 # What is the jerk due to gravity with ascent? Acceleration mars6svhym 2022-05-13 Answered
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?
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The answer depends on the ascent rate $\stackrel{˙}{h}$. Differentiating gravitational acceleration with respect to time yields
$\frac{d\phantom{\rule{thinmathspace}{0ex}}g\left(h\left(t\right)\right)}{dt}=\frac{d}{dt}\left(\frac{{\mu }_{E}}{\left({R}_{E}+h\left(t\right){\right)}^{2}}\right)=-\frac{2{\mu }_{E}}{\left({R}_{E}+h{\right)}^{3}}\stackrel{˙}{h}=-\frac{2g\left(h\right)}{{R}_{E}+h}\stackrel{˙}{h}$
where ${\mu }_{E}$ is the Earth's standard gravitational parameter, ${\mu }_{E}=G{M}_{E}$. It's better to use the gravitational parameter than the product $G{M}_{E}$ because of the large uncertainties in $G$ and ${M}_{E}$
For points close to the surface of the Earth, $2g\left(h\right)/\left({R}_{E}+h\right)\approx 2g/{R}_{E}\approx 3.086×{10}^{-6}{\text{s}}^{-2}$. This is the free air correction to Earth gravitation.
###### Not exactly what you’re looking for? Spencer Lutz
If something is in freefall, starting at $v=0$ at height $h$, then
$\begin{array}{rl}a& =-\frac{GM}{{r}^{2}}\\ v\phantom{\rule{thinmathspace}{0ex}}\stackrel{˙}{v}& =-\frac{GMv}{{r}^{2}}\\ v\phantom{\rule{thinmathspace}{0ex}}\stackrel{˙}{v}& =-\frac{GM\stackrel{˙}{r}}{{r}^{2}}\\ \frac{1}{2}{v}^{2}& =\frac{GM}{r}-\frac{GM}{h}\\ v& =\sqrt{2GM\left(\frac{1}{r}-\frac{1}{h}\right)}\end{array}$
Then,
$\begin{array}{rl}J& =\frac{da}{dt}\\ & =\frac{2GM}{{r}^{3}}\stackrel{˙}{r}\\ & =\frac{2GM}{{r}^{3}}\sqrt{2GM\left(\frac{1}{r}-\frac{1}{h}\right)}\end{array}$
but this doesn't really have any physical significance.