Time to gravitationally collapse ( Derivation)

[ibysaiyan]

Homework Statement

Hi, I have been given a question where they want me to find out the time at which gravitational collapse occurs for a given radius. So for reference I had a read through my notes... most of it I understand with the exception of a bit which I will highlight below. Thanks

Homework Equations

The Attempt at a Solution

Assume an object with all of it's mass concentrated at the center ( center of mass) of radius 'R'.
So using Newton's law we get the following relationship:

F = m$r^{..}$ = -GmM/r^2

=> $r^{..}$ = -Gm/r^2

Here's the bit which confuses me ( from bold line on wards).
=>{ $2r^{.}$$r^{..}$ = $2r^{.}$-Gm/r^2 } *

Integrate both sides w.r.t ( no idea , could it be r' ? )

$r^{.}$ = -2Gm/r + C , and d(r)/dt = -1/r^2 * ($r^{.}$

Can someone shed some light on this ? I have found another way of deriving the 'time' but I would love to know the above as well.

Thanks!

 

[ibysaiyan]

P.S(besides the derivation ):
The time I get for Earth to collapse to a radius of 100m is about 352 seconds which seems absurd.. the formula I used is:

t =$\frac{2}{3}$ * $\frac{R^3/2}{\sqrt{2GM}}$
where M = 5.9*10^-24 kg , radius = 6378km*10^3 - 100m...

I have also found another derivation, presumably of the same thing http://burro.cwru.edu/Academics/Astr221/LifeCycle/collapse.html

What am I doing wrong.

 

[ibysaiyan]

Can anyone lead me into the right direction. Thanks

 

[ibysaiyan]

Anyone ?

 

[vela]

You have
$$\ddot{r} = -\frac{GM}{r^2}$$The problem is the derivative is with respect to time. Unfortunately, you can't integrate with respect to time because you don't know what r(t) is yet. The trick is to multiply both sides by $\dot{r}$ to get
\begin{align*}
\int \dot{r}\ddot{r}\,dt &= -\int \frac{GM}{r^2}\dot{r}\,dt \\
\int \dot{r}\frac{d\dot{r}}{dt}\,dt &= -\int \frac{GM}{r^2}\frac{dr}{dt}\,dt \\
\int \dot{r}\,d\dot{r} &= -\int \frac{GM}{r^2}\,dr
\end{align*}