Computing semimajor axis from ecentricity and period

[Gary Roach]

Homework Statement

Compute the semimajor axes "a" of Halley's Comet.
Given:
orbital eccentricity e = 0.9673
period P =76 days (2.39674E9 seconds)
Gravitational Constant G = 6.67428E-11
solar mass M = 1.9891E30 Kg.

Also used in equations are:
L = angular momentum of center of mass
$$\mu$$ = reduced mass
r = distance from focus to comet mass
b= semiminor axis of elipse.

Homework Equations

I used:

$$\frac{dA}{dt} = \frac{L}{2\mu} = \frac{A}{P}$$

$$A=\pi a b$$

$$b^2=a^2(1-e^2)$$

$$L=\mu\sqrt{GMa(1-e^2)}$$ from text

The Attempt at a Solution

$$L=\frac{2\pi \mu a^2 (1-e^2)}{P} = \mu \sqrt{GMa(1-e^2}$$

$$a = \sqrt[1/3]{\frac{GMP^2}{4\pi^2 (1-e^2)}}$$

Unfortunately "a" turns out to be 6.6964E12 meters = 44.76 AU .
Since the data is for Halley's Comet, a should be 17.8 AU

Where did I go wrong

 

[gabbagabbahey]

Looks like you've got an extra factor of $(1-\epsilon^2)^{-1/3}$ in your final result...I think the $1-\epsilon^2$ you get from $b^2=a^2(1-\epsilon^2)$ should cancel with the $1-\epsilon^2$ you get from $L=\mu\sqrt{GMa(1-\epsilon^2)}$ when you substitute everything in properly...

 

[D H]

The eccentricity is not involved in Kepler's third law, suggesting you made an error. And you did. What is the area of an ellipse?

 

[Gary Roach]

OK Stupid error. Its not the first time. After getting that straightened out, on rereading the problem I wonder if it is possible to solve for the semimajor axis "a" with only the eccentricity, period and gravitational constant.(ie no solar mass) . Any comments?

Thanks for your help

Gary