Time Comet Remains Within Earths Orbit

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1. Apr 11, 2014

cpburris

1. The problem statement, all variables and given/known data

A comet is going in a parabolic orbit lying in the plane of the earth’s orbit. Assuming that the earth’s orbit is a circle of radius a. The points where the comets orbits intersects the earth’s orbit are given by:

cos θ = −1 + 2p/a

where p is the perihelion distance of the comet deﬁned as θ = 0.

Use this to show that the time that the comet remains within the earth’s orbit is given by:

$$\frac{2^{1/2}}{3\pi}(2p/a+1)(1-p/a)^{1/2}\tau_E$$

where $$\tau_E$$ is the period of the earth's orbit (i.e. 1 year).

3. The attempt at a solution

Well I used an expression for the angular momentum per unit mass and integrated to get theta in terms of t, set it equal to the intersection points, solved for t for each, and then took the difference. That yielded an answer, but not one that even remotely resembled what was asked for. I am just not sure what to do now.

2. Apr 11, 2014

tman12321

I haven't worked this problem entirely, but the terms in my scratch work seem to be approaching the answer given. I'm not sure, but I think I know what mistake you may have made. If I understand you correctly, you claim to have found θ(t) explicitly. However, this is an enormously complicated function. All that you can really hope for is an implicit function relating θ and t. To see this, recall that dθ/dt=L/(m*r^2), where L is the angular momentum, m is approximately the mass of the comet, and r(θ) = 2p/[1+cos(θ)]. This reveals that your θ integral will involve 1/[1+cos(θ)]^2. Once you have managed to integrate this, you will get a mess of trigonometric functions relating θ to t. What you must then do is plug in your condition on θ when the comet crosses a, the radius of the earth's orbit. (You'll also need to relate this to sin(θ).) This will tell you the time it takes for the comet to go from θ=0 to the point at which it crosses a. Recall that the orbit has time reversal symmetry to find the total amount of time that the comet is within the earth's orbit. You must then relate this to the period of the earth's orbit using Kepler's third law.

3. Apr 12, 2014

cpburris

asf

I used Shaum's Table of Integrals and yes it was extremely messy, and yes you are correct I simply got an implicit function. Perhaps with a final step of relating this result to the period of the earth's orbit using Kepler's third law I could arrive at the answer, but given the incredible mess of trig functions as you stated I have trouble believing that would be all I would need. I am thinking that there is an entirely different approach to this problem that I haven't explored. Perhaps giving the expression I found for the time would help (l is the angular momentum per unit mass).

$t=\frac{4p^2}{l} ({\frac{1}{2}} tan{\frac{\theta}{2}}+\frac{1}{6} tan^3{\frac{\theta}{2}} )$

Plugging in the intersection point I get a total time of:

$t=\frac{8p^2}{l} ({\frac{1}{2}} tan{\frac { cos^{-1}({ -1+\frac{2p}{a}) } } {2}}+\frac{1}{6} tan^3{\frac { cos^{-1}({ -1+\frac{2p}{a}) } } {2}} )$

Now you see what I mean. I can't imagine using Kepler's Third Law to relate that to the period of the earth's orbit would make that mess equal to the expression given in the problem.

Last edited: Apr 12, 2014
4. Apr 12, 2014

tman12321

You should not be using arccos. Tan can be expressed as sines and cosines, and you already know that cos(x) = 2p/a-1, where x are the angles at which the comet crosses earth's orbit. Use the Pythagorean theorem to find sin(x) in terms of square roots and 2p/a etc. Your trig functions will go away.

5. Apr 12, 2014

cpburris

Oh alright that makes sense I'll give that a try thanks

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