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). :v(t)=v_{0}-gt\, and :y(t)=v_{0}t+y_{0}-\frac{1}{2}gt^2 , where :v_{0}\, is the initial vertical component of the velocity (m/s). :v(t)\, is the vertical component of the velocity at t\,(m/s). :y_{0}\, is the initial altitude (m). :y(t)\, is the altitude at t\,(m). :t\, is time elapsed (s). :g\, is the acceleration due to
gravity (9.81 m/s2 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: v(t)=-gt and y_{0}-y(t)=\frac{1}{2}gt^2. Moreover, because
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.
Uniform gravitational field with air resistance ) at different initial velocities of 35, 25 and 15 km/s. Air pressure and air density are height-dependent. This case, which applies to skydivers, parachutists or any body of mass, m, and cross-sectional area, A, 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, v, has an equation of vertical motion in Newton's regime :m\frac{\mathrm{d}v}{\mathrm{d}t}=mg - \frac{1}{2} \rho C_{\mathrm{D}} A v^2 \, , where \rho is the
air density and C_{\mathrm{D}} 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? 1/ \rho=RT/p), the solution is: : v(t) = v_{\infty}\tanh\left(\frac{gt}{v_{\infty}}\right), where the
terminal speed is given by :v_{\infty}=\sqrt{\frac{2mg}{\rho C_D A}} \, . The object's speed versus time can be integrated over time to find the vertical position as a function of time: :y = y_0 + \frac{v_{\infty}^2}{g} \ln \cosh\left(\frac{gt}{v_\infty}\right). Using the figure of 56 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 y_0 \ll R, g/g_o=1-2y_0/R. 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 g0).
HALO jumps, including
Joe Kittinger's and
Felix Baumgartner's record jumps, also belong in this category.
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
eccentricity (
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: :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), where :t is the time after the start of the fall :y is the distance between the centers of the bodies :y_0 is the initial value of y :\mu = G(m_1 + m_2) is the
standard gravitational parameter. Substituting y = 0 we get the
free-fall time :t_{\text{ff}}=\pi\sqrt{y_0^3/(8\mu)} and t/t_{\text{ff}}=2/\pi\left(\sqrt{y_r\left(1-y_r\right)}+\arccos{\sqrt{y_r}}\right). The separation can be expressed explicitly as a function of time :y(t)=y_0~Q\left(1-\frac{t}{t_{\text{ff}}};\frac{3}{2},\frac{1}{2}\right) ~, where Q(x;\alpha,\beta) is the quantile function of the
Beta distribution, also known as the
inverse function of the
regularized incomplete beta function I_x(\alpha,\beta). This solution can also be represented exactly by the analytic power series :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) =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) +{\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 Evaluating this yields: :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 =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] =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], where 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}. ==In general relativity==