planet     actual diameter         scale diameter\nmercury       4,8880                    1 1/4 in\nvenus         12,104                     1 1/2 in\nearth          12, 756                    1 1/2 in\nmars           6,794                       1 1/4 in\njupiter      142,984                       4 in\nsaturn        120,536                    3in\nuranus       51,118                  2 1/2 in\nneptune     49,528                      2 in\npluto       2300                      1 in
203202
here is a calculator for determining planet and orbit sizes to scale.  i hope this helps.\n\nhttp://www.exploratorium.edu/ronh/solar_system/
39562
Diameter will be 1 units\n\nCircumference =piD\nif Circumference =pi\nthen D=1
57791
http://www.seds.org/billa/tnp/pluto.html
645780
Hi: \n\nFirst off let start by explain what a scale is  a Scale is a ratio of two sizes . It is way to put something that is too big or too small in the real world to a size that we  humans can relate to for example Let say a Saturn V rocket is 309 feet tall to make a 1/12  scale model of it. first we know the real rocket is 309 feet. on our model scale  1inch equal 12 inches or 1 foot so:\n\n309/12 = 25.75 or 25 and 3/4 inches\n\nour scale model of the Saturn V rocket is 25.75 inches or slightly larger then 2 feet tall (actually 2 feet 1 3/4 inches tall) \n\nSomethings are too small in the real world  ( like pocket watch gears or electronic parts) So we  use a enlarging  scale and it written  like this  4/ 1 or 1 inch  equals 4 inches  for example a gear is 1/8" in diameter in the real world would be drawn on a blueprint drawing with the scale 4 to 1 the gear is drawn at 1/2 inch  diameter  or 4 times it's actual size \n\n1/8 * 4/1 = 4/8= 1/2 inches\n\nthe way to remember a scale is what place is the 1 units  in relation to the scaling value or division sign or  " :" symbol  if the 1  is in front of the division sign or ":" symbol  your dealing with a reduction factor with means that the model or drawing is going to be smaller than the actual object itself\n\nIf the 1 unit is in back of the division or  ":" symbol with number greater than 1 in the front your deal with a enlarging factor meaning the drawing is bigger the the actual size of the object itself  \n\n\nOkay let's see what the answer is, for your problem:\n\nlet say your measuring a real rod that is 4.7 mm and your scale on a drawing of this rod is 2.5 :1  what is the drawing scale lenght of this rod is:\n\nrule 1: where is the 1 in relation to the scale value? \n\nanswer: behind it. So we are dealing with a enlargement drawing \n\nSo set the problem as follows:\n\n4.7 *2.5 =  11.75 mm\n\nThe drawing scale measured rod lenght is 11.75 mm or the rod lenght on the drawing is 11.75 mm
529846
Below is just a sample of much information regarding the planets on the site listing below.\n\nTypical planet temperatures- not record breaking temps\n\nPluto - 230 \nUranus -223 \nNeptune – 220 \nMercury – 200\nSaturn – 160 \nJupiter – 145 \nMars – 140\nEarth – 89 \nVenus 464 \nSun 5500
501038
This is a simple ratio problem. The equation would be Dgalaxy/Dearth=Dearth/x so x=(Dearth)^2/Dgalaxy. Asssuming a parsec is 3.6 light years, and a light year is 5.87 trillion miles and the diameter of the Earth is 8,000 miles, then the Earth would be 0.001 millimeter or one micron in size, which is the size of an average bacterium, maybe? Very cool. Oops! I re-did my calculation and I see that I made a factor of ten error. The correct answer is about 0.1 microns, or the size of some viruses. But tell me how you would make such a model of the Earth?
43722
Surprisingly, there is no total agreement about the size of our Solar System. Some say it's defined by the orbit of the outermost planet (used to be Pluto), while others say it's where the Sun's heliosphere ends. Then there's the definition that I personally favor -- the Oort Cloud of comets.\n\nHere's a breakdown of the three major sizes favored:\nPluto orbit - 5,900^6 km\nHeliosphere - 2.2^10 km \nOort Cloud  - 1.5^13 km (1.6 light years)
28662
use nasa's calculator:\nhttp://www.exploratorium.edu/ronh/weight/
86951
The distance between the two planets will varry depending on where each is in their orbit around the sun.\n\nMercury has an orbital radius of about 5.795 E7 km,\nVenus has an orbital radius of about 1.0811 E8 km.\nSo once in a while, when Mercury and Venus are aligned, the minimum distance between the two planets is 5.016 E7 km.  One the other hand, if both planets are on opposite sides of the sun, the maximum distance would be 1.6606 E8 km.\n\nI dont know why you are thinking that a distance would be measured in kilograms (kg) [possibly just a typo I guess].
