Free fall

Free fall is motion under gravity alone: there is no supporting normal force, engine thrust or appreciable drag force. In the Newtonian mechanics model, a test particle of mass $m$ obeys $m\ddot{\mathbf{r}}=m\mathbf{g}(\mathbf{r})$, hence $\ddot{\mathbf{r}}=\mathbf{g}(\mathbf{r})$. The mass cancels. Bodies released at the same position with the same velocity follow the same path in this model, even if their gravitational weights differ. ^[ref]

A dropped stone is one example; a coasting satellite is another. Free fall does not require motion straight down or decreasing distance to the attracting body. An ideal circular orbit has constant radius while its velocity continually turns under centripetal acceleration. The circular orbit as continuous free fall connects the familiar drop with orbital motion.

Near a surface, treating the gravitational acceleration as constant gives a useful first prediction. Over larger heights, the inverse-square law changes the acceleration during the fall. Radial free fall from rest uses conservation of mechanical energy to account for that change exactly, within the specified Newtonian gravitational field. The contrast matters: the simplest formula is an approximation to a physical model, rather than a law with the same numerical $g$ everywhere.

History also makes the idealization tangible. Galileo's Two New Sciences (1638), Third Day, Theorem II, derives distance proportional to the square of time for uniformly accelerated motion from rest; its first corollary gives successive distances in the ratios $1,3,5,7,\dots$ over equal time intervals. This is a kinematic statement, predating the gravitational inverse-square law. ^[ref]

On Apollo 15's final lunar surface excursion, David Scott released a hammer and a feather together. The near-vacuum removed the large difference in drag force familiar on Earth. NASA's report describes their arrival together within the accuracy of the release; this demonstration illustrates the free fall prediction without establishing infinite experimental precision. ^[ref]

The gravitational collapse scale $t_{\mathrm{ff}}\sim \sqrt{R^3/(GM)}$. Radial fall from rest at $R$ to a fixed central point mass has the exact time $\pi \sqrt{R^3/(GM)}/(2\sqrt{2})$.

Take downward displacement $s$ as positive, assume a constant gravitational acceleration $g>0$, and release a point mass from rest at $s=0$. Integrating $\ddot{s}=g$ twice givesFor a downward distance $h>0$, the arrival time and speed areFor the model values $h=20\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, these are approximately $2.0193\mathrm{s}$ and $19.8091\mathrm{m}{\mathrm{s}}^{-1}$. Doubling the height multiplies both quantities by $\sqrt{2}$; it does not double them. The same speed follows from the work-energy theorem, $mgh=m{v}^2/2$.

For intervals of length $\Delta t$, the successive downward displacements areThus equal spacing in time produces increasingly separated positions. This is the odd-number pattern in Galileo's result. It assumes constant acceleration and release from rest; a nonzero initial velocity adds a linear term to $s(t)$.

The approximation can be estimated for a spherical source of radius $R$ and standard gravitational parameter $\mu$. Writing $g_0=\mu /R^2$, the gravitational magnitude at height $h$ isFor $h/R\ll 1$, a Taylor expansion gives $g(R+h)/g_0=1-2h/R+O((h/R{)}^2)$. Small height relative to the source's radius makes nearly constant gravity plausible; negligible drag force remains a separate requirement.

Let a test particle start at rest at radius $r_0$ outside a fixed spherically symmetric source. The spherical shell theorem gives the exterior Newtonian gravitational potential $\Phi (r)=-\mu /r$, where $\mu =GM$ is the source's standard gravitational parameter. Zero initial transverse velocity makes the motion radial. Conservation of mechanical energy per unit mass givesThe minus sign selects inward motion; the positive square root is its speed. For a fall from $r_0=2R$ to the surface $r=R$, that speed is $\sqrt{\mu /R}$. In the limit of release from rest infinitely far away, the familiar free-fall speed is $\sqrt{2\mu /r}$, equal in magnitude to the local escape velocity. A finite starting radius gives a smaller speed.

The arrival time follows from the same first integral. Set $r=r_0{\cos }^2\eta$. ThenFor arrival at $R>0$ this is finite, and the field formula is used only outside the source. For $r_0=2R$, $\eta =\pi /4$ givesKeeping the smaller initial gravitational magnitude $g(r_0)=\mu /(4R^2)$ constant would predict $\sqrt{8}\sqrt{R^3/\mu }$, about $10.0\%$ longer. The actual acceleration increases during the fall.

The curves compare a shallow fall, $h/R=0.02$, with a fall from $r_0=2R$, so $h/R=1$. Each comparison holds the release height and initial gravitational magnitude fixed. The constant-gravity model is close for the shallow fall and increasingly inaccurate for the larger one. This is also a practical use of nondimensionalization: the normalized trajectories depend on $h/R$, while $\sqrt{R^3/\mu }$ supplies the dimensional time scale.

In an Earth-fixed rotating reference frame, the horizontal displacement of a freely falling body receives contributions from Coriolis acceleration and centrifugal acceleration. A short-fall approximation substitutes the unperturbed downward velocity into the Coriolis term and the Earth's surface position into the centrifugal term. If gravitational acceleration excludes centrifugal acceleration, both contributions must be included explicitly.

A body released from rest relative to Earth at the equator has unperturbed downward velocity $v_U=-gt$. The Coriolis acceleration in local Earth coordinates gives eastward acceleration $a_E=-2{\Omega }_E{v}_U=2{\Omega }_{E}gt$. Integration gives $x_E={\Omega }_{E}g{t}^3/3$ and impact time $t_f=\sqrt{2h/g}$, proving the title formula. This is first order in Earth's rotation, neglecting air resistance and the height dependence of the effective local gravitational acceleration. For a short fall at latitude $\phi$, its leading eastward term is multiplied by $\cos \phi$.