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Consider a test mass starting at rest from radius R to a point planet of mass M. What is the time taken for the test mass to reach the planet?

I saw somewhere on the internet (don't ask where please :|) that one can simply use Kepler's law in this situation - i.e. time taken is simply 1/4 of the period of an elliptical orbit with half-major axis length R:

$$ T = \frac{\pi}{2}\sqrt{\frac{R^3}{GM}}$$

BUT if I manually integrate the kinetic energy equation,

$$ \frac{1}{2}\dot{r}^2 = GM\left(\frac{1}{r} - \frac{1}{R}\right) $$ $$ \rightarrow T\sqrt{2GM} = \int^{R}_{0}{\frac{\mathrm{d}r}{\sqrt{\frac{1}{r} - \frac{1}{R}}}}$$ $$ = \space...\space = R\int^{R}_0{\sqrt{\frac{1}{r} - \frac{1}{R}}\space\mathrm{d}r} = \space... \space = R\int^{\infty}_0{\frac{2u^2\space\mathrm{d}u}{\left(u^2+\frac{1}{R}\right)^2}} $$ $$ = \space...\space = R\int^{\infty}_0{\frac{\mathrm{d}u}{u^2+\frac{1}{R}}} = \frac{\pi}{2}R^{\frac{3}{2}}\space \therefore \space T = \pi\sqrt{\frac{R^3}{8GM}}$$

Why is there a $\sqrt{2}$ factor discrepancy? Is it because Kepler's law is invalid when the minor axis length is zero? Thanks for any insight.

Qmechanic
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meowgoesthedog
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1 Answers1

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The approximation is not quite right. First, let's look at a typical elliptical orbit of a blue planet (test mass) around a yellow star (of mass $M$): elliptical orbit I'm using $R$ as the starting distance from the star to distinguish it from the major axis of an elliptical orbit. These are only the same for a circular orbit. For a collision, we'll approximate this as when the planet reaches a vertical line parallel to the minor axis and going through the sun. As you can see from the above diagram, this will take more than a quarter of the orbital period.

Shrinking the orbit down even further: very elliptical orbit The sun remains at one of the foci of the ellipse, which move closer to the edges of the ellipse as the eccentricity increases. So, the sun is more and more off-center from the ellipse, which means that the major axis shrinks compared to the planet-Sun distance.

Finally, in the limit of zero transverse velocity: linear "orbit" we have an "orbit" with zero minor axis length and a semimajor axis length of half the starting distance. By comparing with the previous orbit, we can also see that we want half an orbital period to reach collision.

So, the time to collision is \begin{align} t_{collision} = \frac{1}{2}T &= \frac{1}{2}\left(2\pi\sqrt{\frac{r^3}{GM}}\right) \\ &= \pi\sqrt{\frac{(R/2)^3}{GM}}\\ &= \pi\sqrt{\frac{R^3}{8GM}} \end{align} where $r$ is the semimajor axis of the degenerate ellipse. This agrees with your direct calculation.

Mark H
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  • AHHH OF COURSE so obvious, thank you very much facepalm – meowgoesthedog Feb 01 '17 at 08:22
  • @willywonkadailyblah I first saw this problem years ago in grad school and ended up taking the integral long-way-round, too (with help from Wolfram Alpha). It's only obvious after you see it. Before breaking out the equations to solve a problem, take some time to doodle first. – Mark H Feb 01 '17 at 09:14
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    I was just being stupid and somehow forgot the planet is at the focus not the center... serves me right for doing this at 1am. Anyways thanks again – meowgoesthedog Feb 01 '17 at 11:56
  • One more question though - doesn't the ellipse solution $ r_0 = r(1+e\cos{\theta}) $ blow up at $ r = 0 $? how come Kepler's law still holds in this limit? – meowgoesthedog Feb 01 '17 at 12:00
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    @willywonkadailyblah Everything blows up at $r=0$ since the gravitational force becomes infinite there. That's why we take a limit, so $r\neq 0$. Another thing to consider is that the angular momentum of the test mass is constant, so the procedure above takes the limit as angular momentum goes to zero. Kepler's laws work for any non-zero angular momentum, so taking the limit gives a valid result. – Mark H Feb 01 '17 at 20:49