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Say a particle in one dimension experiences acceleration inversely proportional to the square of displacement. What is its displacement as a function of time?

$$ a=\frac{d^2x}{dt^2}=\frac{k}{x^2} \\ \text{a second-order nonlinear differential equation, apparently} $$

For context, the particle could be gravitationally attracted to a fixed mass $m$, in which $k$ would be $Gm$, where $G$ is the gravitational constant.

I can’t seem to find a solution to this apparently simple problem. Is $x(t)$ very complex? What about $v(t)$, or $a(t)$? Which of these can be expressed simply in terms of $t$?

P.S. It seems that the particle’s acceleration would be infinite when $x=0$, but only for an infinitesimal instant. Does that mean the velocity spikes to infinity, or does it reach a maximum?

Jollywatt
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1 Answers1

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First let $a = \frac{dv}{dt}$, thus we have,

$a = \frac{dv}{dt} = \frac{k}{x^2}$

Now using the chain rule, $\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$

Therefore,

$v\frac{dv}{dx} = \frac{k}{x^2}$

Re-arranging and integrating gives,

$\int v\ dv = \int \frac{k}{x^2} dx$

and therefore,

$\frac{v^2}{2} = -\frac{k}{x} + C$

Or,

$v^2 = \frac{-2k}{x} + C$

for some constant $C$ dependent on initial conditions.

I'll leave as a simple exercise for the reader to now set $v = \frac{dx}{dt}$ and solve for $x(t)$

Kenshin
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