Kepler's 2nd Law: Relation of dA/dt to L

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The discussion focuses on the relationship between the rate of change of area (dA/dt) and angular momentum (L) in orbital mechanics. It establishes that dA/dt can be expressed as L/(2m), where L is the angular momentum vector and m is mass. The participants clarify that while r (radius) and v (velocity) are not always perpendicular, the calculation of dA/dt requires using the perpendicular component of velocity. The conclusion emphasizes that for circular orbits, r and v are indeed perpendicular, but this is not a general case.

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

So where's the problem?
m is a constant; r x v is constant.:wink:
 
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:
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.
 
Originally posted by StephenPrivitera


edit:
Sorry, found the problem. |r x v|=RVsinA and since r and v are always perpendicular.
Right?

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.

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

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 :wink:
 
Last edited:

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