Angular momentum and true anomaly

[Deadstar]

Homework Statement

The angular momentum per unit mass of an object orbiting the Sun is given by $$h = na^2 \sqrt{1 - e^2}$$, where $$n$$ , $$a$$ and $$e$$ are the mean motion, semi-major axis and eccentricity of the object, respectively. Use $$h$$ to obtain an expression for the angular velocity of the object as a function of the true anomaly, $$f$$.

The Attempt at a Solution

Literally the only formulas involving f that we have in our notes are...

$$\.{r} = \frac{na}{\sqrt{1-e^2}}\sin(f)$$

and

$$r = \frac{a(1-e^2)}{1 + e \cos(f)}$$

and since I seem to think $$\.{r}$$ is the angular velocity is the first equation there just the answer?!?

 

[anathema]

Thank you for your question. The formula you have mentioned, \.{r} = \frac{na}{\sqrt{1-e^2}}\sin(f), is indeed the expression for the angular velocity of an object in orbit around the Sun. This can be seen by rearranging the formula to solve for \.{f}, which represents the angular velocity:

\.{f} = \frac{\.{r}}{\frac{na}{\sqrt{1-e^2}}}

Using the definition of angular velocity, \.{f} = \frac{d\theta}{dt}, we can rewrite this as:

\.{f} = \frac{d\theta}{dr}\frac{dr}{dt}

The first derivative, \frac{d\theta}{dr}, represents the rate of change of the angle with respect to the distance from the Sun, and can be calculated using the formula for the eccentric anomaly, E = 2\tan^{-1}(\sqrt{\frac{1-e}{1+e}}\tan(\frac{f}{2})).

The second derivative, \frac{dr}{dt}, represents the rate of change of the distance from the Sun with respect to time, and can be calculated using the formula for the radial velocity, \.{r} = \frac{na}{\sqrt{1-e^2}}\sin(f).

Therefore, the full expression for the angular velocity as a function of the true anomaly is:

\.{f} = \frac{2na\sqrt{1-e^2}}{1+e\cos(f)}\frac{\sqrt{\frac{1-e}{1+e}}}{1+\sqrt{\frac{1-e}{1+e}}\tan(\frac{f}{2})}\sin(f)

I hope this helps. Please let me know if you have any further questions.