How is the distance between the Moon and the Earth is measured ?
87056
I've heard of powerful lasers that can be used to measure the distance.
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)
1089940
With the average distance between the Earth and the Moon being 238,857 miles...the total distance of 600 round trips between the Earth and the Moon is approximately 286,628,400 miles.
239105
to use the gravitational pulls of the earth and the moon
368075
The average distance from the Earth to the Moon is 384,403 kilometers.  Naismith's Rule of Thumb is 5 kilometers/hour.  Assuming the trip from the Earth to the Moon is level, the trip would take 76,880.6 hours (384,403/5) or 3202.4 days or 8.8 years if you traveled 24 hours per day.  If you allow for 6 hours of sleep a day, it would take 4271.1 days or 11.7 years.\n\nIf one was to count the entire distance to the moon as an ascension, this would add 507,338 hours (304,403 *1000 to convert to meters. 304,403,000/300 = 1,014,676 - the number of 1/2 hours to add.  1,014,676/2=507,338 - number of hours).\n\nThis would make the total trip time 584,218 hours or 24,342 days or 66.7 years.\n\nBetter pack a lunch...
969045
Are you taking into account the radius of the moon? The equations you've provided treat "r" as the distance between the centers of mass of each object. Therefore:\n\nr = radius of the moon + distance from it's surface\n\nr = 1785 km + 111 km\n\nr = 1,896,000 m\n\nSubstituting this into your equations provides:\n\nT = 2*pi*sqrt((1896000)^3/(6.674x10^-11 * 7.3x10^22))\n\nT = 7431.6 s = 123.9 min\n\nAdditionally:\n\nV = sqrt((6.674x10^-11 * 7.3x10^22) / 1896000)\n\nV = 1603 m/s\n\nHope that clears things up a bit! :)
272130
When the moon is full, it is on the side of the Earth opposite the sun. It takes the moon about a month to go around the Earth once. So in the course of a single day, it remains in almost the same position.
537957
The Earth is between the moon and the Sun.  \n\nSomeone standing on the moon would see a solar eclipse.  \n\nSolar eclipse  Sun   Moon   Earth\n\nLunar eclipse  Sun   Earth  Moon
909859
We have the ability now to measure the orbit of the Earth, and all other planets, to very high precision, and it is not getting closer to the Sun.\n\nYou can do a very rough measurement yourself. If the Earth gets closer to the Sun, the Sun will appear bigger in the sky. Measure it in photographs taken by the same telescope a year apart. \n\nAnother consideration is this. Currently, the Moon and Sun appear the same size in the sky, which produces spectacular total eclipses of the Sun. If the Earth got just a little closer to the Sun, the Moon would no longer be able to completely cover the Sun. So if total solar eclipses continue as usual, that is evidence the Earth is not getting any nearer to the Sun.
778505
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.
679383
The average distance from the Moon to the Earth is 384,403 kilometers (238,857 miles); it is slowly moving away at about the same rate as your fingernails grow (about 4 cm a year).
28127
You can see about 4 miles to a flat horizon on the Earth. So from one horizon to the other would be 8 miles. (On the Moon, it would only be 3 miles-- the horizon is only a mile and a half away.)
985857
GREAT QUESTION!\n\nThe answer is: In several ways.  First of all:\n\n1. Earth rocks!\n\nInstead of following a smooth Keplerian orbit around the Sun, Earth's own centroid performs an oscillatory motion about this "mean path."  By careful observations of the nearest PLANETS (but not the Moon --- why not?!) the amplitude of this Moon-induced "rocking" can be measured: Earth's centroid wobbles back and forth by about 3/4 of the Earth's own radius.  From this alone, one can deduce that the Moon's mass is ~ 1/80th that of the Earth.  \n\n[How?: From cunning arguments, the Greeks worked out long ago that the Moon was about 60 Earth radii distant; but when things are orbiting around one another with an equal but opposite force attracting them, it's the centroid of the COMBINED system that stays fixed, and the sizes of the "orbits" of both objects about their common centroid are inversely proportional to their masses --- just as though there's a giant Archimedes lever out there in space!  So that's how the "60" and the "3/4" combine to tell you that M_moon ~ M_earth/80.]  \n\nFurthermore, THAT is how it was already known in Victorian times that the gravitational acceleration at the Moon's surface must be ~1/6th that of our familiar 'g' at the Earth's surface.  \n\n[The ACTUAL PHYSICAL distance to the Moon is also needed to work this out, so that one can obtain its physical from its angular size.  You then plug the Moon's mass and physical size into Newton's Law of Gravity, applied at the Moon's surface.  Since the Greeks knew the Moon was ~60 Earth radii away, that actually meant measuring the Earth's size well!  \n\nWhen Newton made his own first "Moon check" on his law of gravity, it didn't seem to fit.  He was away home from Cambridge, in Lincolnshire, avoiding the Great Plague of 1666, and thus away from up-to-date books and other information.  But some time later, perhaps on returning to Cambridge, he learned of a new measurement of the Earth's size.  He then went back to his Moon test again, and found that when he'd corrected his earlier calculation, "it answered pretty nearly."  \n\nThat's how he first confirmed his Law of Gravity.  It tells you something about perseverance.]\n\nSo in the end, things like the acceleration at the Moon's surface were already known long before an astronaut demonstrated the "strength" of it by dropping objects on the surface of the Moon, on live television. \n\n2. The tides.\n\nAnother effect of the Moon's gravity on the Earth is of course the tides, though this is a more subtle and less direct (scientific) effect, as it's then the DIFFERENTIAL TIDAL FORCES (derived --- chuckle --- from differentiating the Moon's gravitational force at different places all over the Earth/Oceans etc.) that provide the forcing effect.  However, as we are riding on the surface of the Earth itself, we can ourselves more directly experience and observe this more subtle effect, through the tides, than we can the "rocking" produced by the stronger, main part of the attraction.\n\nI hope you get to study this some day; it's fascinating stuff!\n\n[Here's a funny thing that has always intrigued students at Starfleet Academy.  As viewed in a frame of reference in which the Sun is at rest --- the frame you use when you want to show the planets orbiting the Sun --- what do you think the path of the Moon looks like in space?  Does it spiral a lot, in overlapping circuits, around the Earth's path, or does it slowly and majestically make a gradually curving sinusoidal like motion, first curving in, then curving out (relative to the solar direction) about the essentially circular path of the Earth?  Would you believe that it always CURVES IN TOWARDS THE SUN?!!\n\nIf you know about gravity, acceleration, and the two separate orbital velocites
288730
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.
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
364790
The average distance from the Moon to the Earth is 384,403 kilometres (238,857 miles).
296979
The Moon is self explanatory, just look skyward at a full Moon to see it is spherical. The Earth takes some proving, but all you need is a space photo or two for people to get the idea. Oh, and the Earth is technically an oblate spheroid.