# Kepler's Second Law with Angular Momentum

icecats

## Homework Statement

I am working on the derivation of Kepler's Second Law based on torque and angular momentum. I understand that the vector "L" is equal to the mass (m) times the cross product of the vector "r" and the vector "v." The source I am following then states that
L = mrvtheta. I do not understand how the source is deriving this from the cross product. This is shown in the second to last step on this web page.

L= r cross p

## The Attempt at a Solution

I am able to get to the point of computing the cross product, but no further. In other words, I have completed the algebra to arrive at L= m(r cross v).

## Answers and Replies

Science Advisor
Homework Helper
Gold Member
2022 Award
L = mrvtheta.
Actually it says L = mrvθ.
##|\vec x\times\vec y| = |x||y||\sin(\theta)|##, where θ is the angle between the vectors.
We are given that this θ is small, so we can use the small angle approximation sin(θ)≈θ (in radians).

Staff Emeritus
Science Advisor
Homework Helper
Gold Member
Actually it says L = mrvθ.
##|\vec x\times\vec y| = |x||y||\sin(\theta)|##, where θ is the angle between the vectors.
We are given that this θ is small, so we can use the small angle approximation sin(θ)≈θ (in radians).
I am pretty sure it says ##L = mrv_\theta## with ##v_\theta## being the tangential component of the velocity. There is no reason for the angle to be small, what is small is ##d\theta##.

Since the radial component of velocity is parallel to the radius, it will not affect the cross product. Either way, you do not really need this if you are aware of the interpretation of a cross product as an area of a parallellogram spanned by two vectors. The area swept in a small time dt is then half the absolute value of ##\vec r \times \vec v \, dt## (it is a triangle, not a parallellogram).