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I want to figure out the motion of a comet that can somehow pass freely through the sun (mass much, much greater than the comet's so that its movement is negligible) if it starts out stationary relative to the sun. This will be oscillatory motion, so I wish to calculate its period.

First Approach

It can be seen that the period will be $4T$, where T is the time it takes for the comet and the sun to collide. My reasoning for this is that it will start from $R_{max}$ and then reach the sun and then go back out to $R_{max}$ (now on the other side), and it repeats this. Because of the conservation of energy, the time of oscillation is $4T$. It is well known that the time of collision of two mutually-attracting bodies is $T = \frac{\pi R_{max}^\frac{3}{2}}{2\sqrt{2GM}}$, so the period of this motion is $\frac{2\pi R_{max}^\frac{3}{2}}{\sqrt{2GM}}$

Second Approach

This is effectively Kepler's problem with angular momentum $l = 0$.

Because this is a Keplerian problem, the comet will orbit in an ellipse (we are assuming a bounded orbit). If we take the limit of $l$ as it goes to $0$, one finds that $2a = R_{max}$, where a is the semi-major axis of the orbit. Therefore, according to Kepler's Third Law of Planetary Motion, the period of this orbit (and thus of this motion) is $\tau = \frac{\pi R_{max}^\frac{3}{2}}{\sqrt{2GM}}$.

The two answers I achieved are off by a factor of $2$. Also, the first answer states that the motion is a line of length $2R_{max}$ and the sun is at the center. But this is a Kepler problem, so shouldn't the sun be at the end of the a line with length $R_{max}$, like in the second answer? I am wondering why the second approach is incorrect.

Edit:

I am wondering why the first event is the actual thing that occurs. Why won't the point-mass comet start from $R_{max}$ and then travel to the singularity, stop suddenly (make a "hairpin" turn), then shoot back?

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2 Answers 2

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The definitions of $R_{max}$ are different in your two cases.

In the first you count the maximum distance from the central sun.

The second is twice the $R_{max}$ of the first , because it has to be counted over twice the distance defined in the first as the sun is at the focus of the elipse and in a limiting case at 2$R_{max}$ of the first.

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The second approach is incorrect, because for a Kepler elliptical orbit the sun must be at one focus, which for an ellipse that approaches a straight line would be at one end of the ellipse. The problem you are trying to solve, has the sun is at the center of the ellipse, so in fact, it is a different problem

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  • $\begingroup$ How can we show that the first approach is, in fact, the physical occurrence? How can we show that it is the case without "guesswork?" $\endgroup$
    – Hypernova
    Commented Feb 20, 2020 at 2:48
  • $\begingroup$ What about the first approach do you consider a guess? Your T is correct, and 4T is correct for the 4 legs of motion that make up a period. $\endgroup$
    – Bill Watts
    Commented Feb 20, 2020 at 3:17
  • $\begingroup$ I understand why the first scenario physically occurs. It is quite easy to see, via conservation of energy, that energy and displacement of this problem are finite. However, if one tries to compute the velocity at the singularity, it is infinite (both with the conservation of energy and the second-order nonlinear ODE say this). However, if the comet had infinite velocity at any one point, then there would be no force to stop it. However, one could argue that the gravity force is infinite at $r = 0$, but I do not like doing $\infty - \infty$. $\endgroup$
    – Hypernova
    Commented Feb 20, 2020 at 4:34
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    $\begingroup$ It is an integrable infinity. Check out physics.stackexchange.com/questions/14700/… $\endgroup$
    – Bill Watts
    Commented Feb 20, 2020 at 4:38
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    $\begingroup$ If you work the problem so the comet really enters the interior of the sun, the velocity never gets infinite, because the force at the center of the sun is zero. I have assumed you are approximating point masses. $\endgroup$
    – Bill Watts
    Commented Feb 20, 2020 at 4:48

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