Gravitational Collapse

Classical Mechanics Level 5

Four identical co-planar point masses are arranged in a symmetrical cross formation around the origin, as shown in the diagram. They are initially at rest, with the masses on the vertical axis being twice as far from the origin as those on the horizontal axis. The masses are free to move.

The system collapses under mutual gravitational attraction $($ there is no ambient $g$ -field $).$ Let $x$ be the distance of the masses on the horizontal axis from the origin, and let $y$ be the distance of the masses on the vertical axis from the origin.

When $x$ is one tenth of its original value, what is $\large{\frac{y}{x}}$ (to two decimal places)?

The answer is 12.26.

2 solutions

Mark Hennings
Jun 21, 2017

We are asked to solve the paired differential equations $\begin{aligned} m\ddot{x} & = -\frac{Gm^2}{4x^2} - 2\frac{Gm^2}{x^2+y^2} \times \frac{x}{\sqrt{x^2+y^2}} \\ m\ddot{y} & = -\frac{Gm^2}{4y^2} - 2\frac{Gm^2}{x^2+y^2} \times \frac{y}{\sqrt{x^2+y^2}} \end{aligned}$ with $x(0) = 1$ , $y(0) = 2$ , $x'(0) = y'(0) = 0$ . Rescaling the time variable, these equations become $\begin{aligned} \ddot{x} & = -\left(\frac{1}{4x^2} + \frac{2x}{(x^2+y^2)^{\frac32}}\right) \\ \ddot{y} & = -\left(\frac{1}{4y^2} + \frac{2y}{(x^2+y^2)^{\frac32}}\right) \end{aligned}$ with the initial conditions $x(0) = 1$ , $y(0) = 2$ , $x'(0) =y'(0) = 0$ .

Solving these equations numerically, we see that $x(t) = 0$ when $t \approx 1.781$ (after which the above model is no longer valid), but that $x(t) = 0.1$ when $t = 1.75112$ , at which point $y = 1.22567$ . The ratio $\tfrac{y}{x}$ at this time is therefore $\boxed{12.257}$ .

Sir, I have a question, how can you solve a equation numerically, do you use any tools to do this?

Kelvin Hong - 3 years, 10 months ago

There are lots of techniques. The simplest approach to solving the differential equation $y'(x) \; = \; f(x,y) \hspace{2cm} y(0) = a$ is to introduce a small number $h$ and consider the recurrence relation $y_0 \; = \; a \hspace{2cm} y_n \; = \; y_{n-1} + hf\big(x_{n-1},y_{n-1}\big)$ Then $y_n$ is an approximation to $y(nh)$ . Making $h$ small should make for a better approximation, but there are disadvantages, since there are more calculations that have to be performed, and every calculation introduces rounding errors.

There are ways of analysing the error bounds for this technique, and there are more sophisticated algorithms than this which are inherently more accurate. This whole business can be expanded to handle higher order differential equations of more than one variable, such as we have here.

I am not particularly into the details of numerical solutions of problems, so I tend to let Mathematica do the detailed numerical work for me.

Mark Hennings - 3 years, 10 months ago

I get it, thanks for patiently explain!

Kelvin Hong - 3 years, 10 months ago

Yes, in fact, that is how I did it. I used Wolfram Mathematica 11.3 I discovered thhe singularity at t=178.11410810514622 also.

A Former Brilliant Member - 2 years, 2 months ago

A Former Brilliant Member
Mar 20, 2019

Here is a numerical solution done in Wolfram Mathematica 11.3 with a bit of meta programming to assure that I did not make mistakes in coding.

Gravitational Collapse

The answer is 12.26.

2 solutions

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