Mortus Posted July 19, 2011 Share Posted July 19, 2011 My current velocity is 11464.0m/s. I'm currently at 13,000k and still climbing. My velocity has been unchanged for about 4 minutes now. Theoretically, does this mean I'm to the point where I'm just going to travel on forever?Also, is there any sites (or people willing to teach) the mathematics involved in escape velocity and orbit and whatnot? I want to learn what I'm doing when I do this, not just do it blindly (even if it does usually end in quite spectacular explosions). Link to comment Share on other sites More sharing options...
StrayEagle Posted July 19, 2011 Share Posted July 19, 2011 My current velocity is 11464.0m/s. I'm currently at 13,000k and still climbing. My velocity has been unchanged for about 4 minutes now. Theoretically, does this mean I'm to the point where I'm just going to travel on forever?Yes.Also, is there any sites (or people willing to teach) the mathematics involved in escape velocity and orbit and whatnot? I want to learn what I'm doing when I do this, not just do it blindly (even if it does usually end in quite spectacular explosions). Here is a community made tool for you to use, it’s a executable jar file. Works quite well.Link: http://kerbalspaceprogram.com/forum/index.php?topic=562Hope this helps Link to comment Share on other sites More sharing options...
Mortus Posted July 19, 2011 Author Share Posted July 19, 2011 Thanks x2. That means I don't have to run it through the night to see where I peak. Link to comment Share on other sites More sharing options...
StrayEagle Posted July 19, 2011 Share Posted July 19, 2011 Should probably linked you to this thread:http://kerbalspaceprogram.com/forum/index.php?topic=647.0As there are a few different options as far as getting the speeds. One is just a table, one is web based and one is a Jar. Link to comment Share on other sites More sharing options...
RangerDanger75 Posted July 19, 2011 Share Posted July 19, 2011 Here's a good guide in the wiki on transfer orbits and stuff. Very useful. Sometimes I'll make it a race against the clock and try to figure out the math by hand before I reach my apokee/perikee. i2 < 3 mathhttp://kerbalspaceprogram.com/wiki/index.php?title=Tutorial:OrbitingExcept the site is down right now for some reason, soooooo...hang in there! ;D Link to comment Share on other sites More sharing options...
DrGast Posted July 19, 2011 Share Posted July 19, 2011 At 11 km/s you'd almost have been able to escape EARTH 8)By the way, maybe OT, anyone knows what the g of Kearth is? Density, mass? I read somewhere the radius was about 600 km. Link to comment Share on other sites More sharing options...
Ice Posted July 19, 2011 Share Posted July 19, 2011 WIKIPEDIA GO!Nope, just kidding, I hate wikipedia. Good thing I got this book!*reads* 'The Handy Space Answer Book'.Okay, I got it down from the bookshelf, now I cant lift it. I think 530 pages is a little too much to be handy.Anyways, taken from my memory of what I read somewhere else a few days ago: Diameter of earth: 13000. Circumstance of earth: 40 000. Mass of earth: 50 000 something. Link to comment Share on other sites More sharing options...
DrGast Posted July 19, 2011 Share Posted July 19, 2011 Yeah, I meant the Kearth ones ;P I guess they'll be scaled somehow. Link to comment Share on other sites More sharing options...
Supraluminal Posted July 19, 2011 Share Posted July 19, 2011 Not sure which thread I saw it in, but apparently Kearth's radius is indeed 600km, and its mass (and hence surface gravity) is the same as Earth's. Density is obviously much higher because of that. Link to comment Share on other sites More sharing options...
cjameshuff Posted July 19, 2011 Share Posted July 19, 2011 Not sure which thread I saw it in, but apparently Kearth's radius is indeed 600km, and its mass (and hence surface gravity) is the same as Earth's. Density is obviously much higher because of that.Its mass is much lower...about 5.29e22 kg, 1/113th that of Earth. If it were the same, surface gravity would be 113 g. (remember, A = G*m/r^2)Density is still extremely high, however: http://kerbalspaceprogram.com/forum/index.php?topic=66.msg279#msg279 Link to comment Share on other sites More sharing options...
