[HN Gopher] How to figure out the size of the moon yourself
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How to figure out the size of the moon yourself
Author : cwillu
Score : 71 points
Date : 2022-02-19 08:25 UTC (2 days ago)
(HTM) web link (profmattstrassler.com)
(TXT) w3m dump (profmattstrassler.com)
| mnw21cam wrote:
| 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-sh...
|
| https://occultations.org/
| typhonic wrote:
| 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.
| blippage wrote:
| 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.
| marcodiego wrote:
| 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.
| greeneggs wrote:
| 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
| = 3.986004418x10^14 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_orbi...
|
| [2]
| https://en.wikipedia.org/wiki/Standard_gravitational_paramet...
|
| [3]
| https://www.wolframalpha.com/input?i=%28%28%5C%5BMu%5D+T%5E2...
| lordnacho wrote:
| 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 wrote:
| 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
| ineedasername wrote:
| 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.
| Thomashuet wrote:
| 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.
| scatters wrote:
| 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 wrote:
| 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 wrote:
| > 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?
| daniel-cussen wrote:
| No, it's not that. So a lot of mathematical research is
| serial, so an extra millennium has worth that multitudes
| of mathematicians (in the last 150 years, for instance)
| lack. It's also very hierarchical, and Maya were the
| civilization that most put emphasis on mathematics.
| Although yes, there's practically nothing left, the
| length of year figure they had, very little else.
| labster wrote:
| We still don't have vertical basketball hoops.
| https://en.m.wikipedia.org/wiki/Maya_Ballgame
| daniel-cussen wrote:
| Those were the real ballers of the new world. That would
| have made great television!
| adolph wrote:
| A timekeeping mechanism of sufficient quality shared among
| the locations is also necessary right? Also the civilization
| would need to have their latitudinal territory in alignment
| with the specific time of the transit.
|
| _In 1663 Scottish mathematician James Gregory had suggested
| in his Optica Promota that observations of a transit of the
| planet Mercury, at widely spaced points on the surface of the
| Earth, could be used to calculate the solar parallax and
| hence the astronomical unit using triangulation. Aware of
| this, a young Edmond Halley made observations of such a
| transit on 28 October O.S. 1677_
|
| https://en.wikipedia.org/wiki/Transit_of_Venus#1761_and_1769
| metacritic12 wrote:
| 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/speci
| a....
| metacritic12 wrote:
| 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 wrote:
| 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 wrote:
| 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).
| epakai wrote:
| I searched duckduckgo for the surface area of the moon and it
| reports the magnitude for square miles, but gives the units as
| "m2".
| alberth wrote:
| 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)
| andrewflnr wrote:
| 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 wrote:
| 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-
| th...
|
| [1] https://skyandtelescope.org/astronomy-blogs/how-many-
| stars-n...
| ezconnect wrote:
| 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.
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