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Earth/Moon Gravity
Nick Wedd writes:
You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Very cute question! Possible spoiler... 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. First I observe either an annular solar eclipse or the start or end of a total one. Either way, I deduce that the Sun and the Moon have almost the same apparent size, so R and D are very nearly equal. I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. Therefore it's the Moon that's driving the tides and therefore its tidal force is significantly larger. But the tidal forces are in the ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is significantly less than 1, or in other words, the Moon is denser. I don't know if this was the intended solution, though. It seems to me that if I have to observe the behavior of the tides, then I'm not only using my eyes as an instrument, which is allowed, but also the ocean and the land, which isn't. For that matter, observing the eclipse would be a lot safer if I was allowed to use a solar filter, but that might be excluded as well -- and I don't have so much dedication to rec.puzzles that I want to risk my eyesight for it. (A pinhole projection system, of course, would be Right Out, even if the pinholes are naturally formed by tree leaves.) -- Mark Brader "I can say nothing at this point." Toronto "Well, you were wrong." -- Monty Python's Flying Circus My text in this article is in the public domain. |
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Earth/Moon Gravity
In message , Mark Brader
writes Nick Wedd writes: You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Very cute question! Possible spoiler... 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. First I observe either an annular solar eclipse or the start or end of a total one. Either way, I deduce that the Sun and the Moon have almost the same apparent size, so R and D are very nearly equal. I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. Therefore it's the Moon that's driving the tides and therefore its tidal force is significantly larger. But the tidal forces are in the ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is significantly less than 1, or in other words, the Moon is denser. I don't know if this was the intended solution, though. It seems to me that if I have to observe the behavior of the tides, then I'm not only using my eyes as an instrument, which is allowed, but also the ocean and the land, which isn't. For that matter, observing the eclipse would be a lot safer if I was allowed to use a solar filter, but that might be excluded as well -- and I don't have so much dedication to rec.puzzles that I want to risk my eyesight for it. (A pinhole projection system, of course, would be Right Out, even if the pinholes are naturally formed by tree leaves.) Very good answer. I wouldn't have bothered waiting for an eclipse, I would just have held my hand out at arm's length in front of the setting sun, and observed that the last joint of my little finger subtends about the same angle as the sun does. Then wait for a full moon, and do the same, showing that it's pretty much the same apparent size. Not as accurate as your method, but good enough for our purposes, as it happens. Then go and stand in the sea for a month or two, noting how far up your body each tide comes. Don't try this in the Bay of Fundy. Nick -- Nick Wedd |
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Earth/Moon Gravity
"Mark Brader" wrote in message ... Nick Wedd writes: You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Very cute question! Possible spoiler... 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. First I observe either an annular solar eclipse or the start or end of a total one. Either way, I deduce that the Sun and the Moon have almost the same apparent size, so R and D are very nearly equal. I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. Therefore it's the Moon that's driving the tides and therefore its tidal force is significantly larger. But the tidal forces are in the ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is significantly less than 1, or in other words, the Moon is denser. I don't know if this was the intended solution, though. It seems to me that if I have to observe the behavior of the tides, then I'm not only using my eyes as an instrument, which is allowed, but also the ocean and the land, which isn't. For that matter, observing the eclipse would be a lot safer if I was allowed to use a solar filter, but that might be excluded as well -- and I don't have so much dedication to rec.puzzles that I want to risk my eyesight for it. (A pinhole projection system, of course, would be Right Out, even if the pinholes are naturally formed by tree leaves.) Or you just post a link to this page. http://www.math.sunysb.edu/~tony/tides/sun-moon.html Saves all that tedious typing donchaknow -- Dave Baker Puma Race Engines www.pumaracing.co.uk Camp USA engineer minces about for high performance specialist (4,4,7) |
