# Planets orbit around the sun

1. Nov 30, 2009

### Kaxa2000

If you were to measure the area of a sector that a planet would sweep out in one week around the sun. It would be the same no matter what time of the year it was. What conservation principle is this example demonstrating? Linear, angular or both? and why?

2. Nov 30, 2009

### mgb_phys

Conservation of energy is probably the simplest explanantion

3. Nov 30, 2009

### Kaxa2000

conservation of energy is not one of the answer choices in the problem

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)

4. Nov 30, 2009

### cepheid

Staff Emeritus
It's Kepler's 2nd Law of planetary motion, and it is used to conclude that planets travel faster when closer to their parent stars and they travel slower when farther away.* That, I think, is a huge hint as to what is being conserved.

*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.

5. Nov 30, 2009

### D H

Staff Emeritus
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

$$\Delta A = \frac 1 2 r^2 \Delta \theta$$

Dividing by the length of the time interval $\Delta t$ and taking the limit $\Delta t \to 0$ yields

$$\dot A = \frac 1 2 r^2 \dot \theta$$

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

$$\vec l = \vec r \times \vec v$$

Representing the position vector as $\vec r = r \hat r$, the velocity vector is

$$\vec v = \dot r \hat r + r\dot \theta \hat{\theta}$$

Thus

$$\vec l = r^2\dot \theta \hat z$$

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

6. Nov 30, 2009

### tiny-tim

Hi Kaxa2000!

It's conservation of angular momentum.

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

So the area swept out by a planet is ∫ r x (r + vdt), = ∫ r x vdt

7. Dec 1, 2009

### Kaxa2000

Wouldn't angular momentum be conserved since the Sun doesn't apply any torque to the planet?

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. Dec 1, 2009

### tiny-tim

Yes, that's right … physics tells us that zero torque means (as D H and I said) that angular momentum is conserved.

And maths (geometry) tells us that conservation of angular momentum means equal areas swept out in equal times.

9. Dec 1, 2009

### D H

Staff Emeritus
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.

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.