Kepler's second law derivation from angular momentum conservation

In summary, the conversation discusses the concept of angular momentum in an elliptic orbit, with specific focus on the equation $$ L = r m v = m r^2 \frac {d \theta} {dt} $$ and whether the equation $$ v = r \frac {d \theta} {dt} $$ is valid at every point. The conversation then explains how this concept applies to the Bohr quantum model and the conservation of angular momentum in central potentials. The concept is further illustrated with a graphical representation using LaTex.
  • #1
mr_sparxx
29
4
TL;DR Summary
Many texts state that Kepler's second law can be derived from conservation of angular momentum, however all demonstrations I have found make an assumption that is not clear for me
Many texts state that in an elliptic orbit you can find angular momentum magnitude as

$$ L = r m v = m r^2 \frac {d \theta} {dt} $$

I wonder if
$$ v = r \frac {d \theta} {dt} $$

is valid at every point. I understand this approximation in a circumference or radius r but what about an arc of ellipse?

Thank you.
 
  • Like
Likes malawi_glenn
Physics news on Phys.org
  • #2
I usually explain it like this (dr vector is exaggerated for the sake of the picture)
1657818834123.png

Where I used that the area of the entire parallellogram formed by the r and dr vectors is equal to the magnitude of their cross product. Here we have half a parallellogram, hence the facor 1/2

Here it is with LaTex:
##\mathrm{d}\vec r = \vec v \mathrm{d}t##
## \mathrm{d} A = \frac{1}{2}| \vec r \times \mathrm{d}\vec r| = \frac{1}{2}| \vec r \times \vec v|\mathrm{d}t =\frac{1}{2}| \vec L |\mathrm{d}t ##

If L is conserved then ## \dfrac{\mathrm{d}A}{\mathrm{d}t}## is constant and thus by integrating it we get the area swept out during a certain time interval is constant along the path
 
Last edited by a moderator:
  • Like
  • Love
Likes Delta2, mr_sparxx and vanhees71
  • #3
Just to point out that this is not particular to the Kepler potential. It is true for any central potential as angular momentum ##L = mr^2\dot\theta## is conserved due to the rotational symmetry.
 
  • Like
Likes mr_sparxx, vanhees71 and malawi_glenn
  • #4
It was also important for the Bohr quantum model of the hydrogen atom. Since the potential is spherical symmetric, angular momentm is conserved. Idea was to hypothesize that electrons bound to proton only exhibit certain angular momenta, namely an integer multiple of Plancks constant / 2 (and was it a factor of π as well somewhere?)
 
  • #5
The true spirit of the Bohr-Sommerfeld model is that for (quasi-)periodic motion the action is quantized in multiples of ##h=2 \pi \hbar##. This implies a quantization of angular-momentum components as integer multiples of ##\hbar## since angular momentum is the canonical conjugate to angle variables.
 
  • Like
Likes malawi_glenn
  • #6
drmalawi said:
I usually explain it like this (dr vector is exaggerated for the sake of the picture)
View attachment 304176
Where I used that the area of the entire parallellogram formed by the r and dr vectors is equal to the magnitude of their cross product. Here we have half a parallellogram, hence the facor 1/2

Here it is with LaTex:
##\mathrm{d}\vec r = \vec v \mathrm{d}t##
## \mathrm{d} A = \frac{1}{2}| \vec r \times \mathrm{d}\vec r| = \frac{1}{2}| \vec r \times \vec v|\mathrm{d}t =\frac{1}{2}| \vec L |\mathrm{d}t ##

If L is conserved then ## \dfrac{\mathrm{d}A}{\mathrm{d}t}## is constant and thus by integrating it we get the area swept out during a certain time interval is constant along the path

I guess you mean ( mass is missing)
## \mathrm{d} A = \frac{1}{2}| \vec r \times \mathrm{d}\vec r| = \frac{1}{2}| \vec r \times \vec v|\mathrm{d}t =\frac{1}{2m}| \vec L |\mathrm{d}t ##

Nice way of explaining it. Just what I was looking for. Thank you.
 
  • Like
Likes Delta2, malawi_glenn and vanhees71
  • #7
mr_sparxx said:
I guess you mean ( mass is missing)
## \mathrm{d} A = \frac{1}{2}| \vec r \times \mathrm{d}\vec r| = \frac{1}{2}| \vec r \times \vec v|\mathrm{d}t =\frac{1}{2m}| \vec L |\mathrm{d}t ##

Nice way of explaining it. Just what I was looking for. Thank you.
Yes I forgot mass.
 
  • Like
Likes Delta2 and vanhees71

Related to Kepler's second law derivation from angular momentum conservation

1. What is Kepler's second law?

Kepler's second law, also known as the law of equal areas, states that a line joining a planet and the sun sweeps out equal areas in equal intervals of time. This means that a planet moves faster when it is closer to the sun and slower when it is farther away.

2. How is Kepler's second law derived from angular momentum conservation?

Kepler's second law can be derived from the conservation of angular momentum, which states that the total angular momentum of a system remains constant unless acted upon by an external torque. In the case of a planet orbiting the sun, the angular momentum of the planet is conserved as it moves closer to and farther from the sun, resulting in equal areas being swept out in equal time intervals.

3. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object around a fixed point. It is calculated by multiplying the object's moment of inertia by its angular velocity. In the case of a planet orbiting the sun, the planet's angular momentum is conserved as it moves in its elliptical orbit.

4. How does Kepler's second law relate to the shape of a planet's orbit?

Kepler's second law states that a planet's distance from the sun changes as it moves in its orbit, resulting in equal areas being swept out in equal time intervals. This results in a planet's orbit being elliptical in shape, with the sun at one of the foci.

5. Is Kepler's second law applicable to all planetary orbits?

Yes, Kepler's second law applies to all planetary orbits, as long as the planet is orbiting a central body and the orbit is not significantly affected by other forces such as gravitational interactions with other planets. This law also applies to other types of orbits, such as satellites orbiting a planet.

Similar threads

Replies
5
Views
870
  • Classical Physics
Replies
30
Views
2K
Replies
2
Views
981
  • Introductory Physics Homework Help
Replies
8
Views
1K
Replies
7
Views
1K
  • Classical Physics
Replies
4
Views
1K
Replies
8
Views
1K
Replies
6
Views
1K
Replies
1
Views
411
Replies
10
Views
1K
Back
Top