# What is the jerk due to gravity with ascent? [closed]

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?

• Well yes, it depends on $R^2$ as Newton's law $g(h) = \frac{GM_E}{(R_E + h)^2}$ where $M_E$ and $R_E$ are Earth's mass and radius. Is that what you mean? – rmhleo Oct 29 '14 at 10:24
• @rmhleo: that is the acceleration due to gravity at the height 'h', my question basically asks what the rate of change of gravity is? – Quantum Sphinx Oct 29 '14 at 11:10
• You don't define "rate of change of acceleration". For a given force, solve $F = ma$, yielding $x(t)$. The third derivative w.r.t. time is the rate of change of acceleration. It is always defined. – ACuriousMind Oct 29 '14 at 11:30
• @QuantumSphinx so in your question, what is the ascent conditions. I think is not clear what you ask, since for an ascending body the jerk will depend on the conditions of movement, while the change in gravitational acceleration will only depend on the height. Since the height will have a dependance with time, I guess that is what you ask, but then that will be determined by the conditions of ascending. – rmhleo Oct 29 '14 at 12:25

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.

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.