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Because stars are >> much greater in mass than planets, you can ignore the planet's mass. If you don't want to ignore it because you like admiring digits far to the right of the decimal point, just add the mass of the planet to the star for the sake of computing. But this is like adding the mass of a mouse to the mouse of an elephant.
circular orbital velocity: v = sqrt (GM/r). For G use 6.67 x 10-11. For M, use kilograms. For r (radius of your orbit) use meters. Here's a calculator that will do it for you. http://orbitsimulator.com/cmc/VelocityCircular.html To solve for mass, just re-write the equation to solve for M. v2 = GM/r v2 r= GM M=(v2 r)/G |
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jlhredshift, I checked out your recomendation here on amazon.com. Most of the reviews were very positive, but one was negative and rather specific. In you humble opinion, is this negative comment accurate? What level of prior knowledge is necessary? I have been looking for a book like this for a while. Being a programmer for 23 years, I'm not at all concerned about understanding the programs, but would my 2nd year undergrad knowledge of physics be a problem?
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Yo Adrian. The Phillies won the Series. |
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His argument is that presenting formulae to calculate numerical values does nothing to teach the fundamental physics.
He'd be correct in his critique if the book was meant to teach astronomy or orbital mechanics, but as it specifically is a collection of numerical algorithms for getting values, he's attacking it for not being something it never intended to be. I have an earlier version of the book and similar knowledge, you have exactly the knowledge needed ![]()
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And the "driving on the freeway on a scooter" analogy still holds true because the pilots are sitting in 7 to 30 ton aircraft o' doom and you are running around them in your very own Meatbody, Mark I. Beep, beep. Big Don Trying to make sense of computers, The Error Log.
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This book doesn't cover everything for orbital calcuation and VSOP87 algorithms, each chapter is rather short than I have thought before purchasing it.
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http://www.yaohua2000.org/ |
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This PDF file has a nifty summary of how to calculate periapsis and apoapsis speed, once the circular velocity is known. It's in the format of a worksheet though:
http://astro.pas.rochester.edu/~aqui...ems/P9_sol.pdf
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"Call me old-fashioned, but I think fire is magic. And it scares me a lot." --The State |
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These might also be useful:
Orbital energy is constant E = 1/2*m(dR/dt)^2 + L^2/(2mR^2) - GmM/R where dR/dt is radial velocity in an elliptical orbit which is 0 at apoapsis, periapsis and in circular orbit L is angular momentum which is constant L = m*R(apoapsis)*v(apoapsis) = m*R(periapsis)*v(periapsis) For circular orbit you get E = 1/2*mv^2 - GmM/R where v is tangential velocity And some formulas with eccentricity e e = d/(2a) where d is the distance between the two focal points and a is the semimajor axis of the ellipse a = (R(apoapsis)+R(periapsis))/2 R(periapsis) = a(1-e) R(apoapsis) = a(1+e) E = -GmM/(2a) e^2 = 1 + 2E/m *(L/(GmM))^2 Orbital velocity v and period T for circular orbit v = (GM/R)^(1/2) T^2 = 4pi^2/(GM)*R^3 T = 2pi/((GM)^(1/2))*R^(3/2) and period for elliptical orbit T^2 = 4pi^2/(GM)*a^3 T = 2pi/((GM)^(1/2))*a^(3/2) |
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r = r_m - r_M Double differentiate by time to get r'' = r''_m - r''_M What are r''_m and r''_M? Let's consider Newton's Law of Universal Gravitation. mr''_m = -GMm/r² r/r => r''_m = -GMr/r³ Mr''_M = GMm/r² r/r => r''_M = Gmr/r³ Therefore r'' = -G(M+m)r/r³ Since we are dealing with a planet orbiting a star, M >> m, therefore, G(M+m) ~ GM = µ Hence, r'' + µr/r³ = 0, which is the universal equation of motion. Now to use this for useful purposes to answer your question. Dot each side by r'. r'.r'' + r'.µr/r³ = 0 First off, r' = v and second off, in general a'.a = aa'. Hence vv' + µrr'/r³ = 0 Now here comes the calculus. You might notice that vv' = d/dt(½v²) and µr'/r² = d/dt(-µ/r) (try it and see). Therefore, d/dt(½v² - µ/r) = 0. Solve and get E = ½v² - µ/r = specific mechanical energy So we have an equation relating the speed and the distance of orbit through the energy. How do we work out the energy? <<Continues to turn pages and realises how much there is to do>> Well just take it that E = -µ/2a, where a is the semi-major axis of the orbit. (if you want to know how to get to that, ask and I'll derive it in another post) So, for a circular orbit, where a = r at all points, -µ/2r = ½v² - µ/r => µ/2r = ½v² => v = sqrt(µ/r) So to answer, your question, the planetary mass is unimportant as long as it is much less than the stellar mass. The stellar mass is part of the gravitional parameter, µ = GM. The orbital speed and the radius is then linked through the above equation.
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2 good 2 need 4 engines |
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I also gained an understanding of how discrete numerical simulations can be useful yet break down under extreme cases. Next step was adaptive sampling intervals and huristics to compensate for errors (e.g. ensure that momentum is concerved). --John --John |