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The softest thing overcomes the hardest thing in the universe. ...Lao Tzu (570-490 BC)
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Author Topic: kepler orbit derivation  (Read 15918 times)
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on: April 01, 2006, 12:11:18 am »

When solving the orbit equation you get the linear second order differential equation:

u¡¦¡¦ + u = k/(h^2) {eq1}

where u¡¦¡¦ is the second derivative of u with respect to theta and u is 1/r where r is the magnitude of the radius vector. Theta is a function of time. h is the angular momentum per unit mass. k= GM, k is the product of the universal gravitational constant G and the mass of the planet or sun M which is at the focus of the ellipse.

Solving this equation gives u = A*cos(theta) + B*sin(theta) {eq2} and one particular solution is u= k/(h^2) {eq3}.

I can manipulate u = A*cos(theta) + B*sin(theta) to get it as a single cosine in the form of R*cos(theta ¡V delta) {eq4} where R is a constant = root(A^2 + B^2) and delta = arctan(B/A).

But in the derivation of the orbit equation it is simply stated that taking eq2 and choosing axis so that B=0 and delta = 0 the solution is u = A*cos(theta) + k/(h^2).

Defining l = (h^2)/k and e= Al and solving for r we get r a function of theta = l/(1 +e*cos(theta) which is the equation of an ellipse in polars with the directrix right of the pole and parallel to the x axis, with e the eccentricity and l the = e*d where d is the distance from the directrix to the pole.

My question is how can I graph eq2 and just pick an axis so that B=0. I want to know how and why this works so that I can apply it to this and other scenarios if appropriate. In other words how do I justify its use.
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The softest thing overcomes the hardest thing in the universe. ...Lao Tzu (570-490 BC)
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