I was trying to derive the position function for a point mass being pulled by a gravitational force from a much more massive object. Without taking into account any form of special relativity or QM, the classical Newtonian equation for gravity has acceleration inversely proportional to position, which when rearranged gives a second order Non-linear diff EQ. I tried to solve it to no avail, and Wolfram gave me some non elementary functions as an answer. So is there a position function for a object being influenced by gravity that consists of elementary functions? (Just a quick note, this all is based on acceleration due to gravity not being constant, which is provided by the Newtonian eq but is commonly simplified to a constant in certain circumstances, i.e. objects on Earth.)

$$F=ma$$ $$F=\frac {Gm_1m_2}{r^2}$$ $$a=\frac {Gm_2}{r^2}$$ $$ar^2=Gm_2$$

  • $\begingroup$ Why not show us the equation? $\endgroup$ – Sean Roberson Oct 27 '18 at 16:44
  • $\begingroup$ Sorry, I asked on my phone and formatting on here is a nightmare. I'll add it $\endgroup$ – Corsair64 Oct 27 '18 at 16:48
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    $\begingroup$ You are trying to (re)derive Kepler's laws, I guess. This is probably already in Laplace's treatise on celestial mechanics. $\endgroup$ – Hans Engler Oct 27 '18 at 16:50
  • $\begingroup$ en.wikipedia.org/wiki/Kepler_orbit $\endgroup$ – Andrei Oct 27 '18 at 17:42
  • $\begingroup$ @HansEngler: Already in Newton's Principia... $\endgroup$ – Hans Lundmark Oct 27 '18 at 18:13

Laplace, in Mécanique céleste vol. 1 §380iv, shows that solutions are conic sections, which are certainly elementary functions.

What it seems you're interested in is the general solution to the system of 2nd order ODEs of the form

$$x″ = F_x,\qquad y″ = F_y,\qquad z″ = F_z,\qquad \text{where}\qquad F = F(r),\qquad r = (x^2 + y^2 + z^2)^{1/2},$$

EqWorld eq. 4.2.8.

  • $\begingroup$ Thank you for the answer, although I am unsure as to the addition of quotation marks around elementary functions. To my knowledge at least, I used the term correctly. Did I not? $\endgroup$ – Corsair64 Nov 9 '18 at 18:08
  • $\begingroup$ @Corsair64 Yes, I'm quoting you. $\endgroup$ – Geremia Nov 9 '18 at 18:51

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