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This is was an exercise on my exam of classical mechanics which I couldn't solve. The question goes as follows:

Imagine a comet moving on a parabolic orbit in the plane of the earth. Take the earth's orbit as a circle with radius R and the sun in the centre. The shortest distance that the comet comes to the sun is $\beta R$ with $\beta<1$. So the comet is on the inside of the earth's orbit for a certain time. Prove that the time the comet spends in the circle is $$ \Delta t = \frac{1}{3\pi}\sqrt{2(1-\beta)}(1+2\beta) \times \text{year} $$

Hints: $$ \int \cos^{-4}x\,dx= \tan{x}+ \frac{\tan^3{x}}{3} + C, $$ $$ \cos{2x}=1-2\sin^2{x} $$

I tried calculating the path of the comet with the polar equations of a hyperbola, but I have no clue how to solve this.

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    $\begingroup$ Imagine a comet moving on a parabolic orbit ... I tried calculating the path of the comet with the polar equations of a hyperbola. A hyperbola is not a parabola. $\endgroup$
    – G. Smith
    Jan 3, 2021 at 17:55
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    $\begingroup$ Wouldn't you have to start with defining the comet's parabola? $\endgroup$
    – Gert
    Jan 3, 2021 at 18:03
  • $\begingroup$ Use a constant. Your end result will then contain it (but that doesn't show in the solution you quoted...) $\endgroup$
    – Gert
    Jan 3, 2021 at 19:00
  • $\begingroup$ Since it is parabolic, there is an extra condition, i.e. total energy E(t) = 0. This condition together with the nearest distance r = beta R is able to find the velocity at the nearest point. The nearest distance and its velocity fix the equation of parabola y = c x^2. $\endgroup$
    – ytlu
    Jan 3, 2021 at 19:00
  • $\begingroup$ @ytlu $y=cx^2$ means the comet crashes into the sun. $\endgroup$
    – Gert
    Jan 3, 2021 at 19:06

2 Answers 2

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Interesting excercize. Kepler's Third Law for the orbit of the Earth gives us $$ T^2_\text{e} = \frac{4\pi^2}{k}R^3\tag{1}, $$ with $k=GM_\odot$ (ignoring the mass of the Earth). The Kepler orbit of a parabola in polar coordinates is $$ r = \frac{h^2}{k}\frac{1}{1 + \cos\theta},\tag{2} $$ where $h$ is the specific angular momentum. The closest approach occurs at $\theta=0, r=\beta R$, from which be obtain $$ h^2 = 2\beta Rk\tag{3}. $$ Let's call $\theta=2x$. At $r=R$, $$ 1 + \cos 2x_\max = 2\cos^2x_\max = 2\beta.\tag{4} $$ From $$ h = r^2\dot{\theta},\tag{5} $$ we get $$ \sqrt{2\beta Rk}\text{d}t = \frac{(2\beta R)^2}{(1+\cos\theta)^2}\text{d}\theta.\tag{6} $$ Eliminating $k$ using $(1)$, and integrating, leads to $$ T = \frac{2T_\text{e}}{\pi}\int_0^{x_\max}\frac{(2\beta)^{3/2}}{(1+\cos 2x)^2}\text{d}x.\tag{7} $$ The rest is straightforward.

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  • $\begingroup$ The quation 1 is irrelavant. You can start with r = c / (1+ cos theta), and fix the constant c at r = beta R. and theta = 0. $\endgroup$
    – ytlu
    Jan 3, 2021 at 19:46
  • $\begingroup$ I shouldn't say eqaution (1) is irrelavant. We need it to convert the unit to year in the final result.. $\endgroup$
    – ytlu
    Jan 4, 2021 at 13:00
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Comet, Earth

I don't have a full solution but would attempt the following in Cartesian coordinates.

  1. determine the coordinates of points $A$ and $B$, from:

$$x^2+y^2=R^2$$ and: $$y=cx^2+\beta R$$

So that:

$$\frac{y-\beta R}{c}+y^2=R^2\tag{1}$$

The two roots of $(1)$ give the coordinates of points $A$ and $B$

  1. determine the path length between $A$ and $B$ (with integration, say $\Delta L$).
  2. determine the velocity-time function $v(t)$ (I'm not sure how to go about this).

Now if the total time traveled between $A$ and $B$ is $\Delta t$, then:

$$\int_0^{\Delta t}v(t)\text{d}t=\Delta L$$

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  • $\begingroup$ Your equation didn't match with the figure. Looking your figure, it should be y = beta R - c x^2. $\endgroup$
    – ytlu
    Jan 3, 2021 at 19:58
  • $\begingroup$ The velocity will be found from the condition that total energy = 0, for case of a parabolic orbital. The total energy is a constant indenpend of time. $\endgroup$
    – ytlu
    Jan 3, 2021 at 20:00

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