278418
If you mean the diameter of the horizon, it's a function of how high you are above the surface.  If you're ON the surface, the diameter is zero.  If you're a distance h above the surface and the radius of the earth is r, the diameter is\n\nd=2*(2rh + h^2)^.5
193662
You have to find the Least Common Multiple.\nFirst we know that it's going to be 84 or more years because Uranus takes 84 years to complete one orbit.\n So we know that the other planets' time length to make one orbit has to go into 84. \nFirst I did Earth: 84/1 = 84. \nNow I need to do Jupiter: 84/12 = 7. \nNext I do Saturn: 84/30 = 2 remainder 24. \nSo now I need to find a common multiple of 30 and 84. \n84 x 2 = 168 (no). 84 x 3 = 252 (no). 84 x 4 = 336 (no). 84 x 5 = 420 (yes).\n\n  14 \n  30)420 \n  -30 \n  --- \n  120 \n  -120 \n  --- \n  0 \n\nSo 420 is the number that goes into all the time lengths for the planets. Therefore we find out out that they will align once in 420 years.\nIf you knew the last time they were all in a line. You add that year + 420 and you get the next year when they will alll be in a line.
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.
304466
Here are some quick facts:\n\nhttp://www.sir-ray.com/ Enter the site, click astronomy then click on Space Dimensions & Planetary Facts (this site is excellent)\nhttp://www.dustbunny.com/afk/planets/\nhttp://www.kidzone.ws/planets/index.htm\nhttp://nssdc.gsfc.nasa.gov/planetary/planetfact.html\n\nThese sites should be everything that you need~the first site should do it but the other three have little bits of information that might be useful~
268281
According to the NASA website, there are 134.*\n\nMercury - 0\nVenus - 0\nEarth - 1\nMars - 2\nJupiter - 62\nSaturn - 31\nUranus - 24\nNeptune - 13\nPluto - 1 \n \nBut there is another website that claims there are 156.**\n\nMercury=0 \nVenus=0 \nEarth=1 \nMars=2 \nJupiter=63 \nSaturn=47 \nUranus=27 \nNeptune=13 \nPluto=3\n\nTake it for what it's worth.
147601
Well, really, we don't actually have a scientific definition of a planet.  Shocking yes.  Embarrassing even more so.  But we are working on it.  The problem we are facing right now is that we have a large collection of "transition objects" out there, things that sit right on the border of "planet or not."\n\nAt the high mass end, it is pretty clear cut, Things that produce internal energy by fusion are stars, things that don't are planets.  The transition between star and planet thus seems to be at about 12 times the mass of Jupiter.\n\nAt the small end things are muddled because of Pluto.  No good definition of planet exists that keeps only 9 planets.  You can make a definition to include Pluto, but then you must include a lot more objects.  This is where we actually focusing on right now.\n\n"gravitationally round" is thrown around a lot.  My professional opinion is that this one one make the cut.  It seems nice, a planet is a planet because it has enough self-gravity to crush itself into a round shape.  The problem with this definition is that it means we would have about 100 planets in out solar system.  The largest asteroid, Ceres is round, as are most of the icy bodies in the Kuiper Belt.\n\nThe one that seems to be picking up steam is "dynamically dominant."  This basically means that a planet "dominates" the space around them.  Take a look at the area around the earth - nothing else similar to it.  Same with Venus, Jupiter, and the rest. (moons and rings are bound to the planet so they don't count).  Ceres (the largest asteroid) is surrounded by many, many other asteroids.  Same goes for Pluto, it is surrounded by many other icy bodies of similar size (including one that is bigger than it).  This definition leaves us with 8 planets.\n\nReally right now the "planets" are just defined by fiat, they are planets because we call them that and nothing more.  As a planetary scientist I feel a little guilty not being able to actually define one of the words in the name of my field, as such I've honestly been spending far more time working on this whole "planet" mess, than on my own research.
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.
68322
Polar Diameter: 7900 mi (12720 KM)\nPolar Circumference:  40008 km\n\nSo from the north pole to south pole it is about 20004 KM or about 12430 miles