Free fall
Newtonian mechanics
Constant acceleration
The simplest form of free-fall is the one where the gravitational acceleration is assumed to be constant, that is,
\begin{equation} \vec{F} = - m g \vec{e}_z \end{equation}This gives us, with Newton's second law, the following equation
\begin{equation} \ddot{\vec{x}}(t) = - g \vec{e}_z \end{equation}Its solution with respect to velocity is
\begin{equation} \dot{\vec{x}}(t) = - g \vec{e}_z t + \vec{v}_0 \end{equation}And, with respect to position,
\begin{equation} \vec{x} = - \frac{g}{2} \vec{e}_z t^2 + \vec{v}_0 t + \vec{x}_0 \end{equation}The trajectory is therefore a parabola. It can be useful, in the modelisation of an object with respect to the ground, to know when it will touch the ground, how long it will remain in the air, and so forth. If we assume that the ground is at $z = 0$, then what we want to know are the roots of
\begin{equation} - \frac{g}{2} t^2 + v_{z,0} t + z_0 = 0 \end{equation}With $\Delta = v_{z,0}^2 + 2 g z_0$, the possible roots are
\begin{equation} t^{\pm} = \frac{v_{z,0} \pm \sqrt{v_{z,0}^2 + 2 g z_0}}{g} \end{equation}We can check that this makes sense : if our object has initial position $z_0 = 0$, then one of the root is indeed $0$, and if $v_{z,0}$ is positive, the object will fall at a future time. If the object is initially at rest, $v_{z,0} = 0$, and the object starts at $z_0 = 0$, both roots are simply $0$, as it is simply a static object. And the stronger $g$ is, the shorter the time interval between the two roots.
Another thing we may want is to know how far an object thrown up will rise. This is the maxima of $z(t)$, which in our case will be the maxima of
\begin{equation} z(t) = - \frac{g}{2} t^2 + v_{z,0} t + z_0 \end{equation} The first derivative being \begin{equation} \dot{z}(t) = - g t + v_{z,0} \end{equation}with critical point $t = v_{z,0} / g$. To check that this is a maximum, we look at its second derivative :
\begin{equation} \ddot{z}(t) = - g \end{equation}It is therefore always a maximum, short of having gravity in the wrong direction. The highest point will therefore be
\begin{equation} z_{\text{max}} = - \frac{g}{2} v^2_{z,0} + \frac{v^2_{z,0}}{g} + z_0 \end{equation}Gravitational field
If we consider the actual gravitational field of earth, where the earth occupies the center of the coordinate system,
\begin{equation} \vec{F} = - G M_\odot m\frac{\vec{x}(t)}{\| \vec{x}(t) \|^3} \end{equation}This corresponds quite simply in spherical coordinates to
\begin{equation} \vec{F} = - G \frac{M_\odot m}{r^2} \vec{e}_r \end{equation}