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If I had a factor of Sinx in the numerator of my integrand, I could solve that integral easily. So I figure that if I exploit the fact that 1=Sin2x+Cos2x, I might be able to solve the sine part of the integral by integrating by parts, but I am left with the intractable cosine-squared part. However, if I could multiply the squared cosine by b2 and then subtract a2, I could factor the resulting binomial, cancel off one of the factors in the denominator, and see what I can do with the resulting simpler integral.
Therefore, I begin by multiplying and dividing the intergral by b2-a2, unpacking the squares of sine and cosine from the b2 term, and simplifying the result:
I can integrate the first integral on the right side of that equation by parts and get
The second term on the right of that equation, the cosine integral, cancels the cosine integral in the second term on the right side of Equation 1, so we get
But we already know the solution of the integral on the right side of that equation, so we have, at last,
Area of an Orbital Segment
We describe the ellipse that represents a planet's orbit by describing the orbit's radius as a function of the longitude measured from the orbit's primary focus, the angle between the radius vector and the line extending from the sun to the perihelion of the orbit. Thus we have
in which equation r0 represents the perihelion radius of the orbit, e represents the orbit's eccentricity, and θ represents the longitude on the orbit. We want to calculate the area that the planet's radius vector sweeps out as the planet goes from perihelion (θ=0) to some arbitrary longitude.
For the minuscule element of area we take an isoceles triangle whose sides have length r and whose base has length rdθ, so we have
We then make appropriate substitutions from Equation 5 and carry out the integration:
If we want to calculate the area in the elliptical segments bounded by longitudes θ1 and θ2, then we need only substitute the coordinates θ1 and θ2 into Equation 7 separately and then subtract the smaller result from the larger one.
Johannes Kepler's second law of planetary motion tells us that the radius vector extending between sun and any given planet sweeps out equal areas in equal times. We can calculate the rate at which the radius vector sweeps out area if we multiply together the length of the radius vector of a given planet's orbit at perihelion r0 and the velocity the planet has at perihelion v0; we get
If we multiply Equation 8 by the
element of time and integrate both sides of the resulting equation, we find that
we can calculate the time that a planet needs to go from perihelion to some
given longitude by dividing Equation 7 by the product r0v0.
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