*facts\n*symbol\n*mass compared with earth\n*denisty compared with water = 1\n*rotation\n*revoultion\n*tilt on axis\n*surface gravity compared with earth\n*temperature\n*moons & rings especially miranda\n*any other info.. etc.
209959
Uranus, the first planet discovered in modern times, was discovered by William Herschel while systematically searching the sky with his telescope on March 13, 1781. It had actually been seen many times before but ignored as simply another star (the earliest recorded sighting was in 1690 when John Flamsteed cataloged it as 34 Tauri). Herschel named it "the Georgium Sidus" (the Georgian Planet) in honor of his patron, the infamous (to Americans) King George III of England; others called it "Herschel". The name "Uranus" was first proposed by Bode in conformity with the other planetary names from classical mythology but didn't come into common use until 1850. \n\nGo to the site below.  It has all kinds of info, even on all of the moons.  I hope this helps :)
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Use your body weight to open the door to a library.\nWalk up to the reference librarian and ask "what is the mass of planet earth"?\n\nNewton showed that, for spherical objects, you can make the simplifying assumption that all of the object's mass is concentrated at the center of the sphere. The following equation expresses the gravitational attraction that two spherical objects have on one another: \n\nF = G * M1 * M2 / R2 \nR is the distance separating the two objects. \nG is a constant that is 6.67259x10-11m3/s2 kg. \nM1 and M2 are the two masses that are attracting each other. \nF is the force of attraction between them. \nAssume that Earth is one of the masses (M1) and a 1-kg sphere is the other (M2). The force between them is 9.8 kg*m/s2 -- we can calculate this force by dropping the 1-kg sphere and measuring the acceleration that the Earth's gravitational field applies to it (9.8 m/s2). \n\nThe radius of the Earth is 6,400,000 meters (6,999,125 yards). If you plug all of these values in and solve for M1, you find that the mass of the Earth is 6,000,000,000,000,000,000,000,000 kilograms (6E+24 kilograms / 1.3E+25 pounds). \n\n1 It is "more proper" to ask about mass rather than weight because weight is a force that requires a gravitational field to determine. You can take a bowling ball and weigh it on the Earth and on the moon. The weight on the moon will be one-sixth that on the Earth, but the amount of mass is the same in both places. To weigh the Earth, we would need to know in which object's gravitational field we want to calculate the weight. The mass of the Earth, on the other hand, is a constant.
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Measurements. The acceleration of gravity is:\ng = G*m/R^2\nWhere G is the gravitational constant 6.67*10^-11 N*m^2/s^2, m is the mass of the planet, and R is the distance between yourself and the planet. For the surface of the Earth, m and R are approximately:\nMe = 5.97*10^24 kg\nRe = 6.38*10^6 m\nUsing this, you should get about 9.8 m/s^2.
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I think the equation W = m x g (W = weight; m = mass; g = acceleration due to gravity) might address your question.\n\nStating they are interchangeable might be misleading if the statement is not qualified by your textbook. The qualification is that when we are measuring objects from an earth reference point, for all intents and purposes, the weight of an object is proportional to its content (i.e., mass). In mathematical terms, the assumption would be, W = m * 1. \n\nHowever, if we are measuring objects in places other than from an earth reference point, then weight and mass are not interchangeable. For example, the force of gravity on the moon is less than on earth. So as an example, the equation for weight on the moon could be, W = m * .5. With this equation, the weight of an object on the moon is less than its weight on the earth. Here is a math example to illustrate:\n\nObject (Mass) = 100 Kg\ng(Earth) =  1 \ng(Moon) = .5\n\nW(Earth) = 100 x 1\n              = 100 Kg (Object's weight on earth)\n\nW(Moon) = 100 x .5\n              = 50 Kg (Object's weight on moon)\n\nWikipedia might explain the notion of interchange between weight and mass better:\n\nhttp://en.wikipedia.org/wiki/Weight#Weight_and_mass
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I am a member of the Jules Verne Forum based in Israel. If you tell me what the questions are, I will pass them on to the forum (which gets contributions from translators, professors and NASA scientists)
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haha, k  Draw a picture and draw a free-body diagram.\n\nI don't feel like busting out a paper/pencil  so I'll try it in my mind\n\nF(sub g) cos Theta  =  Mg cos(20 degrees)\n\nbecause the Normal force is equal to the y-component of gravity.\n\nTwo similar triangles, the incline and one which has the Force of gravity straight down, and the y compentn equal opposite of the normal force, and X component equal opposite to F\n\n\nlots of physics i'm skipping in my explanation tho
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ms = Mass of Starship Enterprise\nvs = Velocity of Starship Enterprise\n\nme = Mass of Earth\nve = Velocity of Earth\n\nm1 = new mass\nv1 = new velocity\n\nm1*v1 = me*ve + ms*vs\n\nsince the velocity of earth is 0, me*ve = 0\n\nTherefore:\n\nm1*v1 = ms*vs\n\nand since m1 = ms + me\n\nwe have \n\nv1 = (ms*vs) / (ms + me)\n\nAs long as the units are consistent, this is your answer.
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By drawing a simple figure and using geometry\n\nD x D = 2 R h.\n\nWhere\nD is the distance\nR is the radius of the earth \nAnd h is the vertical drop.
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Go to http://www.astro.com ~ then click up at the top for free readings. Input data, make sure you have it ALL correct..especially check the city and state, you will see why. Then..click continue....and there is a column that says something like"VIEW ASTRO WHEEL" ~ about 3/4 of the way down....and then your chart will come up degrees and all....I hope that this helps. It is ONLY a Natal Chart..meaning that it's like a Polaroid of the sky at your time of birth. It is FREE so..you can go to an astrology room and have it interpreted for you....OK? I Hope that this helps....
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http://en.wikipedia.org/wiki/Mars
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http://www.windows.ucar.edu/tour/link=/earth/statistics.html
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density,volume,weight,number of electron and proton.
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p = g * h * water density\np = 9.81 * 12 * 1000\np = 117.72 kPa\n\ng = gravity of earth and its constant wihch equal to 9.81
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ok here we go !!!!\n\n \nASTRONOMY QUIZ ANSWERS\n\nHere is the astronomy answers fir the quiz at the millsberry acadomy lab...\n\n1.caloris basin\n2.subsolar point\n3.venus\n4.phobus and deimos\n5.set of thousand rings\n6.metal\n7.venus\n8.northern\n9.thick atmosphere\n10.false\n11.love and beauty\n12.iron\n13.hydrogen and helium\n14.true\n15.true\n16.underground ocean\n17.saturn\n18.-29.5\n19.great red spot\n20.crystallized ammonia\n21.ice\n22.asteroids\n23.a month\n24.venus,pluto,uranus\n25.spiral\n26.gas\n27.false\n28.mercury\n29.meteor\n30.away from the sun\n31.orion
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http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/estatics/u8l2c.html\n\nhttp://www.glenbrook.k12.il.us/gbssci/phys/class/estatics/u8l2b.html\n\nHope it helps. \nBest Wishes,\nAlok Bhatt
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Check out the sites below for in-depth explanations.  \nThe first one is from NASA.
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Dude - its your homework - my help to you is stop screwing around and write it.  However - I would suggest you check out links here http://www.ask.com/web?q=the+earth+is+beautiful&qsrc=145&o=10429&l=dir to give you ideas.\n\nGood Luck!!!