# Time taken for a body to crash into the sun [closed]

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.

• 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? Jun 10, 2021 at 11:07
• I think, you are missing a square in your (1.). Jun 10, 2021 at 11:08
• @CharlesTucker3 I didn't get you very clearly, the mass has no initial velocity. Jun 10, 2021 at 11:13
• "For all ellipses with a given semi-major axis the orbital period is the same, regardless of eccentricity." - en.wikipedia.org/wiki/Orbital_period Jun 10, 2021 at 11:41
• @shadow I'm long out of school, which means I don't have to do integrals any more :-) . Jun 10, 2021 at 14:18

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.

• Wow, that's awesome, never thought about that approach, thank you very much. Jun 10, 2021 at 12:36
• Yes, it's quite a neat hack I found out about from an IPhO problem (Question 3 - IPhO 2012). Jun 10, 2021 at 12:38
• Note that if you use units of au & years, then $\mu=GM=4\pi^2$. Jun 10, 2021 at 12:49
• “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 ... Jun 10, 2021 at 13:35
• @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. Jun 10, 2021 at 15:09

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.

• Wait, is that what you get if you integrate that mess (i.e. (3) of my approach), in a more general form? Jun 10, 2021 at 15:38
• @shadow, yes. Take a look at the linked page Jun 10, 2021 at 16:01
• @PM2Ring That would take work, and I'm deathly allergic! The wikipedia page source might have the mathjax code Jun 10, 2021 at 16:01
• @Shadow That integral is... unpleasant, although Wolfram Alpha, etc, can do it. And it's easy enough to verify by taking the derivative. Jun 10, 2021 at 16:25
• @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 $\cos^2\theta$. Jun 11, 2021 at 9:39