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mgb_phys

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Conservation of energy is probably the simplest explanantion

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I think it's demonstrating both linear and angular momentum conservation but not sure how to explain it.

We know angular momentum is L = r x (mv)

and mv = p (linear momentum)

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cepheid

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*If you can't see how this is so, draw a picture of a short, wide sector (swept out by the planet when it is close to its star) and a long, narrow sector, each having equal area. It takes the planet the same amount of time to sweep out each sector, but the planet travels a much greater distance (the arc) along the fat one than it does along the thin one.

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Angular momentum.

From the perspective of a Sun centered system, linear momentum is *not* conserved. You have to look at things from the perspective of the center of mass and consider both the motion of the planet and the Sun in order to find that linear momentum is conserved. Even then, conservation of linear momentum is trivially conserved.

Over a sufficiently small period of time, the angular rate of the planet's motion with respect to a non-rotating, Sun-centered frame and the distance between the Sun and the planet are more or less constant. The area swept out by the planet during this short period of time is thus

[tex]\Delta A = \frac 1 2 r^2 \Delta \theta[/tex]

Dividing by the length of the time interval [itex]\Delta t[/itex] and taking the limit [itex]\Delta t \to 0[/itex] yields

[tex]\dot A = \frac 1 2 r^2 \dot \theta[/tex]

Kepler's second law says this is zero. Now look at the specific angular momentum of the planet. This is

[tex]\vec l = \vec r \times \vec v[/tex]

Representing the position vector as [itex]\vec r = r \hat r[/itex], the velocity vector is

[tex]\vec v = \dot r \hat r + r\dot \theta \hat{\theta}[/tex]

Thus

[tex]\vec l = r^2\dot \theta \hat z[/tex]

The magnitude of this vector is twice the areal velocity, the quantity that Kepler's second law says is constant.

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tiny-tim

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Hi Kaxa2000!

It's conservation of

The area of a triangle is 1/2 absinθ, = 1/2

So the area swept out by a planet is ∫

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dL/dt = tau = r cross F

r is vector from sun to planet

F is directed from planet to sun

so cross product would be 0

resulting in constant angular motion

- #8

tiny-tim

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Yes, that's right …Wouldn't angular momentum be conserved since the Sun doesn't apply any torque to the planet?

And

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Technically, yes. From a sun-centric point of view, forces do act on a planet, so *none* of the conservation laws apply. However, that was not one of the choices in this (apparently) multiple-guess question.Wouldn't angular momentum be conserved since the Sun doesn't apply any torque to the planet?

Now move over a tiny bit from the center of the Sun to the center of the Sun and planet. From a barycentric point of view, the Sun+planet form an isolated system; the tiny forces from the rest of the galaxy can be ignored. Linear momentum is trivially conserved in the center of mass frame. Angular momentum explains Kepler's equal area law.

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