Martin Posted March 14, 2005 Share Posted March 14, 2005 the fact is that if you take the mass of the sun in kilograms and multiply by the metric value of the newtonian const. G then you get GM which is a quantity in metercube/secondsquare which you can re-write as a quantity in the more convenient units km times (km/s)^2 and that is actually ORBIT RADIUS TIMES SQUARE OF ORBIT SPEED for any smallmass satellite in circular orbit at any radius (this is standard freshman physics I am just saying it very roughly) and so, like GM is constant for the whole solar system, and we have all these relatively small things in approximately circular orbits so we can apply this and what this says is IF YOU DOUBLE THE RADIUS THEN YOU HAVE TO CUT THE SQUARE OF THE ORBIT SPEED BY A FACTOR OF TWO, so the speed goes down by sqrt(2) and if you triple the radius of the orbit then you have to cut the square of the speed by a factor three, and the speed goes down by sqrt(3). and you can PROVE this by v^2/R ah, finally, I said what I wanted. the nice thing about v^2/R is that it is incredibly versatile, you get from it fact after fact after fact, and this thing about jupiter speed is just one of those nice facts Link to comment Share on other sites More sharing options...
Johnny5 Posted March 14, 2005 Author Share Posted March 14, 2005 the fact is that if you take the mass of the sun in kilograms andmultiply by the metric value of the newtonian const. G then you get GM which is a quantity in metercube/secondsquare which you can re-write as a quantity in the more convenient units km times (km/s)^2 and that is actually ORBIT RADIUS TIMES SQUARE OF ORBIT SPEED for any smallmass satellite in circular orbit at any radius Ignoring the rest of the universe' date=' for any two body system in which M>>m GM = (orbital radius)(orbital speed)^2 Interesting. That is not something I have memorized you know. I am mostly concerned about frame switching, and making sure I am performing the analysis in the right frame. For circular motion: [math'] V = R \omega [/math] In which case: [math] GM = R^3 \omega ^2 [/math] OK SO... GM is a constant for our whole solar system, and everything else has a small mass in comparison to M. So... Suppose you DOUBLE THE ORBITAL RADIUS. You have to keep GM constant, therefore, the orbital speed has to decrease to compensate. Mathematically, if you divide the orbital speed by root 2, you keep GM constant. Yep. Link to comment Share on other sites More sharing options...
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