How can I measure the distance of the moon from the earth?
142062
I don't know how you will do it with no equiptment, maybe try the net, leave it to the pro's
126629
It appears as a 3/4 moon.  It helps to make a drawing to visualize this.  The distance from earth to moon is a very small fraction of the distance from either earth or moon to the sun, so it's fair to approximate the light rays hitting earth and moon as parallel.  First, place the moon such that the earth-moon line is at 90 degrees to the earth-sun line.  In this setup we on earth who can see the moon at all see a half moon and moonlings who can see earth see a half earth.  Next move the moon further away from the sun so the earth-moon line is at 45 degrees to the earth-sun line.  Now the moon sees only 1/4 of the sunlit part of earth, a crescent.  (This is the situation described in the question.)  From earth, however, we see 3/4 of the sunlit part of the moon.  Whether the moon phase is waxing or waning depends on which "side" of the sun-earth-sun line the moon is on relative to the moon's earth-orbital rotation direction.
239488
Predictions suggest that the range will increase until the Earth and Moon become double synchronised, that is, both are tidally locked to one another. The Earth-Moon distance would then be about 550000km. This won't occur for something like 50 billion years. Before then, though, our Sun will have died, and both the Earth and Moon almost certainly won't even exist.\n\nThis website for answer confirmation and more info  -->http://www.madsci.org/posts/archives/apr2002/1020171327.As.r.html\n\nGod is spelled B-I-G B-A-N-G\nPBrax@netscape.com
740323
Assume a flag is 1 meter across, and the moon is 3.6x10^8 m distance at perigee.  Your telescope needs a resolving power of 1/3.6x10^8 radians.  \n\nThe resolving power of a typical telescope is ~1.22lambda/D radians where lambda is the wavelength of light being observed and D is the diameter of the telescope.  \n\nSolve for D;  D= 3.6x10^8*1.22*lambda  \n\nAssume a visible light telescope sensitive at 600nm = 6x10^-7 m\nand your telescope must be 263m in diameter.  You will want to orbit that puppy or phase errors from turbulence will kill you, with current adaptive optics technology.
462962
The average distance from earth to moon is 238855 miles.
178225
Try this.\nhttp://www.calculatorcat.com/moon_phases/moon_phases.phtml
928563
What is meant by the moon's altitude is its height above the horizon and is measured in degrees. Then it is directly overhead, its altitude is said to be 90deg. As it is rising or setting it is said to have an altitude of zero deg.\n\nSo, to calculate the altitude you need to draw an imaginary line from the moon to the nearest point on the earth's horizon and then measure the angle between your position and the moon and your position and the point on the horizon.
312096
Yes. Assuming you're exactly between the Moon and the Earth (no angles), then you can figure out the weight.\n\nThe equation is F = G*(m1*m2)/r^2 where F is the force (in this case, weight), G is the gravitational constant (6.6742e-11), m1 and m2 are the masses of the objects, and r is the distance between the two objects.\n\nSo, let's first figure out the force just on Earth. Use the Earth's mass and your mass as the two objects, and for r, use the distance between the center of the earth and the surface. Using an average person's mass (75 kg, or ~165 lbs), I get a weight of 739.31 Newtons, or 166 pounds, which fits with the weight of the guy.\n\nNow do the same equation for the moon, using the mass of the moon, and for r, use the distance from the center of the moon to the Earth's surface. I get a value of 0.0024669 Newtons, or around 0.000556 pounds.\n\nTherefore, when the moon is directly above you, you only weigh 0.000556 pounds less than you would if the moon didn't exist. When the moon is closer on its orbit, it really won't make a difference. When the moon is closest to the Earth, you'll weigh 0.000062 pounds less than the average.\n\nSo, in short, the difference is very, very negligible.
304985
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.
823026
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.
370062
Well, for one, you need the sun. After that, I'd get some sidewalk chalk, then make a marking on the ground. Then place the object whose shadow you are measuring in a position where the top of the shadow meets the marking. Then make another marking at the bottom of the shadow. Then, using a ruler, measure the distance.\n\nIf you're looking to do this mathematically, you would need to know the height of the object and the angle at which the sun is (which you can approximate based on the time of day).
