# Kepler problem: Time spent by a comet inside the orbit of the Earth [closed]

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.

• 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. Jan 3, 2021 at 17:55
– Gert
Jan 3, 2021 at 18:03
• Use a constant. Your end result will then contain it (but that doesn't show in the solution you quoted...)
– Gert
Jan 3, 2021 at 19:00
• 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.
– ytlu
Jan 3, 2021 at 19:00
• @ytlu $y=cx^2$ means the comet crashes into the sun.
– Gert
Jan 3, 2021 at 19:06

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.

• 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.
– ytlu
Jan 3, 2021 at 19:46
• I shouldn't say eqaution (1) is irrelavant. We need it to convert the unit to year in the final result..
– ytlu
Jan 4, 2021 at 13:00 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:

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

• Your equation didn't match with the figure. Looking your figure, it should be y = beta R - c x^2.
– ytlu
Jan 3, 2021 at 19:58
• 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.
– ytlu
Jan 3, 2021 at 20:00