Proof of Kepler's 1st Law

  • I
  • Thread starter Forcino
  • Start date
  • #1
3
0

Summary:

Using Newton's Universal Law of Gravity to prove an elliptical orbit.

Main Question or Discussion Point

When proving that Newton's Universal Law of Gravity will produce an elliptical orbit, the radial acceleration is confusing to me. Can someone please explain how the radial acceleration is equal to d^2/dt^2 - r(d x theta / dt)^2. Could someone please detail how that acceleration is derived? Thank you. Here is a link to the proof http://farside.ph.utexas.edu/teaching/301/lectures/node155.html
 

Answers and Replies

  • #2
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,693
6,186
Summary:: Using Newton's Universal Law of Gravity to prove an elliptical orbit.

When proving that Newton's Universal Law of Gravity will produce an elliptical orbit, the radial acceleration is confusing to me. Can someone please explain how the radial acceleration is equal to d^2/dt^2 - r(d x theta / dt)^2. Could someone please detail how that acceleration is derived? Thank you. Here is a link to the proof http://farside.ph.utexas.edu/teaching/301/lectures/node155.html
That's just the general form of acceleration in polar coordinates. It's the equivalent of ##a_x = \ddot x , \ \ a_y = \ddot y## in Cartesian coordinates. But not so simple!

See here for example:

https://ocw.mit.edu/courses/aeronau...fall-2009/lecture-notes/MIT16_07F09_Lec05.pdf
 
  • #3
3
0
Thank you. I have looked at that site before.

Can you explain why this is true der = dθeθ, and that deθ = −dθer (from the top of page 2)? I can't really see that from the diagram.
 
  • #4
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,693
6,186
Thank you. I have looked at that site before.

Can you explain why this is true der = dθeθ, and that deθ = −dθer (from the top of page 2)? I can't really see that from the diagram.
I prefer the alternative, "more mathematical" approach on the next page.
 
  • #5
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,312
6,708
A very elegant trick is to know that the Kepler problem has an additional dynamical symmetry, leading to the conservation of the socalled Runge-Lenz vector, which immediately leads to the conclusion that the orbits are conical sections (ellipses for ##E<0##, parabolas for ##E=0##, and hyperbolas for ##E>0##). The Runge-Lenz vector is given by
$$\vec{Q}=\frac{1}{\gamma m_1 m_2} \dot{\vec{r}} \times \vec{L} -\frac{\vec{r}}{r}.$$
Here
$$\vec{L}=\mu \vec{r} \times \dot{\vec{r}}$$
with ##\mu=m_1 m_2/(m_1+m_2)## is the angular momentum in the center-of-mass frame.
 
Last edited:
  • #6
3
0
I prefer the alternative, "more mathematical" approach on the next page.
Thank you for writing. I am working on understanding the acceleration in polar form. I see that der = dθeθ. 1. Am I correct in saying that means that a small change in the unit vector in the r direction is equal to a small change in the angle in the theta direction? I see that deθ = −dθer. 2. Am I correct in saying that means that a small change in the angle vector equals a small change in the angle in the angle in the r vector direction? 3. Why is the negative sign there?
 
  • #7
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
13,693
6,186
Thank you for writing. I am working on understanding the acceleration in polar form. I see that der = dθeθ. 1. Am I correct in saying that means that a small change in the unit vector in the r direction is equal to a small change in the angle in the theta direction? I see that deθ = −dθer. 2. Am I correct in saying that means that a small change in the angle vector equals a small change in the angle in the angle in the r vector direction? 3. Why is the negative sign there?
That's not how I would put it. The correct terminology is "derivatives of the radial and angular unit vectors".

See the derivation on the page under the note: Alternative calculation of the unit vector derivatives.

What you have above is these derivatives in differential form.

The negative sign arises naturally from the differentiation. It's not something the author has added as an afterthought!
 
  • #8
vanhees71
Science Advisor
Insights Author
Gold Member
2019 Award
15,312
6,708
Ok, just let's do the simple math. Let's assume we argued from Kepler's 2nd Law, which is nothing else than the conservation of angular momentum that the orbit is in a plane perpendicular to the angular-momentum vector, and now we introduce polar coordinates. The Cartesian coordinates are given by the polar coordinates as
$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} r \cos \varphi \\ r \sin \varphi \end{pmatrix}.$$
From this we calculate the velocity vector,
$$\begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} \dot{r} \cos \varphi - r \dot{\varphi} \sin \varphi \\ \dot{r} \sin \varphi + r \dot{\varphi} \cos \varphi \end{pmatrix},$$
and the acceleration
$$\begin{pmatrix} a_x \\ a_y \end{pmatrix} = \begin{pmatrix} \dot{v}_x \\ \dot{v}_y \end{pmatrix} = \begin{pmatrix} \ddot{r} \cos \varphi -2 \dot{r} \dot{\varphi} \sin \varphi -r \dot{\varphi}^2 \cos \varphi -r \ddot{\varphi} \sin \varphi \\ \ddot{r} \sin \varphi + 2 \dot{r} \dot{\varphi} \cos \varphi +r \ddot{\varphi} \cos \varphi -r \dot{\varphi}^2 \sin \varphi \end{pmatrix}.$$
Now it's easy to project the vectors to the curvilinear orthonormal basis,
$$\vec{e}_r=\partial_r \vec{x}=\begin{pmatrix} \cos \varphi \sin \varphi \end{pmatrix}, \quad \vec{e}_{\varphi} = \frac{1}{r} \partial_{\varphi} \vec{x}=\begin{pmatrix} -\sin \varphi \\ \cos \varphi \end{pmatrix}.$$
You get
$$a_{r} = \vec{e}_r \cdot \vec{r} = \ddot{r}-r \dot{\varphi}^2, \quad a_{\varphi}=r \ddot{\varphi} + 2 \dot{r} \dot{\varphi}.$$
The force is
$$\vec{F}=-\frac{\gamma m M}{r^2} \vec{e}_r,$$
and thus the equations of motion read
$$\ddot{r}-r \dot{\varphi}^2=-\frac{\gamma M}{r^2}, \quad r \ddot{\varphi} + 2 \dot{r} \dot \varphi=0.$$
Of course, we already now that the angular momentum is conserved, i.e.,
$$\ell=m (x \dot{y}-y \dot{x})=m r^2 \dot{\varphi}=\text{const}.$$
Indeed, taking the derivative of this equation wrt. time, you get the 2nd equation of motion (the ##\varphi##-component).

I think the rest of the calculation is really very nicely explained in the linked manuscriped of posting #1, though I think it's much more elegant to use the Runge-Lenz vector ;-).
 

Related Threads on Proof of Kepler's 1st Law

  • Last Post
Replies
3
Views
10K
Replies
2
Views
2K
Replies
1
Views
381
  • Last Post
Replies
13
Views
4K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
17
Views
4K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
7
Views
4K
  • Last Post
Replies
11
Views
2K
  • Last Post
Replies
3
Views
3K
Top