-sky-does-th... [1] https://skyandtelescope.org/astronomy-blogs/how-many-stars-night-sky-09172014/ https://skyandtelescope.org/astronomy-blogs/how-many-stars-n...

typhonic 5y ago

In 1973, a high school student from another state told me that he became interested in astronomy because of a newsletter and activity he participated in. It seems there was a grad student somewhere trying to crowdsource astronomical observations. Periodically, he would identify a time and place where it was expected that one could observe a particular star as it was blocked from view and then reappeared from behind the moon. Participants would go to those locations in groups of three, one on the spot, one some distance to the East and another the same distance to the West. The teams of three would then send in the timing of their observations so the project leader could make corrections to astronomical data. The grad student, if I remember correctly, used the mailing list to continue the newsletter for some time after the project end date. I always liked that story and I imagined it was not the only effort of its kind.

okamiueru 5y ago
The distance to the sun was first calculated in very much the same way. When Venus passes in front of the sun, it crosses over at different places on the sun disk. Using Kepler's laws for orbits, we knew the distance to Venus. Together with the distance between the observations and angle of crossing, the only unknown was the distance to the sun. They were off by 0.8% in 1769. The story itself is pretty cool. The idea was presented in 1663, and it wasn't until 1761 and 1769 that the planets would align. And, this also doesn't happen all that often, although it comes in pairs. 1761 and 1769, 1874 and 1882, 2004 and 2012. And it won't happen again until 2117. So, this was considered so important, and the transit of Venus had to be observed from distances far away on earth. In spite of wars going on, these expeditions were given safe passage.

alberth 5y ago

Completely random: I love the site design. I guess it shouldn't surprise me that a site dedicated to scientific information would be dense in well, information (not trendy with flashy graphics, little content nor confusing color schemes where you can't figure out what is a link or not)

ineedasername 5y ago

1) Close 1 eye, hold thumb up to moon 2) Adjust distance of thumb from face until it just barely occludes vision of the moon. 3) Holding very still, use accurate calipers to measure thumb. Under many repetitions with many different experimenters, this should consistently yield a result showing that the width of the moon is one thumb wide.

stavros 5y ago
What does the calipers step achieve?
- Doxin 5y ago
  It's fairly effective at briefly making you think it's a serious and effective method of measuring the size of the moon.

westcort 5y ago

When the moon is half full, the sun is just under 90 degrees away in the sky. This is how we know the moon is close by. You can use the law of sines to find out how many moon-earth distances are between Earth and then sun. Then, if you have a few friends who know what day it is in different countries, you can find out how large the moon's shadow is on the Earth. The rest is just math.

Tepix 5y ago

Two reasons why the ancient Greeks didn't use this method are: 1. They didn't know how far away the stars were. The stars might as well be illuminated dots on a sphere that is surrounding the earth not too far away. 2. Also you'd have to measure the time very precisely at several quite distant points on the Earth's surface. Not an easy feat.

Tepix 5y ago

The idea of crowdsourcing the occultation data is used by the eVscope telescope, a small (11cm mirror) telescope that replaces the optical viewfinder with a Sony IMX224 sensor. Despite being expensive, it had successful kickstarter campaigns and via its app you can join campaigns with your telescope and contribute to science. Kinda neat.