So how did they suss out the diameter of the moon?\n\nSpecifics please!  \n\nIf there's a proportional thing (eg hold a coin until it blocks the moon) then you'll need to tell me how they worked out how far the moon is from the Earth...
127482
This is a good question that made me think for a while.  First of all, you need to know how big Earth is.  Eratosthenes did that in ancient times by measuring the altitude of the sun at two different places on the Earth at the same time on the same day (presumably he had a friend helping him).  I've linked to a site detailing this measurement.\n\nOnce you know that, just wait for a lunar eclipse.  When the Earth is partially obstructing the moon, you will be able to see an arc of the Earth at the same time as an arc of the moon.  Then you can get a relative difference in curvatures between the Earth and the Moon (maybe by using coins of different sizes, for instance).  So once you know the ratio of the Moon's circumference to the Earth's circumference, just multiply by the Earth's cirumference to get the answer.
378807
I dont know why you think anyone on a genric site like 'yahoo answers' would really know if there was a secret conspiracy in the US government/NASA.\n\nALL the so called 'inconsistancies' in the 'moon hoax conspiracy theory' have been answered time and time again, but the lunatic fringe insist on trotting out the same old (non) evidence\nthis site answers most of their rubbish:\nhttp://www.badastronomy.com/bad/tv/foxapollo.html\n\nI say "Most of the rubbish", because you can never really tell what new 'fact' a hoax-nutter is going to pull out of their a*se next.
599285
Thats why they asked theoretically.\n\nsame radius as the solar system\n1267650600228229401496703205376 kms\n\nOf which you allready knew, otherwise u would not be asking
590716
QUOTE-----"The photo must be by telescope from the Earth to the moon! If no one can find any recent photos then the moon landings are a massive hoax!"\n\nTo remind u and other readers...we CAN'T see the lunar rover or whatsoever using our telescope on earth.No matter how BIG the telescope is, we still CAN'T see it because of atmospheric blurring.\n\nEven HUBBLE can't see it...because that landing site (flag and all that) is so SMALL.... and MOON is MOVING so fast.\n\nIt's just like trying to see a COIN when you are travelling with an AEROPLANE high in the SKY\n\nDID U GET IT? OR U NEED ME TO REPEAT IT AGAIN?
257499
Moon/Earth Comparison\nBulk parameters\n\n                                   Moon         Earth      Ratio (Moon/Earth)\nMass (1024 kg)                    0.07349       5.9736     0.0123    \nVolume (1010 km3)                 2.1958      108.321      0.0203\nEquatorial radius (km)           1738.1        6378.1    0.2725      \nPolar radius (km)                  1736.0        6356.8    0.2731\nVolumetric mean radius (km)        1737.1        6371.0    0.2727\nEllipticity (Flattening)          0.0012        0.00335    0.36    \nMean density (kg/m3)               3350          5515      0.607      \nSurface gravity (m/s2)            1.62          9.80       0.165    \nSurface acceleration (m/s2)       1.62          9.78       0.166    \nEscape velocity (km/s)            2.38         11.2        0.213    \nGM (x 106 km3/s2)                 0.0049        0.3986     0.0123 \nBond albedo                       0.11          0.306      0.360\nVisual geometric albedo           0.12          0.367      0.330    \nVisual magnitude V(1,0)          +0.21         -3.86          -\nSolar irradiance (W/m2)            1367.6        1367.6    1.000      \nBlack-body temperature (K)          274.5         254.3    1.079      \nTopographic range (km)               16            20      0.800      \nMoment of inertia (I/MR2)         0.394         0.3308     1.191\nJ2 (x 10-6)                     202.7        1082.63        0.187  \nRef The foll link
562808
