Can Conservation of angular momentum be used?

[PhysicsKid703]

Homework Statement

if the distance between the Earth and the sun were to be cut in half, what would be the number of days in the year?

Homework Equations

The Attempt at a Solution

I can solve this question using simple centripetal force = gravitational force of attraction and then halving the radius, thus finding the new velocity and then new time period. That's is easy, and I get the right answer of 129 days.
My problem lies in a different approach. Seeing as there is no external torque, why can't we conserve angular momentum? I tried to do so and ended up with 91.25 days instead. I used the m(r x v) formula for angular momentum as it revolves around the sun and I can't seem to find out why it can't be conserved. Any help is appreciated. Thanks!

 

[Andrew Mason]

You have asked a very good question.

Total angular momentum will be conserved. But not necessarily orbital angular momentum. The sun's rotation about its axis has to be included. Changes in the orbital angular momentum of the Earth will affect the rotation of the sun about its axis. As the Earth orbit is reduced, the sun's rotation speed will increase. It is not unlike the increase in the rate of rotation of a figure skater as he/she brings his/her arms in during a spin.

Similarly, the gradual increase of the moon's orbital radius as it orbits the Earth reduces the Earth's rotation speed. See: http://www.physlink.com/Education/AskExperts/ae695.cfm

AM

 

[PhysicsKid703]

Ohh wow, that is extremely interesting and something I never would have intuitively thought of. Thank you so much!

 

[Jediknight]

m(r v x)? isn't mrv?- thiss super interesting, I feel like if we don't consider the sun spinning in any other calculations it shouldn't affect these, but maybe in others it was just always safe to assume it satyed constant, imma try to work this out then ill come edit this post, cool thread though

 

[Andrew Mason]

m(r v x)? isn't mrv?-

For a point mass rotating about a fixed intertial point, L = m (r x v) where the x operator is the cross product. r is the radius vector from the fixed point. (r x v) = |r||v|$\sin\theta$. For circular orbit, $\theta = \pi/2$ so the orbital angular momentum is L = mrv $\hat{r} \times \hat{v}$.

But that is just the orbital angular momentum. The angular momentum of the sun is $L = I \omega$ where I is the moment of inertia of the sun and $\omega$ is its angular speed of rotation about its axis.

Because the sun has a finite size, there are tidal effects caused by the fact that the gravitational tug of the Earth and the other planets on the sun varies slightly across the sun's surface. This causes a slight torque on sun which affects its rate of rotation. The Earth is actually moving ever so slightly away from the sun. This post provides a more detailed explanation.

AM

 

[Andrew Mason]

Ohh wow, that is extremely interesting and something I never would have intuitively thought of. Thank you so much!

Of course to actually move the Earth to a distance of half of its current orbital radius about the sun, you would have to slow it down. It is not as if the Earth is on a string that can be just reeled in. If there is no external torque (we are assuming this is a two-body problem - there is just the Earth and the sun), this can only be done by the sun applying a torque to the earth. This necessarily would have the effect of transferring some of the Earth's orbital momentum to the sun. If there is an external torque (e.g. a large body hits the Earth and slows its orbital speed) then angular momentum (sun-earth) is not conserved.

AM

 

[PhysicsKid703]

That's what I had initially thought, that there must be some torque to move the sun, but I've completely understood it. Thanks!