Free Fall Timescale

1. Oct 31, 2009

Stratosphere

I've seen this equation in a book once but I can't seem to find a good explanation of how it was derived.

$$\ t_{ff}=\sqrt{\frac{3\pi}{32g\overline{\rho}}}$$

Last edited: Oct 31, 2009
2. Nov 1, 2009

Staff: Mentor

See: http://en.wikipedia.org/wiki/Free-fall_time" [Broken]

Last edited by a moderator: May 4, 2017
3. Nov 1, 2009

Stratosphere

In the book I have it derives it from the acceleration due to gravity equations. There is only one part about it I dont get though.

$$-\frac{GM}{r^{2}}$$ Then it says to do this.

$$\int^{R}_{t=0} -\frac{GM}{r^{2}}dr$$

The boundary conditions are r=R and when t=0, I don't understand what to do for the definite integral. I can integrate but I don't know what to plug in. The question you're suppose to get is different than the one I originally posted.

$$\frac{1}{2}\sqrt{\frac{R^{3}}{GM}}$$

4. Nov 1, 2009

Staff: Mentor

It's a bit of a calculus problem. You are trying to find the time it takes for the outer shell to collapse under the influence of the gravity of the mass below it. You have the acceleration of gravity (thus d2r/dt2) as a function of r:
Given that, you need to solve for the time it takes to go from r = R to r = 0, starting from rest.

5. Nov 1, 2009

Stratosphere

$$\int^{R}_{0} -\frac{GM}{r^{2}}dr$$

Like this? If I integrate this I get stumped at this,

$$-GM\left[-\frac{1}{r} \right]\right|^{R}_{0}$$

What do I plug in for r?

6. Nov 1, 2009

Staff: Mentor

No, you need to solve this equation:

$$\frac{d^2r}{dt^2} = -\frac{GM}{r^2}$$

One way to solve it is in stages, starting with the trick of using a = dv/dt = v dv/dr.

7. Nov 1, 2009

Stratosphere

Sorry I have to ask but, how do I solve it?

8. Nov 1, 2009

Staff: Mentor

It's a bit of a pain. I'll start you off:

$$\frac{dv}{dt} = v \frac{dv}{dr} = -\frac{GM}{r^2}$$

First solve that for v in terms of r. Then use v = dr/dt, and solve that to get the time.

9. Nov 1, 2009

Stratosphere

I've got that

$$\frac{dv}{dt}=-\frac{GM}{r^{2}}$$

$$v=\frac{\sqrt{GM}}{r}$$

$$\frac{dv}{dr}=-\frac{\sqrt{GM}}{r}$$

$$\frac{dr}{dt}=\frac{\sqrt{GM}}{r}$$

How do I get time?

10. Nov 2, 2009

Stratosphere

Did I do something wrong?

11. Nov 2, 2009

Staff: Mentor

How did you go from the first equation to the second? Please explain each step so I can see what you're trying to do.

12. Nov 2, 2009

Stratosphere

$$\frac{dv}{dt}=-\frac{GM}{r^{2}}$$

$$\int -\frac{GM}{r^{2}}dr$$

$$=v=\frac{\sqrt{GM}}{r}$$

$$\frac{dv}{dr}=-\frac{GM}{r^{2}}\cdot\frac{r}{\sqrt{GM}}$$

$$\frac{dv}{dr}=-\frac{\sqrt{GM}}{r^{2}}$$

13. Nov 3, 2009

Staff: Mentor

Sorry, but I'm still struggling to understand what you are doing here.
This is the starting point. OK.

Not sure what you mean here. Are you taking the integral of both sides with respect to dr? (Where's the equation?)

How did you get here?

14. Nov 3, 2009

Stratosphere

I took the integral of $$\frac{dv}{dt}$$ shouldn't that give me v? Maybe that's were I'm confused. $$\int-\frac{GM}{r^{2}}dr=\frac{GM}{r}$$.

It looks like $$\frac{dr}{dr}=GMLog(r)$$ to get that I did:

$$\int -\frac{GM}{r^{2}}dr=\frac{GM}{r}$$

$$\int\frac{GM}{r}dr=GMLog(r)$$

So $$v=\frac{GM}{r}$$,

15. Nov 3, 2009

Staff: Mentor

If you were integrating with respect to t that would be true, but you're integrating with respect to r.

$$\frac{dv}{dt} dr = v dv$$

See post #8.

16. Nov 3, 2009

Stratosphere

So you're saying that I need to integrate with respect to r? How do I do that?

$$-\frac{GM}{r^{2}}\cdot GMlog(r)$$

Is that right? Does $$dv=GMlog(r)$$?

17. Nov 3, 2009

Staff: Mentor

From an earlier post:

$$\frac{dv}{dt} = v \frac{dv}{dr} = -\frac{GM}{r^2}$$

Integrate both sides with respect to dr:

$$\int \frac{dv}{dt} dr= \int v \frac{dv}{dr} dr = \int -\frac{GM}{r^2} dr$$

$$\int v dv = \int -\frac{GM}{r^2} dr$$

(This is just the beginning. This is a several step calculus problem.)

18. Nov 3, 2009

Stratosphere

I will try to solve it but now I'm completely confused as to what I'm trying to do. Could you restate the problem?

19. Nov 3, 2009

Redbelly98

Staff Emeritus
The question is: How long does it take the radius to collapse from a value of R to a value of zero?

So, ultimately you are looking for a time as the answer.

20. Nov 4, 2009

Staff: Mentor

To add to Redbelly98's statement, take a look at post #4.

Think of it like this: Imagine you are standing on a planet that has just collapsed, so nothing is holding you up. Assume that all the mass has collapsed to a point. How long would it take for you to fall (you're in free fall, thus the name) from your initial position on the surface at r = R to the center of the collapsed planet at r = 0?