Well, this is an interesting lesson in geometry, eh? Since the earth is a sphere (more or less) that rotates on an axis - the N & S points were chosen as the intersecting points for longitudinal lines of measurement - which all go back to the length of a day and segments thereof - as well as 360 degrees which were derived roughly from the concept of a 360 day year (it is actually 365+ but back in time the 360 measurement worked wonders for calculationsand thus it fit into geometry of all kinds .. including the longitudinal slices of the earth in 1 degree increments). If you chose to - you could select another two points each an oposite 90 degrees from the N and S poles (which made sense becauseof the magnetic N and the rotational axis). The other two points would be arbitrary. Somebody would have to mark those spots. A secondary longitudinal series could then overlay the primary longitudinal lines and create a kind of variable grid based on an arbitrary selection of the two secondary points. Actually ONE of those lines would lie on top of one of the primary (current) longitudianal lines and it would be a double use line or a kind of disection of hemispheres.  \nBUT\nIt would not be measuring anything very well because the secondary points are arbitrary and not related to anything else.  \nSO\nThe reason the latitude disectors came into play with their resulting grids over the face of earth'ssphere, was becuase they help measure something.  Since stars, the sun and the moon are  always - based on measured timepieces - in the same place and the same time repeatedly, you could measure the distance above the horizon to a star and figure out how far you were N or S of the equator - a halfway point from N to S.  It became easy this way to use the North Star to detect your position N or S by the angle of THAT star above the horizon becuase it was always amost right on top of the N Pole.  So the slices of LATITUDE N and S were critical to navigation. Once the clock was perfected to hold accurate time within a few seconds a month, the accuracy of E and W postion on earth relative to the stars also became possible.  If people did not need to know were they were on earth (especially ships at sea) there would have been no need for a good grid system.
17544
I decided to go ahead and post here, despite the numerous posts already.\n\nI'm not going to use my own value for radius, I'm going to choose one that's already here. Namely 3963.19 mi. (This will be "R")\n\nLet:\n\nUnaltered circumference (C) = 2(pi)R\n\nAltered Circumference (c) = 2(pi)r\n\nThe length you need is: c - C\n\n==> c - C = 2(pi)r - 2(pi)R\n==> 2(pi)[r - R]\n\nr - R is the difference in radii, which is what you stated.....1".\n\nSo, r - R = 1"\n\nTherefore: 2(pi)(1") = 2(pi) inches!\n\n-or-\n\n6.28 inches.\n\nThis is how much you'd need to get that 1" gap between the earth and the string.
7493
I want to add more:\n1) Due to the distance to the Moon relative to the size of our planet, it doesn't really matter from which point on Earth you are looking at our natural satellite. You will see virtually the same part of the surface.\n2) The reason the Moon rotates around its axis at about the same time it rotates around the Earth is because it is tidally locked with the Earth. Tidal lock worked for billions of years to achieve the present effect, but it's still not perfect: the Moon slightly advances or gets behind in its axial rotation relative to its revolution around the Earth. So, over time, we are actually able to see around 57% of its surface, rather than just 50%.\n3) The Earth is not close enought to the Sun to be tidally locked by its gravity (even Mercury, the closest planet to the Sun, isn't). So that's why the Sun-Earth relation is different.
433590
Being so far away, telescopes don't have the resolution needed to see the equipment. It's a bit like trying to use Google Earth and figuring out what the make and model of each car is. We can see the area of the moon where they touched down, but the equipment is just too small.\n\nBeing so far away, the light reflected off the metal facets of the hardware will rarely be reflected at just the right angle to be seen. The moon orbits Earth at a 5 degree inclination. Plus, the landers aren't at the moon's equator, which complicates things even more. The equipment will need to be at the right place at the right time, and our telescopes will need to be looking in the right place at the right time, just for a chance flash of something. It's a lot of variables that almost never come together right.
