# Two-body equation of motion resolution

• greg_rack
In summary: Got it, but what is exactly meant by "partial solution"? Is it merely a way of solving the equation which gives the sought radius as a scalar, and not as ##{\textbf{r}(t)}##?Yes, a partial solution is a solution that only gives part of the solution, in this case the size and shape of the orbit. It is not a function of time, so it does not describe the movement. Finding a complete solution would involve finding a function ##{\textbf{r}(t)}## that describes the movement over time.
greg_rack
Gold Member
Hi guys,

I have started studying differential equations on my own, sticking to my last high-school year's textbook, along with a few physics applications of ODEs.
Online I came across the n-body problem, which then took me to the basic two-body problem!

I'm here to ask you a few things about the two-body vector differential equation of motion:
$$\ddot{\textbf{r}}+\frac{\mu}{r^3}\textbf{r}=0$$

First: which type of DE is it? My textbook covers(approximately) linear DEs, explaining just a few resolution methods for first and second order homogeneous and non-homogeneous ones.
Is this non-linear, and so not solvable with "usual" and easier methods? If yes, why is it so?

Second: the textbook I found says: "though this equation looks easy, its complete solution is not".
Why?

Lastly, after the aforementioned consideration, it provides the steps to find a "partial solution that will tell us the size and shape of the orbit", which is a function ##r##(rel. position of the two bodies) of the form of a conic section: this "partial solution" thus tells us -only- the shape and size of the orbit, but doesn't provide us with a function of time for the movement.
How is that? What is a "partial solution", and what would be needed to find a "complete" one?

Online resources look all very confusing and over-complicated, and I'm having a really hard time deciphering those... hope you can help me, and my questions aren't too silly :)

Hi,

greg_rack said:
First: which type of DE is it?
It took me a bit of searching to find your textbook. Next time, perhaps you can insert the link in your post ? Saves others time !

And on top of page 8 you missed the answer:

The assumptions that follow and that lead to ##\ddot{\textbf{r}}+\frac{\mu}{r^3}\textbf{r}=0## don't change that.

Is this non-linear, and so not solvable with "usual" and easier methods? If yes, why is it so?

Second: the textbook I found says: "though this equation looks easy, its complete solution is not".
Why?
I could shrug this off with 'ours is not to reason why' or 'I really don't know ' . Why things are as they are is not physics. There are no tools in our toolkit to solve this analytically.

Suffice it to point to Kepler's equation ##M = E - e\sin E\ ## that remains after a whole lot of chewing.
One could argue that it's so difficult because the effective one-dimensional potential in which the motion takes place is such an unpleasant function of the radial distance ( ##-\displaystyle{a\over r} + \displaystyle{b\over r^2}\ ## ). Does that help you ?

The partial solution is 1-41 or 1-42 ( ##r = \displaystyle{p\over 1+e\cos\nu} \ ## ). The complete solution would be ##{\textbf{r}(t)}##, for which the book refers to Chapter 4, over a hundred pages further. You have your work cut out for you !

greg_rack said:
all very confusing and over-complicated, and I'm having a really hard time deciphering those... hope you can help me, and my questions aren't too silly :)
Your questions are not silly and I admire your courage and stamina to tackle something like this. But I think you are trying to bite off more than you can chew. I can't help in making it easier. The authors do a good job and I can't rewrite their book.

If your angle is physics stick with Mechanics_of_planar_particle_motion and Kepler's laws
And if it's differential equations, find an undergraduate calculus book you like and do the exercises.

##\ ##

Last edited:
greg_rack and Ibix
BvU said:
Next time, perhaps you can insert the link in your post ? Saves others time !
Definitely will do! Haven't really thought about it before posting

BvU said:

The assumptions that follow and that lead to ##\ddot{\textbf{r}}+\frac{\mu}{r^3}\textbf{r}=0## don't change that.
Great, and what does "defied solution" mean? I came across, I got the sense, but wasn't really able to get the precise meaning... I'm not English, and no translator gave me a satisfying translation for "defied solution"

BvU said:
One could argue that it's so difficult because the effective one-dimensional potential in which the motion takes place is such an unpleasant function of the radial distance ( ##-\displaystyle{a\over r} + \displaystyle{b\over r^2}\ ## ). Does that help you ?
That does definitely help, thanks a lot!

BvU said:
The partial solution is 1-41 or 1-42 ( ##r = \displaystyle{p\over 1+e\cos\nu} \ ## ).
Got it, but what is exactly meant by "partial solution"? Is it merely a way of solving the equation which gives the sought radius as a scalar, and not as ##{\textbf{r}(t)}##?

greg_rack said:
defied solution
It's a part of the verb "to defy". "It has resisted our attempts at finding a solution" is an expanded translation. Or "we can't solve it".

greg_rack
greg_rack said:
what is exactly meant by "partial solution"
Is is only a part of the solution, namely the radial distance as a function of an angle that varies in time.
So it gives the shape of the trajectory.
For a complete solution you would still need how that angle varies in time.

##\ ##

greg_rack
Because the equation of the two body problem is what is known as a transcendental equation, and those often do not have a closed form solution. https://en.wikipedia.org/wiki/Transcendental_equation.

It is solvable though, but only using iterative calculation or geometry. If you just want to create some neat orbits then i suggest you use the Euler method instead of Keplers. Since the latter is probably a bit too difficult to start with.

Last edited:
greg_rack
greg_rack
Thank you so much guys!

## 1. What is the two-body equation of motion?

The two-body equation of motion is a mathematical model that describes the motion of two objects under the influence of their mutual gravitational attraction. It is used to predict the position and velocity of both objects at any given time.

## 2. How is the two-body equation of motion derived?

The two-body equation of motion is derived from Newton's laws of motion and the law of universal gravitation. It involves solving a set of differential equations that take into account the masses, positions, and velocities of the two objects.

## 3. What are the assumptions made in the two-body equation of motion?

The two-body equation of motion assumes that the two objects are point masses, that there are no external forces acting on them, and that the gravitational force between them is the only significant force.

## 4. How is the two-body equation of motion used in real-world applications?

The two-body equation of motion is used in many fields, including astronomy, spaceflight, and satellite orbit prediction. It is also used in simulations and computer models to study the motion of celestial bodies.

## 5. Are there any limitations to the two-body equation of motion?

Yes, the two-body equation of motion is only accurate for systems where the masses are significantly smaller than the distance between them. It also does not take into account the effects of relativity or other external forces, such as atmospheric drag.

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