These are parametric equations for the Keplerian orbit of a planet: tan(phi/2) =

Question

These are parametric equations for the Keplerian orbit of a planet:

tan(phi/2) = sqrt{(1+e)/(1-e)} tan(psi/2)

r = a ( 1 -e cos(psi))    and     t = ( T/ 2pi) (psi- e sin(psi))

Here (r,phi) are plane polar coordinates, t is time, and psi is an independent variable parameter. Also, T = period of revolution, a = semimajor axis, and e= eccentricity.

(A) Determine the Cartesian coordinates (x,y) and time t for the point P shown on the figure. (Express the answers in terms of a, e, T.)

(B) Determine the Cartesian coordinates (x,y) and time t for the point R shown on the figure. (Express the answers in terms of a, e, T.)

(C) Determine L/m where L is the angular momentum of the planet, and m is its mass. (Express the answer in terms of a, e, T.)

Explanation / Answer

tan(q/2) = sqrt{(1+e)/(1-e)}tan(k/2)

r = a(1-e cosk)

t = (T/2pi) (k-e sink).

We restrict our attention to just one revolution through the orbit, starting from P, so k varies in between [0,2pi].

(A)

Point P is on the positive x-axis and a is semi-major axis.

So P = (a(1-e),0). (distance from center of ellipse to endpoint on major axis is a and the distance from center of ellipse to focus is ae, so distance between endpoint and nearest focus is a-ae = a(1-e)

=> a(1-e) = r = a(1-e cosk)

=> cos k = 1.

Also q = 0 => 0 = tan(q/2) = sqrt{(1+e)/(1-e)}tan(k/2)

=> tan(k/2) = 0.

The two equations above imply k = 0.

So time t = 0 s. (agreeing with our assumption that revolution starts at P).

Also note that when the planet returns to P, we have k = 2pi....(Observation).This is because for P k = 0, 2pi, 4pi,...

(B)

From the figure and the details of the ellipse

R = (0,ae.sqrt{1-e^2}) (for an ellipse with semimajor axis a, eccentricity e,

Thus r = ae.sqrt{1-e^2}

=> ae.sqrt{1-e^2} = r = a(1-e cosk)

=> cos k = (1/e) - sqrt{1-e^2}.

Also q = pi/2

=> 1 = tan q/2 = sqrt{(1+e)/(1-e)}tan(k/2)

From the above two we get k is in [0,pi),

k = 2 arctan(sqrt{(1-e)/(1+e)}).

So time t = (T/2pi) (k-e sink).

= (T/2pi) (2 arctan(sqrt{(1-e)/(1+e)}) - e sin(2 arctan(sqrt{(1-e)/(1+e)})).

(C)

We have from our physics that angular momentum is conserved for a planetory motion.

so To find angular velocity we need to divide 2pi, the total revolution angle by its time period T.

So L = m.(2pi/T).

=> L/m = 2pi/T.

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