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A small body starts from rest towards the Sun from the position of Earth, find the total time it takes for the body to crash into the Sun (assume Sun is a point object and no other forces act on the system).

My approach to this question:

  1. $a = \frac{GM}{r^2}$ for the body at any r distance form Sun.
  2. Next I found out velocity, $v = \sqrt{2GM(\frac{1}{r} - \frac{1}{R})} $, where R is initial distance
  3. From here we get $\int_{R}^{0} \frac{dx}{v}$

Now how do I integrate this monstrosity? Is there any easier and faster way than the one above? Any help would be appreciated.

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  • $\begingroup$ But the earth is anyways in a free fall towards the sun, it just has enough tangential speed to miss it and keep an almost circular orbit... So you would need to decelerate your test mass? $\endgroup$
    – Charles Tucker 3
    Jun 10 '21 at 11:07
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    $\begingroup$ I think, you are missing a square in your (1.). $\endgroup$
    – Photon
    Jun 10 '21 at 11:08
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    $\begingroup$ @CharlesTucker3 I didn't get you very clearly, the mass has no initial velocity. $\endgroup$
    – Shadow
    Jun 10 '21 at 11:13
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    $\begingroup$ "For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity." - en.wikipedia.org/wiki/Orbital_period $\endgroup$
    – gandalf61
    Jun 10 '21 at 11:41
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    $\begingroup$ @shadow I'm long out of school, which means I don't have to do integrals any more :-) . $\endgroup$
    – Carl Witthoft
    Jun 10 '21 at 14:18
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This problem is usually tackled using Kepler's Third Law. The trajectory of this object falling into the sun is actually 'half' of a degenerate ellipse with semi-major axis equal to R/2.
From Kepler's Third Law: $ \frac{a^3}{T^2}=\frac{GM}{4 \pi^2} $ $$ T=\sqrt{\frac{4 \pi^2 a^3}{GM}} = \sqrt{\frac{\pi^2 R^3}{2GM}} $$ So the result is: $$ \Delta t = \frac{T}{2} = \frac{\pi}{2} \sqrt{\frac{R^3}{2GM}}$$ As far as I'm aware, it is very difficult to find the time it takes a body to travel between two arbitrary positions on an elliptical orbit (you'd basically have to compute the area swept by the body), so calculating that integral is probably hopeless.

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  • $\begingroup$ Wow, that's awesome, never thought about that approach, thank you very much. $\endgroup$
    – Shadow
    Jun 10 '21 at 12:36
  • $\begingroup$ Yes, it's quite a neat hack I found out about from an IPhO problem (Question 3 - IPhO 2012). $\endgroup$
    – S. Stefan
    Jun 10 '21 at 12:38
  • $\begingroup$ Note that if you use units of au & years, then $\mu=GM=4\pi^2$. $\endgroup$
    – PM 2Ring
    Jun 10 '21 at 12:49
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    $\begingroup$ “Calculating that integral is probably hopeless” ... except that we already know the orbital period of a body in an elliptical orbit with the same semi-major axis as the Earth’s orbit ... $\endgroup$
    – gandalf61
    Jun 10 '21 at 13:35
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    $\begingroup$ @PM2Ring Got it - yes, you are right., semi-major axis is 0.5 au It is the limiting case of a very tight sun-grazing orbit which whips around the far side of the sun and back again. $\endgroup$
    – gandalf61
    Jun 10 '21 at 15:09
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Surprise :-) , Wikipedia to the rescue.

The time $t$ taken for an object to fall from a height $r$ to a height $x$, measured from the centers of the two bodies, is given by: $$t={\frac {\arccos {\Big (}{\sqrt {\frac {x}{r}}}{\Big )}+{\sqrt {{\frac {x}{r}}\ (1-{\frac {x}{r}})}}}{\sqrt {2\mu }}}\;r^{3/2}$$ where $\mu =G(m_{1}+m_{2})$ is the sum of the standard gravitational parameters of the two bodies.

Glad they did it; I would hate to have worked out that antiderivative myself.

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  • $\begingroup$ Wait, is that what you get if you integrate that mess (i.e. (3) of my approach), in a more general form? $\endgroup$
    – Shadow
    Jun 10 '21 at 15:38
  • $\begingroup$ @shadow, yes. Take a look at the linked page $\endgroup$
    – Carl Witthoft
    Jun 10 '21 at 16:01
  • $\begingroup$ @PM2Ring That would take work, and I'm deathly allergic! The wikipedia page source might have the mathjax code $\endgroup$
    – Carl Witthoft
    Jun 10 '21 at 16:01
  • $\begingroup$ @Shadow That integral is... unpleasant, although Wolfram Alpha, etc, can do it. And it's easy enough to verify by taking the derivative. $\endgroup$
    – PM 2Ring
    Jun 10 '21 at 16:25
  • $\begingroup$ @Shadow Actually, the integral isn't too bad, once you realise that a trig substitution is involved. Let $x=r\cos^2\theta$, and then it's pretty straightforward, assuming you know how to integrate $\cos^2\theta$. $\endgroup$
    – PM 2Ring
    Jun 11 '21 at 9:39

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