Deriving Oort Constants From Kepler's Third Law

[sfbsoccer25]

Hello everybody. New here. Hope you are all doing well. I am a sophomore in college, studying physics and astronomy. I am in my first astrophysics class and struggling with how to approach problems. That being said, I need a little homework help if you can afford some time!

1. Starting with Kepler's Third Law, (M1 + M2)p^2=4(pi^2)(a^3), derive expressions for the Oort Constants, A and B, for circular motion around a point mass. (Hint: first derive a relation for V(R) for a star in Keplerian orbit.)

Any help or pushes in the right direction would be greatly appreciated. Thanks for your time.

 

[lippyka]

The Oort constants, A and B, are related to the angular momentum of an object in a circular orbit around a point mass. To derive expressions for these constants, we first need to derive a relation for the velocity of a star in a Keplerian orbit. To do this, we can start with Kepler's Third Law, which states (M1 + M2)p^2=4(pi^2)(a^3). Here, M1 and M2 are the masses of the two objects, p is the orbital period, and a is the semi-major axis. We can rearrange this equation as follows: a^3=(M1 + M2)p^2/4(pi^2).Next, we use the equation for centripetal acceleration, a_c=v^2/r, where v is the velocity and r is the radius. We can combine this equation with the expression for a^3 that we derived earlier to get v^2=4(pi^2)(M1 + M2)p^2/r^3. Now, let's define the angular momentum of the orbiting object, L=mvr, where m is the mass of the object. Substituting this expression into our equation for v^2 gives us L^2=4(pi^2)(M1 + M2)p^2m^2/r^4.Now, if we assume the orbit is circular, then the angular momentum is constant. So, if we solve for r, we can get r^4=L^2/4(pi^2)(M1 + M2)p^2m^2. Substituting this expression into our equation for v^2 gives us v^2=L^2/(m^2r^3). Finally, we can solve for the Oort constants A and B by dividing both sides of the equation by r^2, giving us A=L^2/(m^2r^5) and B=L^2/(m^2r^7).