119943
Why can't we see the flag on the moon?  \nA.    Lets assume that: \nThe height of the flag (a) pole in ft. = 6 \nThe average distance to the Moon (approx) in miles = 221,000 \nWhich means that the distance to the Moon (approx) in feet (b) = 1,166,880,000 \nB.    Given the above: \nThe tangent (t) of the above in radians [tan(a/b)] = 0.0000000051, in radians \nConverted to Degrees [180*(t/pi)] = 0.000000295 degrees \nThis means that a telescope would have to be able to resolve an angle of 2.95 x 10-07 degrees to see the flag left on the moon.\nC.    That means:\nSince one Arc Degree (1/360 of a circle) = 1.000000000 Deg. \nAnd one Arc Minute (1/60 of a degree) = 0.016666667 Deg. \nAnd one Arc Second (1/60 of an arc-minute) = 0.000277778 Deg. \nWe can calculate that:  \nThe Angle to be resolved is: 0.000000295 Deg. \nAnd the Angle to be resolved in Arc Seconds is: 0.001060596 \n\nSo, to see the flag, a telescope must be able to resolve an angle less than an 1/1000th of an arc second.\n\n"So how big of a telescope would that require?", you ask.\nThe Dawes limit calculates how close two objects can be and still be resolved by a telescope. \nThe formula is:  Resolution (in arc seconds) = 4.56/Diameter. \nSo, to determine the required diameter, the formula is:  D = 4.56/Resolution(arc seconds) \n\nDoing the math, that is 4.56/.001060596 arc seconds, to see the flag a telescope would require a mirror 4,299.47 inches wide! That is 358.29 Feet!\nA super-telescope THAT big would have to be in orbit, to eliminate the turbulence of the earth's atmosphere.
217574
No tides (not true) \nNo intertidal zone (not true) \nNo eclipses \nDarker skies for astronomy \nSome animals affected (such as salmon and turtles) \nDifferent night creatures \nNo gravity (not true) \nMore impacts on Earth by space debris \nNo lunar calendar \nNo lunar phases \nWorse surfing \nDarker nights \nDifferent name for "lunatics'' \nNo lunar fables (werewolves and so on) \nNo nearby place for astronauts to go \nA lost vehicle for romance \nA lost theme for love songs \nNo Man in the Moon \nThe day would be eight hours long. \nThe winds would be much stronger.\nhttp://www.astrosociety.org/education/publications/tnl/33/33.html\n\nFYI \nThe Moon is moving away from the earth which is why the question is in theory with scientists..The Moon is receding at 2 inches per year\n\nAs for your unintelligent insult on which you are trying to be supreme, you have no CLUE...\n\nIn 1897 George Darwin, son of Charles, suggested that tides are causing the Moon to spiral AWAY FROM US. His hypothesis was CONFIRMED in 1969 after the Apollo 12 astronauts placed corner reflectors (like the red and orange ones on a car) on the Moon. By firing lasers from Earth to the Moon, off the corner reflectors, and back again, astronomers measured the round-trip time and, hence, the distance to the Moon. Repeating the experiment over several years, they confirmed Darwin's prediction. The Moon is receding at 2 inches per year.\n\nSo instead of trying to sound politically correct find facts to the basis of your answer DORK..\nSo here is an explaination of why our days would be 8 hrs long.Earth loses angular momentum because the high tide closest to the Moon is trying to get back directly underneath the Moon, while the high tide farthest from the Moon is trying to get as far away from the Moon as possible. Consequently, the high tides flow westward, and in so doing, they encounter continents and islands. The water pushes against these land masses, which, because of rotation, are moving eastward. The net result is that the eastward rotation is retarded by the westward motion of the tides -- slowing down the rotation. The day is getting longer by about 0.002 seconds per century. It doesn't sound like much, but over billions of years it adds up. Not that you would understand..\n\nNone of your previous questions have been intelligent..
491117
nice one mate ........i dont think many people got the point.... the answer from scots_lad what a moon man
3904
Here is a link that should be perfect! Scroll down a bit and there is a table with all the data you could possibly want!  http://en.wikipedia.org/wiki/2003J6%28moon%29--_statistics_of_Jupiter%27s_Moon
400520
For any planet with a moon, the planet's mass can be approximated by GmM/r² = mv²/r = mrω² = mr(2π/P)², where G is the gravitational constant, m the mass of the moon, M the mass of the planet, r the distance between planet and moon, and P the period of the moon. The mass of the moon cancels out, leaving\nM = r³(2π/P)²/G\n\nThe masses of moonless planets have to be determined by Perturbation Theory, the effect of one planet's orbit on the orbits of other planets.\n\nThe diameter of another celestial object can be estimated by multiplying the angle subtended by the object by its distance from us.  A closer estimate is obtained by\nd/2 = Dsin(α/2) \nwhere d is the diameter, D is distance to the object, and  α is the subtended angle.
