Deriving Vr and V(sub-theta) from Kepler's second law and

[LordCalculus]

Deriving Vr and V(sub-theta) from Kepler's second law and...

Homework Statement

Beginning with r=[a(1-e²)]/(1+e*cos [tex]\theta[/tex]) and Kepler's second law, derive general expressions for v_r and v_{[tex]\theta[/tex]} for a mass m₁ in an elliptical orbit about a second mass m₂. The final answers should be functions of P, e, a, and [tex]\theta[/tex] only.

P=orbital period
e=eccentricity
a=semimajor axis

That's supposed to be v(subtheta) and cosine of theta!

Homework Equations

A'(t)=L/(2*[tex]\mu[/tex])

 

[diazona]

[itex]v_\theta[/itex], understood. What have you done so far and where exactly did you get stuck?

 

[LordCalculus]

Not far at all. I don't even know where to begin. If you do, then please start me off. Thanks!

 

[diazona]

Well, here would be my first thought: you have some information about position, and you are asked to find velocity. What does that tell you?

 

[paranormal]

= constant
L = angular momentum = r x p = mvr

The Attempt at a Solution

To derive vr and v\theta, we will use the following steps:

1. Start with Kepler's second law, which states that the area swept out by the radius vector r in a given time is constant. This can be written as:

dA/dt = (1/2)r^2 d\theta/dt = constant

2. From the given expression for r, we can rewrite it as:

r = a(1-e^2)/(1+e*cos \theta)

3. Taking the derivative of r with respect to time, we get:

dr/dt = a*e*sin\theta * d\theta/dt

4. Substituting this into the equation for dA/dt, we get:

dA/dt = (1/2) * (a(1-e^2)/(1+e*cos \theta))^2 * a*e*sin\theta * d\theta/dt

5. Simplifying and rearranging, we get:

dA/dt = (1/2) * a^2 * (1-e^2)^2 * e * sin\theta * d\theta/dt

6. Using the fact that A'(t) = L/(2*\mu) = constant, we can rewrite the above equation as:

L/(2*\mu) = (1/2) * a^2 * (1-e^2)^2 * e * sin\theta * d\theta/dt

7. Solving for d\theta/dt, we get:

d\theta/dt = (L/(2*\mu)) * (2/(a^2 * (1-e^2)^2 * e * sin\theta))

8. Now, we can use the expression for angular momentum, L = mvr, to rewrite the above equation as:

d\theta/dt = (mvr/(2*\mu)) * (2/(a^2 * (1-e^2)^2 * e * sin\theta))

9. Finally, we can substitute this expression for d\theta/dt into our original expression for dr/dt to get:

dr/dt = (mvr/(2*\mu)) * (2/(a^2 * (1-e^2)^2 * e * sin\theta)) * a*e*sin\theta

10.