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Earth/Moon Gravity
Nick Wedd and I (Mark Brader) write:
You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Very cute question! Possible spoiler... 40 39 38 37 36 35 34 33 32 31 30 29 28 27 26 25 24 23 22 21 20 19 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2 ... First I observe either an annular solar eclipse or the start or end of a total one. .... I now go to a coastal location and observe the behavior of the ocean over a few days. ... I don't know if this was the intended solution, though. It seems to me that if I have to observe the behavior of the tides, then I'm not only using my eyes as an instrument, which is allowed, but also the ocean and the land, which isn't. For that matter, observing the eclipse would be a lot safer if I was allowed to use a solar filter... Very good answer. Thanks. I wouldn't have bothered waiting for an eclipse, I would just have held my hand out at arm's length in front of the setting sun, and observed that the last joint of my little finger subtends about the same angle as the sun does. Then wait for a full moon... Ah, good point. I thought about using a tree or building or something, but not my finger. Anyway, I already *have* observed some solar eclipses (admittedly with suitable protection), so I'm covered on that part. Then go and stand in the sea for a month or two, noting how far up your body each tide comes. If you like, but this still does not answer my objection: I'm using the ocean and the land as an instrument, not just my body. Don't try this in the Bay of Fundy. Hear, hear! -- Mark Brader, Toronto "...one man's feature is another man's bug." --Chris Torek My text in this article is in the public domain. |
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Earth/Moon Gravity
In article ,
Nick Wedd wrote: In message , Mark Brader writes Nick Wedd writes: You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Very cute question! Possible spoiler... ............................. Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. First I observe either an annular solar eclipse or the start or end of a total one. Either way, I deduce that the Sun and the Moon have almost the same apparent size, so R and D are very nearly equal. I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. Therefore it's the Moon that's driving the tides and therefore its tidal force is significantly larger. But the tidal forces are in the ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is significantly less than 1, or in other words, the Moon is denser. I don't know if this was the intended solution, though. It seems to me that if I have to observe the behavior of the tides, then I'm not only using my eyes as an instrument, which is allowed, but also the ocean and the land, which isn't. For that matter, observing the eclipse would be a lot safer if I was allowed to use a solar filter, but that might be excluded as well -- and I don't have so much dedication to rec.puzzles that I want to risk my eyesight for it. (A pinhole projection system, of course, would be Right Out, even if the pinholes are naturally formed by tree leaves.) Very good answer. I wouldn't have bothered waiting for an eclipse, I would just have held my hand out at arm's length in front of the setting sun, and observed that the last joint of my little finger subtends about the same angle as the sun does. Then wait for a full moon, and do the same, showing that it's pretty much the same apparent size. Not as accurate as your method, but good enough for our purposes, as it happens. Then go and stand in the sea for a month or two, noting how far up your body each tide comes. Don't try this in the Bay of Fundy. Nick One could use the same method do find out that the density of M31 is much smaller than the density of the Sun, or the Moon: M31 has an apparent diameter which is (very approximately) the same as the apparent diameter of the Sun or the Moon. Now, M31 causes no noticeable tides at all -- therefore the mean density of M31 must be much smaller than the density of the Sun, or the Moon. The experiment can be repeated for any object having approximately the same apparent diameter as the Sun or the Moon, since the tidal force can be considered to depend only on the density and the apparent diameter of the object, no matter how far away it is. -- ---------------------------------------------------------------- Paul Schlyter, Grev Turegatan 40, SE-114 38 Stockholm, SWEDEN e-mail: pausch at stockholm dot bostream dot se WWW: http://stjarnhimlen.se/ |
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Earth/Moon Gravity
In sci.astro message , Tue, 5 Dec
