A Is There a Solution for a Particle Falling into a Gravitational Well?

[kairama15]

Hello, I am trying to find a function x(t) that describes a particle falling into a gravitational well from a certain distance. So, for example, I am trying to figure out the differential equation: F = m*(dx^2/dt^2) = -GMm/x^2.

Or simply, dx^2/dt^2=-GM/x^2

or even more simply,
x''(t)=-k/x^2 where k is a constant.

Does anyone know how to solve this or if it is even possible? I cannot find any resources online for this problem, but I feel like such a simple differential equation would have been attempted or successfully solved by now...

Or is there an easier way to solve this problem - finding x(t) in the gravitational well.

Thanks for any help!

 

[mfb]

Wikipedia has formulas and also some description how to get them.
Unfortunately x(t) doesn't have a nice expression. t(x) works better but it is still messy.

 

[jfizzix]

Hello, I am trying to find a function x(t) that describes a particle falling into a gravitational well from a certain distance. So, for example, I am trying to figure out the differential equation: F = m*(dx^2/dt^2) = -GMm/x^2.

Or simply, dx^2/dt^2=-GM/x^2

or even more simply,
x''(t)=-k/x^2 where k is a constant.

Does anyone know how to solve this or if it is even possible? I cannot find any resources online for this problem, but I feel like such a simple differential equation would have been attempted or successfully solved by now...

Or is there an easier way to solve this problem - finding x(t) in the gravitational well.

Thanks for any help!

If I recall correctly, you can use the same techniques that are used to solve the nonlinear pendulum equation. In particular, you can start with the differential equation:

\frac{d^{2}x}{dt^{2}}=-\frac{GM}{x^{2}}

Multiply by \frac{dx}{dt} on both sides:

\frac{d^{2}x}{dt^{2}}\frac{dx}{dt}=-\frac{GM}{x^{2}}\frac{dx}{dt}

and show that these are actually total time derivatives of a more complicated function:

\frac{d^{2}x}{dt^{2}}\frac{dx}{dt} = \frac{d}{dt}\Big[\frac{1}{2}\Big(\frac{dx}{dt}\Big)^{2}\Big]=\frac{d}{dt}\Big[\frac{GM}{x}\Big]

Then, you can integrate the equation and solve it (easier said than done). An approximate solution starting with a stationary particle, and letting it fall, would give a radial coordinate r(t) that goes approximately as:

r(t)\approx r_{0}\Big(1-\Big(\frac{9GM}{r_{0}^{3}}\Big)t^{2}\Big)^{1/3}

The approximation is valid for times small compared to the time it takes to fall to the center of the earth, but should still be better than the parabolic curves taught in elementary kinematics.

 

[hilbert2]

Actually, there doesn't seem to exist a simple solution for any $x''(t)\propto x^{-a}$ with $a>0$.

I remember this was an exercise in some 2nd year classical mechanics course I attended many years ago, probably as a demonstration of how limited our set of named special functions is in describing even some simple physical problems like this.