857464
Unfortunately, work is based on the distance the meteor travels, so you'll need to know how far from the moon it started.\n\nFor the rest:\nW=F*d\n\nW is work\nF is Force\nd is distance travelled\n\nF is the force of gravity and is defined as\nF=(G*m1*m2)/r^2\n\nWhere:\nG is the Universal Gravitational Constant\nm1 & m2 are the mass of the meteor and moon\nr is the distance between the moon and the meteor\n\nI know this doesn't completely answer your question, but hopefully puts you on the right track.
198137
The average distance from the Moon to the Earth is 384,403 kilometers (238,857 miles). The Moon's diameter is 3,476 kilometers (2,160 miles).
115838
The moon is very far away from the Earth.  It may be hard to imagine, but it is possible to see the moon from many different points on the earth at the same time.\n\nAlso realize that all of Earth experiences the same phase of the moon at the same time.  When we see a full moon, the people on the other side of the world are seeing the same full moon.
1041670
The distance to the sun can be computed by a variety of techniques.  Trigonometry and parallax measurements were used by ancient greek and arab astronomers to come up with pretty good estimates for the distances to the sun, moon, and other planets.\n\nThe sun's diameter at its equator is about 865,000 miles, and the mean distance from the earth to the sun is about 93 million miles.\n\nThe moon's diameter at its equator is about 2,160 miles, and the mean distance from the earth to the moon is about 238,900 miles.\n\nThe ratio of the diameters is 865000 / 2160 or about 400.  The ratio of distances is 93,000,000 / 238,900 or about 390.  So the sun is about 400 times larger than the moon, but it's also about 400 times farther away - so the sun and moon appear about the same size in the sky.\n\nThe moon's orbit around the earth isn't a circle, and the lunar distance varies from about 220,000 miles to about 252,000 miles.  When it is close to the earth, the moon appears slightly larger than the sun.  When it is in the further part of it's orbit, the moon appears slightly smaller than the sun.  If an eclipse occurs when the moon is furthest from earth, the moon's disk can't completely cover the sun.  The result is an "annular" eclipse.\n\nIncidentally, the moon gets slightly farther away from the earth every year.  Eventually, in the distant future, it will be too far away to produce a total eclipse of the sun.
10539
I'm not sure about the distance to the moon, but I can explain the rest...\n\nLots of questions here... \n\nFirst, they found out about the curvature of the earth mainly by observing the earth's shadow on the moon during a lunar eclipse. You can easily see the Earth's shadow as curved and not straight.\n\nThe experiment with the sticks, performed by Eratosthenes, was to determine the circumference of the earth, and thus, how big it is.\n\nYou're absolutely right with the argument that the shade produced by a stick on a flat surface would be different than that on a curved surface.\n\nEratosthenes went with the [correct] assumption that the sun was so far away, that its light rays striking the earth would be parallel to each other.\n\nIn truth, this only occurs for an object that is infinitely far away, but the sun is SO far away, that this difference, from two different places on Earth, is negligable.\n\nSo if you have a flat surface, then no matter where you are on the surface, the shadow cast by a stick should be the same. If its noon, and the light rights are parallel, and the Earth is flat, then there should be no shadow no matter where you are on the surface.\n\nNow, Eratosthenes saw that at noon, in two different cities, that in one city, there was no shadow cast by the stick, meaning the sun was directly overhead, and at the same time in the other city, there was a small shadow, and measured the angle the stick made with the shadow. \n\nHe noticed the angle was about 7 degrees, which means its 1/50th of a full circle (360 degrees). Therefore, the two cities were 1/50th the way around the globe. Knowing how far the cities were, he multiplied their distance by 50 and got the distance around the globe. c = 2*pi*r, so he could then calculate the radius of the earth as well!\n\nThat's pretty much it, hoped you understood it. Its really amazing how they could get the size of the earth simply by using two old sticks and a little thought! :)
45815
Mean Distance of Moon from Earth - 238,712 mi (384,400 km)\nGreatest Distance of Moon from Earth (Apogee) - 252,586 mi (406,740 km) \nShortest Distance of Moon from Earth (Perigee) - 221,331 mi (356,410 km)