Kepler's 2nd Law

1. Sep 9, 2003

StephenPrivitera

d2A/dt2=0
or
dA/dt=k=rv/2

L=r x p
L/m=r x v
L/(2m)=(r x v)/2

How does dA/dt relate to L?
In a certain book, it says dA/dt=H/2 where H=L/m=rv
but L is a vector and H is not.

2. Sep 9, 2003

Creator

dA/dt = L/(2m) = constant

So where's the problem?
m is a constant; r x v is constant.

3. Sep 10, 2003

StephenPrivitera

L is a vector. So is dA/dt=L/(2m) where L=|L|?
Besides L/(2m)=(1/2)(r x v). Is it correct to say
|L/(2m)|=(1/2)|r|*|v|=rv/2=dA/dt?

edit:
Sorry, found the problem. |r x v|=RVsinA and since r and v are always perpendicular. |r|*|v|=RVsinpi/2=RV
So, |L/(2m)|=(1/2)|r|*|v|=rv/2=dA/dt
Right?

edit:
"since r and v are always perpendicular"
but wouldn't that only work for circular orbits?

Last edited: Sep 10, 2003
4. Sep 10, 2003

enigma

Staff Emeritus
I don't have my orbital dynamics book on me and some of your variables are using different characters than I'm used to seeing (I'm using Vallado).

What are you defining L , H, and m to be?

r and v are usually _not_ perpendicular. The only cases where they are, are either a) circular orbit or b) sat. is at apoapsis or periapsis.

5. Sep 10, 2003

Creator

Wrong; you are correct that it is the cross product; but r and v are not always perpendicular.
r x v does NOT mean v is always perpendicular to r!!
By taking the cross product you are, in effect, by definition, taking the perpendicular component of v. The calculation requires using the perpendicular component of the velocity.

Correct,Stephen, the orbit would be circular IF and only if v were always perpendicular to r; but it is not.
However, inspite of that, we still always use the perpendicular component of v FOR THE PURPOSE OF CALCULATING dA/dt.
This is necessary because L is defined in the same manner, that is, by taking the perpendicular component of the tangental orbital velocity. L= (r x mv)= (r)*(m)*(v)sin@

Did that make sense??
Creator

Last edited: Sep 10, 2003