Comments on: Q&A: Planet X has five times the diameter and three times? http://www.ufo-watch.com/conspiracies-2/qa-planet-x-has-five-times-the-diameter-and-three-times.html UFO videos, images and reports from all over the world. Wed, 23 May 2012 20:04:39 +0000 hourly 1 http://wordpress.org/?v=3.3.2 By: Ben G http://www.ufo-watch.com/conspiracies-2/qa-planet-x-has-five-times-the-diameter-and-three-times.html#comment-32576 Ben G Sun, 29 Jan 2012 14:19:49 +0000 http://www.ufo-watch.com/conspiracies-2/qa-planet-x-has-five-times-the-diameter-and-three-times.html#comment-32576 Newton said that the gravitation acceleration at distance r from the center of a sphere of mass M is proportional to the mass of the sphere and inversely proportional to r-squared. That means that it is proportional to M/(r-sq) So the mass is three times that of Earth. That's in the numerator. And if the diameter is five times that of Earth, then the radius also is five times that of Earth. Square that to get 25. That is in the denominator ("inversely proportional"). So the ratio is 3/25, or 12/100 = 12% -- much weaker than Earth. Newton said that the gravitation acceleration at distance r from the center of a sphere of mass M is proportional to the mass of the sphere and inversely proportional to r-squared. That means that it is proportional to M/(r-sq)

So the mass is three times that of Earth. That’s in the numerator. And if the diameter is five times that of Earth, then the radius also is five times that of Earth. Square that to get 25. That is in the denominator (“inversely proportional”).

So the ratio is 3/25, or 12/100 = 12% — much weaker than Earth.

]]> By: amar_chaba http://www.ufo-watch.com/conspiracies-2/qa-planet-x-has-five-times-the-diameter-and-three-times.html#comment-32575 amar_chaba Sun, 29 Jan 2012 14:03:06 +0000 http://www.ufo-watch.com/conspiracies-2/qa-planet-x-has-five-times-the-diameter-and-three-times.html#comment-32575 g=GM/R^2, M=mass of the planet,R=radius of the planet g(X)/g(E) =M(X)R^2(E)/M(E)R^2(X) =[M(X)/M(E)][R^2(X)/R^2(E)] [3][25] =75 g=GM/R^2, M=mass of the planet,R=radius of the planet
g(X)/g(E)
=M(X)R^2(E)/M(E)R^2(X)
=[M(X)/M(E)][R^2(X)/R^2(E)]
[3][25]
=75

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