# How long does it take a object captured by a star falling to the center?

If the captured object do not have tangential velocity, it's just the free-fall time. But when it has, it may take longer time to fall in, right ?

The function should be

$\ddot{r} = -GM/r^2 + (v_0r_0/r)^2 / r = -GM/r^2 + v_0^2r_0^2 / r^3$ ,

where v_0 is the initial tangential velocity . After one integration, it becomes

$\dot{r}^2/2=GM(1/r-1/r_0)-v_0^2r_0^2(1/r^2-1/r_0^2)/2$ .

I don't know how to deal with it. But I guess there is a analytic solution.

Anyone knows something about it ?

• Without losing energy the object should just keep orbiting. What is that $(v_0 r_0 / r)^2 / r$ term? Centrifugal acceleration? – kennytm Dec 6 '10 at 11:39
• @KennyTM: yes. The equation comes from the standard Newtonian potential for spherical systems. – Marek Dec 6 '10 at 11:44
• when its tangential velocity is less than the critical velocity, it will fall into the center early or late. And the $(v_0r_0/r)^2/r$ is the part of gravity to capture it. angular momentum conservation is considered. – gerry Dec 6 '10 at 11:47
• @Marek: Ah right. I thought $v_0$ is the initial total speed. – kennytm Dec 6 '10 at 12:40

Now, supposing you already determined that the object will fall into the star, you can solve for the point of crossing star's surface (this is just geometry, intersection of the conic section and the circle) and then solving for the time of arrival into that point. All of this is simplified by noting that you only need to know the $r$-coordinate which is simply the radius of the star.
So this all boils down to finding a solution to your equation. You can do that just by taking a square root, separating variables and computing the integral. That will give you dependence of $t = t(r)$. So then just plug in the radius of the star and you're done.