# Planetary Motion :-)

• May 2nd 2008, 09:16 PM
Kipster1203
Planetary Motion :-)
If a planet moves in an elliptical path with sun located at one focus and the ellipse has a major axis of length 10 and a distance between foci of 6. The radius vector from the sun to the planet sweeps out area at a constant rate of 80. What is the speed of the planet at the point on the ellipse that is farthest from the sun (aphelion)?

Thanks!
• May 4th 2008, 04:36 AM
topsquark
Quote:

Originally Posted by Kipster1203
If a planet moves in an elliptical path with sun located at one focus and the ellipse has a major axis of length 10 and a distance between foci of 6. The radius vector from the sun to the planet sweeps out area at a constant rate of 80. What is the speed of the planet at the point on the ellipse that is farthest from the sun (aphelion)?

Thanks!

We know the following relations between the semi-latus rectum, the eccentricity:
$\displaystyle r_{min} = \frac{p}{1 + \epsilon}$

$\displaystyle a = \frac{p}{1 - \epsilon ^2}$

From the information given we know that $\displaystyle r_{min} = 2$ and a = 5, so we may solve this system for p and $\displaystyle \epsilon$. I get
$\displaystyle p = \frac{16}{5}$
and
$\displaystyle \epsilon = \frac{3}{5}$

Now, Kepler's 2nd Law says
$\displaystyle \frac{dA}{dt} = \frac{L}{2m}$

At $\displaystyle r_{min}$ we know that dA/dt = 50, thus L = 100m. So at maximum r:
$\displaystyle L = 100m = mvr_{max}$

$\displaystyle v = \frac{100}{r_{max}} = \frac{100(1 - \epsilon)}{p} = \frac{25}{2}$

-Dan
• May 5th 2008, 02:35 PM
poorm87
How do you know that r(min) is equal to 2? And what does the variable p represent? Thanks a lot!
• May 5th 2008, 06:12 PM
topsquark
See the file below.

The red line is the major axis and is 2a long. The blue line is the minor axis and is 2b long. The line in pink is parallel to the minor axis and is called the semi-latus rectum. (I have no idea why.)

When one focus of the ellipse is at the origin, the polar equation of the ellipse is
$\displaystyle r = \frac{p}{1 + \epsilon ~cos(\theta)}$
where $\displaystyle \epsilon$ is the eccentricity of the ellipse. ($\displaystyle \epsilon = \sqrt{a^2 - b^2}$.)

Now, you were given the major axis as 10, and the distance between the foci as 6. The foci are situated symmetrically about the center of the ellipse, so the remainder of the major axis, 10 - 6 = 4 is going to be twice the distance from the outer point of the major axis to the focus. Thus the minimum distance of the ellipse to the origin is 4 / 2 = 2.

-Dan