Supraluminal Posted July 19, 2011 Share Posted July 19, 2011 Its mass is much lower...about 5.29e22 kg, 1/113th that of Earth. If it were the same, surface gravity would be 113 g. (remember, A = G*m/r^2)Density is still extremely high, however: http://kerbalspaceprogram.com/forum/index.php?topic=66.msg279#msg279Oh, heh. Never mind then! I must have erroneously imagined the bit about the mass being the same. Pretty sure I read that it's 1 Earth gravity at the surface, though, yes? Link to comment Share on other sites More sharing options...
cjameshuff Posted July 19, 2011 Share Posted July 19, 2011 Oh, heh. Never mind then! I must have erroneously imagined the bit about the mass being the same. Pretty sure I read that it's 1 Earth gravity at the surface, though, yes?Yeah. Though because of its small mass, it falls off faster...1/4 that at 600 km above sea level, instead of 6378 km above sea level in the case of Earth.Similar situation in our solar system:http://en.wikipedia.org/wiki/File:Terrestrial_planet_size_comparisons.jpgMercury, on the far left, is the smallest and lowest mass of the terrestrial planets, but has higher surface gravity (0.38 g) than Mars (0.376 g) because of its higher density (5.2 g/cm^3, second only to Earth), which puts its surface deeper in its gravity well. Link to comment Share on other sites More sharing options...
foamyesque Posted July 20, 2011 Share Posted July 20, 2011 Yeah. Though because of its small mass, it falls off faster...1/4 that at 600 km above sea level, instead of 6378 km above sea level in the case of Earth.That's not so much to do with the mass as it is to do with the radius you start from. Link to comment Share on other sites More sharing options...
RangerDanger75 Posted July 20, 2011 Share Posted July 20, 2011 How do I calculate escape velocity? I know the formula is sqrt(2GM/r), but when I use the numbers G=9.807, M =5.29e22, and whatever altitude I'm at, the result doesn't line up with the ones in velocity calculators/charts. Link to comment Share on other sites More sharing options...
Chromium Posted July 20, 2011 Share Posted July 20, 2011 How do I calculate escape velocity? I know the formula is sqrt(2GM/r), but when I use the numbers G=9.807, M =5.29e22, and whatever altitude I'm at, the result doesn't line up with the ones in velocity calculators/charts.Your problem there is the G. Capital, or 'big' G is the gravitational constant, equal to about 6.67 x 10^-11 m^3 kg^-1 s^-2 (don't mind the wacky units, that just makes equations balance). Lower-case, or 'little' g is the acceleration due to gravity on Earth's surface, which is 9.8 m s^-2. You want to use Big G in this case.Also, remember to always calculate r as the sum of your altitude and the planet's radius. For a 100 kilometer orbit, the r in your equation is actually equal to 700, which is your altitude plus the 600 km radius of Kearth. We do that correction to account for the fact that you orbit the center of gravitational attraction of a body, which lies at the center of a planet. Link to comment Share on other sites More sharing options...
Victor Posted July 20, 2011 Share Posted July 20, 2011 Moving this thread to the How To... forumGeneral Discussion -> How To.../Moderator Link to comment Share on other sites More sharing options...
Entroper Posted July 20, 2011 Share Posted July 20, 2011 BTW, if you don't want to remember or look up the values for G and M all the time, you can use the much easier to remember g (Kearth surface acceleration due to gravity, 9.807 m/s^2) and R (radius of Kearth, 600 000 m). G*M is called the standard gravitational parameter of a body, and you can solve for G*M in the standard gravitation equation F = G*M*m/r^2. G*M = F*r^2/m. At the surface of Kearth, F/m = g, and r = R, so you have G*M = g*R^2. Link to comment Share on other sites More sharing options...
RangerDanger75 Posted July 20, 2011 Share Posted July 20, 2011 BTW, if you don't want to remember or look up the values for G and M all the time, you can use the much easier to remember g (Kearth surface acceleration due to gravity, 9.807 m/s^2) and R (radius of Kearth, 600 000 m). G*M is called the standard gravitational parameter of a body, and you can solve for G*M in the standard gravitation equation F = G*M*m/r^2. G*M = F*r^2/m. At the surface of Kearth, F/m = g, and r = R, so you have G*M = g*R^2.It seems this way isn't quite as accurate, but still useful nonetheless. Link to comment Share on other sites More sharing options...
Entroper Posted July 20, 2011 Share Posted July 20, 2011 It seems this way isn't quite as accurate, but still useful nonetheless.It should actually be more accurate, since g and R should be the exact figures used by the game. Your G and M will vary depending on how many figures are in your reference, how many figures M was calculated to, etc. Link to comment Share on other sites More sharing options...
cjameshuff Posted July 20, 2011 Share Posted July 20, 2011 It should actually be more accurate, since g and R should be the exact figures used by the game. Your G and M will vary depending on how many figures are in your reference, how many figures M was calculated to, etc.It's more accurate in reality as well, because G is only known to a handful of digits, while G*M can be measured directly to high precision. Link to comment Share on other sites More sharing options...
Tim_Barrett Posted July 25, 2011 Share Posted July 25, 2011 After a certain point Kearth stops pulling. Link to comment Share on other sites More sharing options...
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