Planet traversing the half of its orbit closest to the sun?

[Sven Andersson]

Homework Statement

What fraction of its "year" (i.e. the period of its orbit) does a planet spend traversing the half of its orbit closest to the sun? Give the answer in terms of the eccentricity ε of the planet's orbit. This is problem 15 from page 852 of Adam's Calculus 3ed. No detailed solution available.

2. Homework Equations

Partial area integral of an ellipse.

3. The Attempt at a Solution

Just divide (area of ellipse to the "right" of latus rectum)/(total area of ellipse) according to Kepler's law. However I get stuck in "impossible" integrals. The answer is (1/2)-(ε/π).

S.A.[/B]

 

[mfb]

How is that half defined? If it is just half the ellipse, all you need is the area of a triangle.
If it would be 180° as seen by the central star, then the given answer would be wrong.

 

[Buzz Bloom]

Hi Sven:

The length of time corresponding to a portion of an orbit is the area bounded by the portion of the orbit, and two radial lines from the focus (the sun in this case) to the two ends of the portion of the orbit. You want to find the two symmetric ends of the orbit such that this area is 1/2 the total area of the elliptical orbit. The latus rectum does not do that.

Regards,
Buzz

 

[mfb]

You want to find the two symmetric ends of the orbit such that this area is 1/2 the total area of the elliptical orbit.

That would find the points separated by half the orbital period. The problem statement asks for a time for different points.

 

[Buzz Bloom]

Hi mfb:

Sorry, another senior moment. I misread the problem statement. Thanks for the correction.

In Cartesian coordinates with the major axis along x and the minor axis along y, the two points bounding the half orbit closest to the sun are the points where y = +b and y = -b, where b is the semi-minor axis.The corresponding fraction of the year is the fraction of the total area bounded by the half orbit and two radii.

The latus rectum is still not a boundary of the correct part of the orbit.

Regards,
Buzz

 

[LCKurtz]

@Sven Andersson: Has the above discussion led you to the solution yet? You don't need any calculus to solve the problem if you take as given that the area of an ellipse with major semi axis ##a## and minor semi axis ##b## is ##\pi ab##.

 

[Sven Andersson]

Yes, yes, just two triangles and an ellipse and the solution is then obvious. I really misunderstood the question from the beginning. Sorry...