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in relation to the other planets in our solar system... our moon is bigger than pluto but smaller than the other planets.\n\nin relation to other moons of other planets in our solar system... it is the 5th largest out of all the other planetary satellites.\n\nthe exact figures are at the link below:\n-(this other link might help as well)
422247
42 folds to the Moon\n51 folds to the Sun\n53 folds to Jupiter\n56 folds to Pluto\n69 folds to the nearest star\n70 folds to the nearest extra-solar planet\n\nMost people do not believe these answers as they are counter-intuitive, so I will talk you through it in easy stages. The point to grasp is that every ten folds means just over a thousand times increase in the height of the resulting pile of paper.\n\nA ream of ordinary 70 gsm photocopy paper is 500 sheets and about 2 inches thick, \n\nSo let us say that after 10 folds of our theoretically infinitely large (i.e. as large as we need it to be) piece of paper we have 1,024 thicknesses (2^10) and that that is 4 inches thick (a third of a foot).\n\nAfter 20 folds we have 2^20 thicknesses = 4096 inches = 341 feet 4 inches.\n\nThe Empire State Building is 1,250 feet tall, so after 22 folds of the paper it has grown taller than the height of a 102-floors-tall skyscraper,\n\nThere are 5280 feet in a mile, \n\nAfter 30 folds we have 2^30 thicknesses = 349,184 feet 4 inches = 116,395 yards = 66.13 miles. i.e. it was over a mile high after 24 folds and more than 2 miles high after 25 folds\n\nMount Everest is only 5 miles high so the pile of paper was taller than Mount Everest after 27 folds.\n\nThe moon is 238,000 miles away on average (let us not worry about apogee and perigee).\n\nAfter 40 folds we have 2^40 thicknesses = 1,024 x 66.13 miles = 67,720 miles i.e. more than a quarter of the way to the moon.\n\nSo clearly 42 folds more than suffices to take us to the moon. How about the Sun?\n\nThe Sun is on average 93,000,000 miles away (let us not worry about perihelion and aphelion)\n\nAfter 50 folds we have 2^50 thicknesses = 1,024 x 67,720 miles = 69,345,477 miles \n\nSo after 50 folds, we would then be most of the way to the sun. 51 folds more than suffices to get there. So how about Jupiter?\n\nThere are on average 486,587,517 miles to Jupiter. So another couple of folds should do the trick. 53 folds = 8 x 69,345,477 miles = 554,762,813 miles i.e. 53 folds more than suffices to get to Jupiter, And 56 is ample for the short hop to Pluto.\n\nAnd the nearest star?\n\n1 light year = 5.879 ×10^12 miles\n4.2 light years = 24.69 ×10^12 miles\n\nAfter 60 folds we have 2^60 thicknesses = 1,024 x 69,345,477 miles =  71,009,769,448 miles\n\nAfter 70 folds we have 2^70 thicknesses = 1,024 x 71,009,769,448 miles = 72,714,002,890,752 miles (72 x 10^12 miles)\n\nSo clearly 69 folds gets us to Proxima Centauri and half way back again! 70 would take us to Sirius (8.6 light years away) and half way back again. Or, more interestingly, it would easily take us to Epsilon Eridani b, the nearest known extra-solar planet (10.5 light years away),\n\nI will leave you to work out how many folds you would need to get to the nearest galaxy!
369329
During my Time Travels I did do some Intergalactic Exploration. When I was in Grid 946 Quadrilateral 47, the solar system was similar to your solar system. Basically we ( my Friends and myself) observed than any planetary structure is protected from outside forces due to Gravity and Atmosphere. Moons, as you are accustomed to calling them lack those two important aspects, thus, any outside forces bombard those structures and they similar to the moon as larger structures with gravity and atmosphere are similar to Earth. It simple Physics according to Earth's Concept. Once your Scientist understand Dimensions, then you understand the many things you now consider mystery.