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What function/ functions express radial motion of planet by means of non-linear ODE

$$ \ddot r - \frac{A}{r^3} +\frac{B}{r ^2}=0 $$

(The Kepler/Newton constants are: $\,B= a^3 \omega^2\, ; A=B p \,; $

There is no need to remind.. this differential equation is the, at any rate among the.. oldest of differential equations.

If solutions are elliptic first or second type or both, is the form simplest with Jacobi inverse functions?

EDIT 1:

What is time period in terms of $A,B ?$

EDIT 2:

Also is the Kepler's Equation useful in this closed form derivation?

Kepler's Equation

Narasimham
  • 42,260
  • A closed-form solution requires special functions, elliptic integrals iirc.. – Wouter Mar 31 '16 at 15:00
  • What is $x?$ Are $c, C$ same? Can $f,g$ be evaluated if boundary conditions are given like $ \dot r =0 @ ,t=0?$ Apart from the trivial circle which ellipse can be neatly expressed? – Narasimham Mar 31 '16 at 18:51
  • That scary !? Say we define $ R(t) = \sqrt{-Ac+2cB r(t)+c^2r(t)^2} ,, r^{ \prime} = 0 , , r =r_{apogee} $ at start, I believe it does simplify and why not post it as answer? I tried it on Mathematica this part, but is messy but worth trying as never seen it in print. – Narasimham Mar 31 '16 at 20:36
  • About going to chat. I request it be kept current here. The matter I believe is interesting to many on MSE in view of its history and importance. – Narasimham Apr 01 '16 at 11:00
  • Please do. This is such an important and classical problem, with references available everywhere including Wikipedia.I shall also clean up all transient comments and keep clean this comments area. When people see elliptic functions they immediately back-off, falsely assuming they hit a road-block. Hope the moderator co-operates under these mutually agreed conditions to get a classical Winther equation! – Narasimham Apr 01 '16 at 11:41
  • Is related to the above Kepler's equation? If not an elementary function, we can now define a new function rather than write them all out for numerical work. right? – Narasimham Apr 01 '16 at 17:42
  • I am not able to figure out why $A = Bp$. It is clear to me that $B/r^2 = GM/r^2$ and, by Kepler's third law $GM = a^3\omega^2$ (this explains the expression for $B$). However $A/r^3 = L_z/m^2r^3$ where $L_z$ is the angular momentum orthogonal to plane of the orbit, but I can't see pq $Lz/m=Bp$. Here, what does the variable $p$ represent? – Davius Jun 22 '22 at 23:23
  • Semi-latus rectum – Narasimham Jun 22 '22 at 23:27

0 Answers0