How long does it take for a falling particle to reach the halfway point?

[SlickJ]

I had this question in class the other day and need some help:

A particle falls to Earth starting from rest at a great height. Neglecting air resistance, show that the particle requires about 9/11 of the total time of fall to traverse the first half of the distance.

Anything would be appreciated.

Thanks!

 

[Integral]

That is a very hard thing to show... because it is not correct.


The equation of motion for a falling body, starting from height D is:

$$h = D - \frac 1 2 g t^2$$

so the time to reach the ground is

$$t_0 = \sqrt { \frac { 2 D} g }$$

The time to reach the half way point is:

$$\frac D 2 = D - \frac 1 2 g t^2$$

or

$$t_{\frac D 2} = \sqrt {\frac D g }$$

So

$$\frac {t_{\frac D 2}} {t_0} = \frac 1 {\sqrt 2} = .707$$

this is not $\frac 9 {11}$

 

[Tide]

Integral,

When the problem states "from a great height" I interpret that as meaning the inverse square behavior of gravity must be taken into account. Ultimately, this means comparing

$$\int_{1/2}^{1} \frac {dr}{\sqrt{ \frac {1}{r} -1}}$$

with

$$\int_{0}^{1} \frac {dr}{\sqrt{ \frac {1}{r} -1}}$$

which works out to about 9/11.

 

[Integral]

Ok,
That makes sense.

 

[SlickJ]

Thank you both.
I was pursuing Integral's method at first before running into the same problem. I agree with Tide's reasoning but I don't follow how you got to that integral, and even more so how I would solve that integral. Could you help with that?
Thanks

 

[Tide]

Slick,

Energy will be conserved so we can relate speed with position:

$$v = \sqrt { 2GM \left( \frac {1}{r} - \frac {1}{r_0} \right)$$

But $v = \frac {dr}{dt}$ so we can integrate the equation and find

$$t = \frac {1}{\sqrt {2GM}} \int_{r}^{r_0} \frac {dr'}{\sqrt { \frac {1}{r'} - \frac {1}{r_0}}}$$

Use $r = 0$ to get the full time to fall and $r = r_0/2$ to get the time to fall halfway. It is helpful to use a new variable of integration given by $r/r_0$. The rest of the problem reduces to getting approximations for the integrals though the integral from 0 to 1 reduces to something called a Beta Function (with a little effort you can cast it into a form that you can look up in integral tables) and the integral from 1/2 to 1 is reduces to an Incomplete Beta Function and you might have to use a numerical approxiimation to evaluate it. Some graphing calculators will do them both numerically for you.

 

[SlickJ]

Ohhh, ok. Thank you Tide, much appreciated. I just missed that leap from setting up the integral to what you posted I guess. Thanks everyone.