2006 23:29:19, Mark Brader wrote: Nick Wedd writes: You may assume knowledge of Newtonian physics, and of the general structure of the solar system, but not of any of its dimensions. Using only your own body as a measuring instrument, find which is denser, the sun or the moon. Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. Tidal field is proportional to the density of the attractor and to the cube of its angular diameter, only. First I observe either an annular solar eclipse or the start or end of a total one. Either way, I deduce that the Sun and the Moon have almost the same apparent size, so R and D are very nearly equal. Eclipse not needed. The angular size of the Sun can be estimated by pointing the head at the Sun, making a pinhole aperture with the fingers in front of the face, and observing the projected image on a toe. It will probably fit the size, in at least one direction, of at least one toe. Then, later, lie with the head away from the Moon and compare its apparent size with the standard toe. Of course, the Sun can be observed quite comfortably by the eye at sunset, or at other times when clouds co-operate. I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. As stated, not true in *all* locations. Coastal geometry may mean that high tides are doubled, e.g. to the North and West of the Isle of Wight. Therefore it's the Moon that's driving the tides and therefore its tidal force is significantly larger. But the tidal forces are in the ratio M/R^3, which is almost equal to M/D^3. Therefore this ratio is significantly less than 1, or in other words, the Moon is denser. And by using the relative amplitude (measured, of course, in feet) of Spring & Neap Tides, one gets a figure for the density ratio. -- (c) John Stockton, Surrey, UK. Turnpike v6.05 MIME. Web URL:http://www.merlyn.demon.co.uk/ - FAQqish topics, acronyms & links; Astro stuff via astron-1.htm, gravity0.htm ; quotings.htm, pascal.htm, etc. No Encoding. Quotes before replies. Snip well. Write clearly. Don't Mail News. |
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Earth/Moon Gravity
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#18
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Earth/Moon Gravity
Mark Brader:
Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. John Stockton: Tidal field is proportional to the density of the attractor and to the cube of its angular diameter, only. I said tidal force was proportional to M/R^3, which is true. Angular diameter is D/R and density is M/D^3, so yes, tidal force can also be described as proportional to (M/D^3)*((D/R)^3) -- in effect, my posting was deriving that. So what's the "only" supposed to mean? (In the preceding paragraph I'm deliberately being loose as to whether M is a mass or a ratio of masses, and likewise the other variables -- this doesn't affect how they relate to each other.) I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. As stated, not true in *all* locations. Coastal geometry may mean that high tides are doubled, e.g. to the North and West of the Isle of Wight. True, but the cycles will still relate to the Moon's position and not the Sun's. -- Mark Brader | "Unless developers are careful, good software Toronto | attracts so many improvements that it eventually | rolls over and sinks..." --Ben & Peter Laurie My text in this article is in the public domain. |
#19
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Earth/Moon Gravity
In sci.astro message , Thu, 7 Dec
2006 06:28:34, Mark Brader wrote: Mark Brader: Say that the Sun is R times as far away as the Moon, D times the diameter of the Moon, and M times the mass. Then it must have M/D^3 times the density. John Stockton: Tidal field is proportional to the density of the attractor and to the cube of its angular diameter, only. I said tidal force was proportional to M/R^3, which is true. Angular diameter is D/R and density is M/D^3, so yes, tidal force can also be described as proportional to (M/D^3)*((D/R)^3) -- in effect, my posting was deriving that. So what's the "only" supposed to mean? That the tidal field depends on no other variable. The sentence states more compactly what you proved. (In the preceding paragraph I'm deliberately being loose as to whether M is a mass or a ratio of masses, and likewise the other variables -- this doesn't affect how they relate to each other.) I now go to a coastal location and observe the behavior of the ocean over a few days. I find that alternate high tides occur when the Moon is in a similar position between rising and setting, not when the Sun is. As stated, not true in *all* locations. Coastal geometry may mean that high tides are doubled, e.g. to the North and West of the Isle of Wight. True, but the cycles will still relate to the Moon's position and not the Sun's. Yes, but instead of alternate high tides occurring at an interval of 12.5 hours or so, they (alternately) have intervals of (roughly, for Bournemouth, from ancient memory) about 2 and 10.5 hours. The simplest fix is to go to a small mid-ocean island. -- (c) John Stockton, Surrey, UK. / Web URL:http://www.merlyn.demon.co.uk/ - FAQish topics, acronyms, & links. Correct = 4-line sig. separator as above, a line precisely "-- " (SoRFC1036) Do not Mail News to me. Before a reply, quote with "" or " " (SoRFC1036) |
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