Planet Y has 2 times the diameter?
956084
Consider mass m on planet Y\nas also on Earth. We have\nm*g[e]=G*m*M/R^2.........1\nm*g[y]=G*m*[2M]/[2R]^2....2\nDivide eqn.2 by eqn. 1 we have\ng[y]/g[e]\n=2M/M*[1/4R^2]/[1/R^2]\n=2/4\n=1/2
1032173
Depends on how long ago you were in school -- Pluto is so far away that for many years we thought it was larger than it has turned out to, in fact, be.\n\nThe most recent value I have for the diameter of the two planets is:\nMercury = 4879.4 km\nPluto = 2302 km\n\nThen, of course, there's the debate about whether or not Pluto is even truly a planet . . .
338526
diameter x pi\npi=3.14159...\n\nthat's it
147445
Its all relative..usually we use sun as a point of a reference when we talk about relative size of planets which circle it. Pluto is about the same size as mercury.
1175686
This can be solved in several different ways.  One of course is by calculus, by developing a formula for the surface area of a "slice" of the remainder of the sphere and integrating it about a rotated axis, or something similar.  Another is by digging for the rather obscure formula for the "volume of the endcaps of a sphere" and subtracting two of those and the volume of a cylinder from the volume of the original sphere (and discovering that many of the terms will cancel out nicely).\n\nThe easiest method, however, is via the use of simple logic.  Unfortunately, this method also requires one more "given" in order for it to be valid, though this given is not an unreasonable one:  There must be a single, real answer (or more specifically, a unique answer on the real number line).  Such an assumption would be folly in a practical (real-world) context, since without some form of proof, one should never immediately assume that this is the case.  Perhaps there are multiple answers, or there is no answer at all?  But as luck would have it, there actually is one answer, and it may be derived logically due to its unique existence.  (*thumbs nose at any philosophers who may be reading*)\n\nSolution:\n\nLet R be the radius of the cylinder drilled into the sphere.  If and only if there is a single, real answer, then that answer would still be the same no matter the value of R... even if R=0.  Thus, the answer is the volume of a sphere of radius 3 (diameter 6), or 36*pi cubic units.\n\n(Yes, the volume is always the same, whether the original sphere were the size of a basketball, or the size of a planet.)\n\nIt's a surprising answer, along the lines of the more familiar puzzle: "If the Earth were a perfect sphere and a band were placed snugly about its equator, and then one more meter were added to the length of the band and it were stretched so that all points along it were equidistant from the surface of the Earth, how far above the Earth would the band rise?"  Many are surprised to find that the size of the Earth plays absolutely no part in the equation, and that the "rise" is exactly the same if the sphere were basketball-sized or, umm... Earth-sized (with a circumference greater than 40,000,000 meters).  It rises exactly the radius of a circle with a 1-meter circumference: 1 / (2*pi) meters.
265928
1.27 X 1022 KG (0.0021 Earths)
462097
"The equatorial diameter is c.7,926 mi (12,760 km) and the polar diameter 7,900 mi (12,720 km)"\nThe Earth is the fifth largest planet of the solar system. The reason why earth is in a spherical shape, is because of the gravitational pull. The number that appeared most often in my sources was 12,756 km. There are actually two diameters of Earths: equatorial and polar. One of my sources stated that the equatorial diameter is 40 kilometers wider than the polar diameter. The Earth is not a perfect sphere because of a bulge in the equator. Most of the current knowledge of the Earth has been collected in three methods, through satellites, telescopes and outer space exploration.
674138
Very similar to Earth's size.
858571
G'day Dragon Warrior,\n\nThank you for your question.\n\nThe diameter of the earth in miles is 7,296 miles at the equator and 7,900 at the poles.\n\nI have attached sources for your reference.\n\nRegards
871143
Largest known to date is the "puffy planet" HAT-P-1b orbiting the sunlike star ADS 16402 B, one of a binary pair of G0 V stars 450 light yeard away, in the constellation Lacerta. Our sun is of spectral class G2 V i.e. a little cooler than ADS 16402 B.\n\nHAT-P-1b has a radius of 1.36 x the radius of Jupiter. \n\nIt was discovered on 13 September 2006 and is 450 light years away. It has the lowest density of any known extrasolar planet. It also has the name Krais.\n\nIt was announced as follows:\n\n"Using small automated telescopes in Arizona and Hawaii, the HATNet project has detected an object transiting one member of the double star system ADS 16402 AB. This system is a pair of G0 main-sequence stars with age about 3 billion years, at a distance of ~139 parsecs and projected separation of ~1550 AU. The transit signal has a period of 4.46529 days."\n\nNote: Only the radii of transiting planets are known.
501741
Pretty close!\nThe "equitorial" diameter = 3,476.2 km\nand the "polar" diameter = 3,472.0 km\nSo, it is a sligtly "oblate" spheroid.\nabout 0.12% from being a "perfect" sphere.
147303
2 times pi times the radius\nor\npi times the diameter\n\npi = 3.14\n\nThe first person to answer gave you the formula for the area, not circumference.
289682
circumference = pie times diameter\nradius is half of the diameter\n\nso if you have either the circumference or the diameter, you can figure it out.
905767
You are making it needlessly hard on yourself.  The planet has the same mass as the earth, and someone on it weight the same as on earth, therefore the surface of the planet is the same distance from its center as the earth radius, i.e. this planet has the same dimension as earth.\nSo, the radius of the planet is 6,372.797 km, same as earth's.\nYou don't need to compute anything.  The orbital velocity of the planet has nothing to do with it.\nIf the weight was different, you only need to scale it remembering that the weight would vary as an inverse of the square of the radius.
451278
here is a calculator for determining planet and orbit sizes to scale.  i hope this helps.\n\nhttp://www.exploratorium.edu/ronh/solar_system/
118436
The size wise order of the planets in our Solar system is as follows:-\n1) Jupiter, 2) Saturn,3) Uranus or Herschel,4) Neptune\n5) Earth, 6) Venus, 7)Mars,8)Mercury,9)Pluto.\nThe largest is Jupiter and the smallest is Pluto. Our moon is a natural satellite of the earth and not an independent planet. It is smaller than Pluto.
502090
Centre is 1.5,2 and the Radius is 3