What is the jerk due to gravity with ascent? Acceleration is defined as the rate of change of velocity with time. Jerk is defined as the rate of change of acceleration with time. What is the jerk due to gravity with ascent?
 A: The answer depends on the ascent rate $\dot h$. Differentiating gravitational acceleration with respect to time yields
$$\frac {d\,g(h(t))}{dt} = \frac {d}{dt}\left(\frac {\mu_E}{(R_E+h(t))^2}\right) = -\frac {2\mu_E}{(R_E+h)^3}\dot h = -\frac{2g(h)}{R_E+h} \dot h$$
where $\mu_E$ is the Earth's standard gravitational parameter, $\mu_E = GM_E$. It's better to use the gravitational parameter than the product $GM_E$ because of the large uncertainties in $G$ and $M_E$.
For points close to the surface of the Earth, $2g(h)/(R_E+h) \approx 2g/R_E \approx 3.086\times 10^{-6} \text{s}^{-2}$. This is the free air correction to Earth gravitation.
A: If something is in freefall, starting at $v = 0$ at height $h$, then
$$\begin{align}
a &= - \frac{GM}{r^{2}}\\
v\,{\dot v} &= - \frac{GM v}{r^2}\\
v\,{\dot v} &= - \frac{GM {\dot r}}{r^2}\\
\frac{1}{2}v^{2} &= \frac{GM}{r} - \frac{GM}{h}\\
v &= \sqrt{2GM\left(\frac{1}{r} - \frac{1}{h}\right)}
\end{align}$$
Then,
$$\begin{align}
J &= \frac{da}{dt}\\
&= \frac{2GM}{r^{3}}{\dot r}\\
&= \frac{2GM}{r^{3}}\sqrt{2GM\left(\frac{1}{r} - \frac{1}{h}\right)}
\end{align}$$
but this doesn't really have any physical significance.
