Planets orbit around the sun

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?

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mgb_phys
Homework Helper
Conservation of energy is probably the simplest explanantion

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)

cepheid
Staff Emeritus
Gold Member
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.

D H
Staff Emeritus
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?
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.

tiny-tim
Homework Helper
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?
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

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

tiny-tim
Homework Helper
Wouldn't angular momentum be conserved since the Sun doesn't apply any torque to the planet?
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.

D H
Staff Emeritus