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

  • Thread starter StephenPrivitera
  • Start date
  • Tags
    2nd law Law
In summary, the book says that dA/dt=H/2, where H is the L/m=rv/2, but L is a vector and H is not. dA/dt is L/(2m) = constant. So where's the problem? m is a constant; r x v is constant. 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?
  • #1
StephenPrivitera
363
0
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.
 
Astronomy news on Phys.org
  • #2
dA/dt = L/(2m) = constant

So where's the problem?
m is a constant; r x v is constant.:wink:
 
  • #3
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:
  • #4
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
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:

1. What is Kepler's 2nd Law: Relation of dA/dt to L?

Kepler's 2nd Law, also known as the Law of Equal Areas, states that the line joining a planet to the sun sweeps out equal areas in equal times. This law describes the relationship between the speed of a planet in its orbit and its distance from the sun.

2. How is dA/dt related to L in Kepler's 2nd Law?

dA/dt, or the rate of change of area, is directly proportional to L, the angular momentum of the planet. This means that as the planet's angular momentum increases, the rate at which it sweeps out equal areas also increases.

3. What is the significance of Kepler's 2nd Law?

Kepler's 2nd Law helps to explain why planets move at different speeds in their elliptical orbits around the sun. It also provides a mathematical relationship between a planet's distance from the sun and its orbital velocity.

4. How does Kepler's 2nd Law apply to other objects in orbit?

Kepler's 2nd Law applies not just to planets orbiting the sun, but to any object in orbit around a central body. This includes moons orbiting planets and artificial satellites orbiting Earth.

5. Is Kepler's 2nd Law always true?

Kepler's 2nd Law is an idealized law that assumes planets and other objects in orbit follow a perfectly elliptical path. In reality, factors such as gravitational pull from other objects and atmospheric drag can affect the accuracy of this law.

Similar threads

  • Introductory Physics Homework Help
2
Replies
42
Views
3K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Replies
6
Views
1K
Replies
3
Views
445
  • Introductory Physics Homework Help
Replies
1
Views
665
  • Introductory Physics Homework Help
Replies
6
Views
990
  • Programming and Computer Science
Replies
8
Views
1K
  • Special and General Relativity
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top