Editing Free fall (section)

==Free fall in Newtonian mechanics==
{{Main|Newtonian mechanics}}

=== Uniform gravitational field without air resistance ===
This is the "textbook" case of the vertical motion of an object falling a small distance close to the surface of a planet. It is a good approximation in air as long as the force of gravity on the object is much greater than the force of air resistance, or equivalently the object's velocity is always much less than the terminal velocity (see below).{{Citation needed|date=April 2025}}

[[File:Free-fall.gif|right|100px|Free-fall|alt=Free-fall alt.//vel.?|border]]
:<math>v(t)=v_{0}-gt\,</math> and

:<math>y(t)=v_{0}t+y_{0}-\frac{1}{2}gt^2 ,</math>

where
:<math>v_{0}\,</math> is the initial vertical component of the velocity (m/s).
:<math>v(t)\,</math> is the vertical component of the velocity at <math>t\,</math>(m/s).
:<math>y_{0}\,</math> is the initial altitude (m).
:<math>y(t)\,</math> is the altitude at <math>t\,</math>(m).
:<math>t\,</math> is time elapsed (s).
:<math>g\,</math> is the acceleration due to [[gravity]] (9.81 m/s<sup>2</sup> near the surface of the earth).
If the initial velocity is zero, then the distance fallen from the initial position will grow as the square of the elapsed time:

<math>v(t)=-gt</math> and <math>y_{0}-y(t)=\frac{1}{2}gt^2.</math>

Moreover, because [[square number#Properties|the odd numbers sum to the perfect squares]], the distance fallen in successive time intervals grows as the odd numbers. This description of the behavior of falling bodies was given by Galileo.<!-- posthumous publ. --><ref>{{Cite book |last1=Olenick |first1=R.P. |url=https://books.google.com/books?id=xMWwTpn53KsC&pg=PA18 |title=The Mechanical Universe: Introduction to Mechanics and Heat |last2=Apostol |first2=T.M. |last3=Goodstein |first3=D.L. |publisher=Cambridge University Press |year=2008 |isbn=978-0-521-71592-8 |page=18 |language=en}}</ref>

=== Uniform gravitational field with air resistance ===
[[File:MeteorAccGraph.jpg|thumb|Acceleration of a small meteoroid when entering the Earth's atmosphere 80 km high (above which the [[Kennelly–Heaviside layer]]) at different initial velocities of 35, 25 and 15 km/s. Air pressure and air density are height-dependent.]]

This case, which applies to 1. skydivers, parachutists or any body of mass, <math>m</math>, and cross-sectional area, <math>A</math>, 2. with [[Reynolds number]] Re well above the critical Reynolds number, so that the air resistance is proportional to the square of the fall velocity, <math>v</math>, 

has an equation of vertical motion in Newton's regime
:<math>m\frac{\mathrm{d}v}{\mathrm{d}t}=mg - \frac{1}{2} \rho C_{\mathrm{D}} A v^2 \, ,</math>
where <math>\rho</math> is the [[Density of air|air density]] and <math>C_{\mathrm{D}}</math> is the [[drag coefficient]], assumed to be constant (Re > 1000) although in general it will depend on the Reynolds number.

Assuming an object falling from rest and no change in air density with altitude (ideal gas? <math display="inline">1/ \rho=RT/p</math>), the solution is:
: <math>v(t) = v_{\infty}\tanh\left(\frac{gt}{v_{\infty}}\right),</math>

where the [[terminal speed]] is given by
:<math>v_{\infty}=\sqrt{\frac{2mg}{\rho C_D A}} \, .</math>

The object's speed versus time can be integrated over time to find the vertical position as a function of time:

:<math>y = y_0 - \frac{v_{\infty}^2}{g} \ln \cosh\left(\frac{gt}{v_\infty}\right).</math>

Using the figure of 56&nbsp;m/s for the terminal velocity of a human, one finds that after 10 seconds he will have fallen 348 metres and attained 94% of terminal velocity, and after 12 seconds he will have fallen 455 metres and will have attained 97% of terminal velocity. Gravity field is (vertical) position-dependent g(y): when <math display="inline">y_0 \ll R</math>, <math display="inline">g/g_o=1-2y_0/R</math>. Linear decrease with height, small height compared to Earth's radius R = 6379 km.

However, when the air density cannot be assumed to be constant, such as for objects falling from high altitude, the equation of motion becomes much more difficult to solve analytically and a numerical simulation of the motion is usually necessary. The figure shows the forces acting on small meteoroids falling through the Earth's upper atmosphere (an acceleration of 0.1 km/s² is 10 g<sub>0</sub>). [[High-altitude military parachuting#High Altitude Low Opening – HALO|HALO jump]]s, including [[Joe Kittinger]]'s and [[Felix Baumgartner]]'s record jumps, also belong in this category.<ref>An analysis of such jumps is given in {{cite journal|title=High altitude free fall|journal= American Journal of Physics |volume= 64 |issue= 10 |page= 1242 |author1=Mohazzabi, P. |author2=Shea, J. |doi=10.1119/1.18386|bibcode=1996AmJPh..64.1242M|url=http://www.jasoncantarella.com/downloads/AJP001242.pdf|year= 1996 }}</ref>

=== 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>