# Solving Universal law of Gravitation as differential equation

1. Feb 20, 2007

### Treborman

Hi, I'm trying to solve Newton Universal law of Gravitation as a differential equation in one dimension with two objects. One mass is fixed at distance r = 0, whilst the other is some initial distance away. I'm using distance rather than displacement and also ignored the proportionality factor to make the formula simpler. So in essence, this is the differential equation I want to solve:

$$a = \frac{d^2r}{dt^2} = -\frac{1}{r^2}$$

However, I've hit a snag when trying to integrate for the 2nd time (see the image file). I have used two substitutions (a and b), and the last line in the file suggests that I will obtain the inverse function that will contain hyperbolic and polynomial terms, i.e. I won't be able to determine the original function of r with respect to t.

Can anyone help?

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• ###### gravitation.JPG
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Last edited: Feb 20, 2007
2. Feb 21, 2007

### Treborman

I've just discovered the Lagrange inversion theorem - would anybody be able to give me resources for this and limits if possible (preferably free, i.e. the internet).

3. Feb 26, 2007

### gammamcc

Mult. both sides by dr/dt: Integrate.

4. Feb 26, 2007

### Treborman

I don't get it - do you mean integrate with respect to dr/dt? Would you be able to get me started - I was (hoping) to expand this investigation fairly quickly, not get stuck at the first hurdle

5. Feb 26, 2007

### gammamcc

wrt. both sides. Use subst.

6. Feb 26, 2007

### HallsofIvy

gammamcc, I think you are missing the point. The problem is not to integrate $d^2r/dt^2= 1/r^2$. He's alread done that. That results in
$$v= \frac{dr}{dt}= \sqrt{c_1- \frac{2}{r}}$$
The question is how to integrate that.

7. Feb 26, 2007

### gammamcc

ditto

Use separation of variables. (You wind up solving t explicitly, instead of r)

8. Feb 26, 2007

### HallsofIvy

Yes, of course. The whole question is about integrating
$$\int \frac{dr}{\sqrt{c_1- \frac{2}{r}}}dr$$

9. Feb 26, 2007

### gammamcc

Your attachment was not immediately available.......

This is all one expects typically. If you want more fun, put in some angular momentum.

Last edited: Feb 26, 2007
10. Feb 26, 2007

### Matthew Rodman

$$r^{\prime \prime} = - \frac{1}{r^2}$$

times by r' and integrate to get

$$\frac{1}{2}r^{\prime 2} + C = \frac{1}{r}$$

square it to get

$$\frac{1}{r^2} = (\frac{r^{\prime 2}}{2} + C)^2$$

but we can sub r'' for the $$1/r^2$$, so

$$-r^{\prime \prime} = (\frac{r^{\prime 2}}{2} + C)^2$$

Now make the sub

$$r^{\prime} = \sqrt{2C} \tan{v}$$

to give us

$$\sqrt{2C} v^{\prime} \sec^2{v} + C^2 \sec^4{v} = 0$$

or in other words

$$v^{\prime} + \sqrt{\frac{C^3}{2}} \sec^2{v} = 0$$

now multiply by cos^2 and integrate, and you get

$$v + \frac{1}{2} \sin{2v} = \kappa - \sqrt{2 C^3} t$$

(kappa is a constant). So, what you can do is create a function, $$F$$, such that it's inverse
is defined by

$$F^{-1}(p) = p + \frac{1}{2} \sin{2p}$$

and you can express v as

$$v(t) = F(\kappa - \sqrt{2C^3}t)$$

and then

$$r(t) = \sqrt{2C} \int{ \tan{F(\kappa - \sqrt{2C^3}t)} dt}$$

or more elegently written as

$$r(t) = (2 \varphi)^{\frac{1}{3}} \int{ \tan{F(\kappa + \varphi t)} dt}$$

where $$\varphi = -\sqrt{2 C^3}$$.

...as far as I can tell... although it's late and I'm rather tired right now...

Last edited: Feb 26, 2007
11. Feb 27, 2007

### gammamcc

not trivial

This does not seem to make an explicit (or even convenient) solution for r any easier. Know this is not a trivial problem cf. Goldstein's "Classical Mechanics" pp. 98-102 (2nd ed.).

12. Feb 27, 2007

### Matthew Rodman

I've got a feeling it won't be possible to get a completely explicit solution for r(t). It's more a question of getting the most convenient implicit solution. One thing: the function

$$F^{-1}(p) = p + \sin{p}$$

is, as far as I can tell, one-to-one: i.e., the gradient is never negative, and there shouldn't be any ambiguity when inverting it.

I think it makes a good candidate for a named function.

Last edited: Feb 27, 2007
13. Mar 2, 2007

### HallsofIvy

What it reduces to is an "elliptic" integral that cannt be done in terms of elementary functions. I remember seeing ("B.C."- "before computers) a 6-foot shelf of "Values of the elliptic integrals of the first and second kinds" in a university library.

14. Mar 2, 2007

### arildno

I will do tihs for you:
$$\ddot{r}=-\frac{K}{r^{2}}\to\frac{1}{2}\dot{r}^{2}=\frac{K}{r}+C$$
Or, not bothering about the sign right now, or the name on the constants, we have:
$$\dot{r}=\sqrt{\frac{2K+Cr}{r}}\to\int\sqrt{\frac{r}{2K+Cr}}dr=t+A$$
Now, make the following substitution:
$$u=\sqrt{\frac{r}{2K+Cr}}\to{r}=\frac{2Ku^{2}}{1-Cu^{2}}$$
Therefore, we have:
$$\frac{dr}{du}=\frac{4Ku(1-Cu^{2})+4KCu^{3}}{(1-Cu^{2})^{2}}=\frac{4Ku}{(1-CKu^{2})^{2}}$$
Therefore, you get the integral in the equation transformed to:
$$\int\frac{4Ku^{2}du}{(1-KCu^{2})^{2}}=A+t$$
In order to perform the integration, we note that:
$$\int\frac{4Ku^{2}du}{(1-KCu^{2})^{2}}=\frac{2u}{C(1-KCu^{2})}-\frac{2}{C}\int\frac{du}{1-KCu^{2}}$$

You will therefore get an implicit equation for u as a function of t, and hence, r will also be only implicitly determined as a function of time.
The indefinite integral remaining will either be an arctan- or artanh-expression, depending on the sign of the product KC.

Last edited: Mar 2, 2007
15. Mar 3, 2007

### CPL.Luke

hmm maple can produce an explicit formula for r, however the formula is quite long and involves a number of "Rootof" operations. So you most likely will onlly have an eact solution for a couple of select points.

16. Mar 27, 2009

### vasil georgie

solutidn a=1/r^2 is r=b t^2/3