Editing Free fall (section)

=== Inverse-square law gravitational field ===
It can be said that two objects in space orbiting each other in the absence of other forces are in free fall around each other, e.g. that the Moon or an artificial satellite "falls around" the Earth, or a planet "falls around" the Sun. Assuming spherical objects means that the equation of motion is governed by [[Newton's law of universal gravitation]], with solutions to the [[gravitational two-body problem]] being [[elliptic orbits]] obeying [[Kepler's laws of planetary motion]]. This connection between falling objects close to the Earth and orbiting objects is best illustrated by the thought experiment, [[Newton's cannonball]].

The motion of two objects moving radially towards each other with no [[angular momentum]] can be considered a special case of an elliptical orbit of [[Orbital eccentricity|eccentricity]] {{nowrap|''e'' {{=}} 1}} ([[radial elliptic trajectory]]). This allows one to compute the [[free-fall time]] for two point objects on a radial path. The solution of this equation of motion yields time as a function of separation:
:<math>t(y)=\sqrt{\frac{{y_0}^3} {2\mu}} \left(\sqrt{\frac{y}{y_0}\left(1-\frac{y}{y_0}\right)}+\arccos{\sqrt{\frac{y}{y_0}}}\right),</math>

where
:<math>t</math> is the time after the start of the fall
:<math>y</math> is the distance between the centers of the bodies
:<math>y_0</math> is the initial value of <math>y</math>
:<math>\mu = G(m_1 + m_2)</math> is the [[standard gravitational parameter]].

Substituting <math> y = 0</math> we get the [[free-fall time]]

:<math>t_{\text{ff}}=\pi\sqrt{y_0^3/(8\mu)}</math> and <math>t/t_{\text{ff}}=2/\pi\left(\sqrt{y_r\left(1-y_r\right)}+\arccos{\sqrt{y_r}}\right). </math>

The separation can be expressed explicitly as a function of time <ref>{{cite journal |last1=Obreschkow |first1=Danail |title=From Cavitation to Astrophysics: Explicit Solution of the Spherical Collapse Equation |journal=Phys. Rev. E |date=7 June 2024 |volume=109 |issue=6 |page=065102 |doi=10.1103/PhysRevE.109.065102 |pmid=39021019 |url=https://link.aps.org/doi/10.1103/PhysRevE.109.065102|arxiv=2401.05445 |bibcode=2024PhRvE.109f5102O }}</ref>

:<math>y(t)=y_0~Q\left(1-\frac{t}{t_{\text{ff}}};\frac{3}{2},\frac{1}{2}\right) ~,</math>

where <math>Q(x;\alpha,\beta)</math> is the quantile function of the [[Beta distribution]], also known as the [[inverse function]] of the [[regularized incomplete beta function]] <math>I_x(\alpha,\beta)</math>.

This solution can also be represented exactly by the analytic power series

:<math>y(t)=\sum_{n=1}^{\infty}{\frac{x^{n}}{n!}}\cdot
\lim_{r\to0}\left(\frac{\mathrm{d}^{\,n-1}}{\mathrm{d}r^{\,n-1}}\left[r^n\left(\frac{7}{2}\bigl(\arcsin(\sqrt{r}) -\sqrt{r-r^2}\bigr)\right)^{-\frac{2}{3}n}\right]\right)</math>

<math>=x/\lim_{r\to0}[(\frac{7}{2}\bigl(\arcsin(\sqrt{r})-\sqrt{r-r^2}\bigr))^{\frac{2}{3}}]' +{\frac{x^2}{2!}}\lim_{r\to0} \left(\frac{\mathrm{d}^{1}}{\mathrm{d}r^{1}} \left[
r^2\left(\frac{7}{2}\bigl(\arcsin(\sqrt{r}) -\sqrt{r-r^2}\bigr)
\right)^{-\frac{4}{3}}\right]\right)</math>

<math> +{\frac{x^3}{3!}}\lim_{r\to0} \left(\frac{\mathrm{d}^{2}}{\mathrm{d}r^{2}} \left[
r^3\left(\frac{7}{2}\bigl(\arcsin(\sqrt{r}) -\sqrt{r-r^2}\bigr)
\right)^{-2}\right]\right)+\cdots
</math><ref><math>x^1/1! \cdot\lim_{r\to0} [Num(r)/Den(r)] =x\cdot [0/(7/2\cdot(0-0))^{2/3}]</math> = <math>x\lim_{r\to0}[\operatorname{d}\!{Num}/\operatorname{d}\!r:\operatorname{d}\!{Den}/ \operatorname{d}\!r] =x\lim_{r\to0}[\operatorname{d}\!{r^1}/\operatorname{d}\!r:{Den}'] </math> </ref>

Evaluating this yields:<ref>{{cite journal|doi=10.1088/0143-0807/29/5/012|title=From Moon-fall to motions under inverse square laws|journal=European Journal of Physics|volume=29|issue=5|pages=987–1003|year=2008|last1=Foong|first1=S K|bibcode=2008EJPh...29..987F|s2cid=122494969 |doi-access=free}}</ref><ref>{{cite journal|doi=10.1119/1.3246467|url=https://apps.dtic.mil/sti/pdfs/ADA534896.pdf|title=Radial Motion of Two Mutually Attracting Particles|journal=The Physics Teacher|volume=47|issue=8|pages=502–507|year=2009|last1=Mungan|first1=Carl E.|bibcode=2009PhTea..47..502M}}</ref>
:<math>y(t)/y_0=x-\frac{1}{5}x^2-\frac{3}{175}x^3-\frac{23}{7875}x^4-\frac{1894}{3,031875} x^5-\frac{3293}{21,896875}x^6-\frac{2,418092}{62,077,640625}x^7-\cdots
</math>
<math> =x-\frac{1}{5}x[x+(\frac{3}{7}x^2+\frac{23}{315}x^3+\frac{1894}{121,275}x^4+\frac{3293}{875,875}x^5+\frac{2,418092}{2,483,105625}x^6+\cdots)/5]
</math>

<math>=x-\frac{1}{5}x[x+(\frac{3}{7}+(\frac{23}{63}+ \frac{1894}{24,255}x+\frac{3293}{175,175}x^2+\frac{2,418092}{480,621125}x^3)x/5+\cdots)x^2/5],</math>

where

<math>x=\left[\frac{3}{2}\left(\frac{\pi}{2}-t\sqrt{\frac{2\mu}{{y_0}^3}}\right)\right]^{2/3}=[3\pi/4\cdot(1-t/t_{\text{ff}})]^{2/3}.</math><ref>At t=0 <math display="inline">x=(3/4\cdot\pi)^{2/3}</math> and y=y<sub>0</sub>, at <math display="inline">t=t_{ff}</math> x=0 and y=0.</ref>