Find time, given acceleration as a function of position.

1. Jan 2, 2013

tent

Hello guys, the following problem I've had is from Beer and Johnston, Dynamics:

1. The problem statement, all variables and given/known data
At a distance R above the Earth's surface, a particle at rest is let go. Given the gravitational acceleration: $a =-g\frac{R^2}{r^2}$ where r is the distance from the center of the Earth to the particle, R is the Earth radius and g is the gravitational acceleration at the surface, calculate (a) the velocity of the particle when it hits the Earth, (b) the time it takes to fall to the surface. (Earth radius = $R = 6.380 km$, $g=9.81 m/seg^2$) Results: V=7910 m/s, t=34.5 m = 2070 s

2. Relevant equations
1.$a =-g\frac{R^2}{r^2}$
2.$a = vdv/dr$
3.$v = dr/dt$
4.$a = dv/dt$

3. The attempt at a solution
Okay, so for (a) what I did was integrating the acceleration using 2. and get velocity as a function of r or the distance Earth center - particle from $r$ to $r_{0}$, these being R to 2R, $\int-gR^2/r^2 dr$ and I got $V(r)=\sqrt{2gR^2(\frac{1}{R} -\frac{1}{2R})}$ since $V_{0}$ is 0 because the particle is initially, at r=2R and t=0, at rest. For r=R, it gave me 7911 m/s which is the solution the book gives. Now as for the time, I tried integrating the indefinite integral of the acceleration again, to get the time, by using 3. dt=dr/v, which is $\int-gR^2/r^2 dr$ and gives $V(r)=\sqrt{2gR^2/r}+C$ that didn't work, because when I integrate V, I have the constant, and to get that constant I did V(r=2R)=0, because it is at rest, that gave me C=-7911, and that's is impossible because I tried V(r=R) which should be 7911, but it gave me 3277, so I don't know what I'm doing wrong. I tried the other way, by doing V(r=R)=7911, but the constant didn't correspond to V(r=2R)=0.
The result is 34.5 m or 2070 s. I also tried integrating the acceleration over time, by using 4., but that got me nowhere, since I would have to use double integrals because r varies. Though I'm not really sure about this last part.

2. Jan 2, 2013

Staff: Mentor

How do you get two different values for V(r)?
You can use your first result.

dr/dt = f(r) => dt/dr = 1/f(r)
Integrate, and you are done.

3. Jan 2, 2013

tent

What I did was integrate a over r to get the velocity as a function of the position, then use the position 2R, the initial one, and R, the point where it hits the surface, as limits in a definite integral. That gives the the answer of the velocity. Now the only thing I could do with the time is use V=dr/dt because a=dv/dt won't get me anywhere as I'll have to integrate acceleration over time, and they give it to me as a function of the position. So the only way I see to do this is V=dr/dt.

I'm not sure how to explain, basically you indefinitely integrate acceleration over position, then you find the constant, but using the data they give you to determine the constant, first that V(r=2R)=0 and then using the one you got as a solution V(r=R)=7911, don't give the same velocity at a given position, because each constant I get from those two known situations gives me a different constant. In order to integrate the velocity in dt=dr/v(r), I need that constant. Or maybe I don't, I'm not sure about that.

4. Jan 2, 2013

MrWarlock616

Okay, so integrating $dt=\frac{dr}{v(r)}$, we get $\int_0^t\,dt=\int\frac{1}{v(r)}dr$ (the limits on RHS are actually from 2R to R, I couldn't do it on LateX). This should give you the answer, since you already have the velocity equation in terms of r..

5. Jan 2, 2013

tent

But, how could I integrate V(r) if I don't have the constant? Don't I need it? There's something I'm not getting here.

6. Jan 2, 2013

MrWarlock616

You don't need the constant as it's a definite integral from 2R to R.

7. Jan 2, 2013

tent

It is a definite integral when I do dt=dr/V(r), but in order to get V(r), what do I have to do? Because the first velocity I got was a definite integral to the final point from the starting point $\int^{R}_{2R}a(r)dr$, and I can't integrate the result of a definite integral which is a value, so I need an indefinite integral of the acceleration to get the velocity, and then find the constant of that indefinite integral in order to definitely integrate it in dt=dr/V(r),$\int^{R}_{2R}dr/V(r)$. Either that or I'm not quite grasping the concept of integrals.

8. Jan 2, 2013

MrWarlock616

The equation $\frac{v^2}{2}=\int \frac{-gR^2}{r^2}dr$ yields a constant which can be found by putting v=0 and r=2R. Then you get the final velocity equation in terms of r, with no arbitrary constants.

9. Jan 3, 2013

tent

So, I tried that, I put those two and got C=-7911 and then tried to integrate it in dt=dr/v(r) between upper limit R and lower 2R, to the final point from the starting one, but I got a problem, how do I integrate that$\int^{R}_{2R}\frac{dr}{\sqrt{\frac{2gR^2}{r}}-7911}$ so I first tried searching immediate ones, but couldn't find any, then I separated the sqrt in two and multiplied 7911 by sqrt(r) and then passed sqrt(r) to the dividend where dr is. Then I tried doing that, but got nowhere and used wolfram alpha, and what I got was just ridiculous, the result was billions and billions and fractions the order of $10^{-9}$. I mean it can't be that complicated, I don't know why am I getting this kind of answers, maybe the constant isn't that one, and I should try putting V=7911 for r=R, but then I'd get a different result, and I don't know why that happens, the constant is a constant, V(r) is a function and as such, knowing that there are 2 values of r that satisfy it, they should give the same constant. I think I completely misunderstood something and it's now messing up the whole calculations.

10. Jan 3, 2013

MrWarlock616

Yes, I know. I'm sorry I cannot help you with the integral. We just have to wait for a more experienced user to comment on this.

11. Jan 3, 2013

rcgldr

12. Jan 3, 2013

Staff: Mentor

You have the result for the velocity already. For arbitrary r,

$$V(r)=-\sqrt{2gR^2(\frac{1}{r} -\frac{1}{2R})}$$

Notice the minus sign, since the velocity is in the negative r direction. So,
$$\frac{dr}{dt}=-\sqrt{2gR^2(\frac{1}{r} -\frac{1}{2R})}$$

13. Jan 3, 2013

MrWarlock616

The math for finding the time is really annoying.

14. Jan 3, 2013

Staff: Mentor

I don't think so. Integrating the equation I gave in my previous post is pretty straightforward using a couple of substitutions.

Chet

15. Jan 7, 2013

tent

Sorry I'm posting so late, but I couldn't do this earlier. Now I tried 2 different websites that compute integrals, and also tried it myself but got nowhere. I computed 3-4 different expressions of that integral and substituted r for R, but I either got -358 I think or 3307, which is about 55 minutes. I also looked at the link that rcgldr posted, but it led me nowhere since I couldn't understand how he did the last integration. If anyone knows how to integrate that, please give me some steps or some link that explains how to integrate this step by step. Thank you.

16. Jan 7, 2013

Staff: Mentor

$$\frac{dr}{dt}=-\sqrt{2gR^2(\frac{1}{r} -\frac{1}{2R})}$$
Rewrite right hand side with least common demoninator:

$$\frac{dr}{dt}=-\sqrt{gR\frac{(2R-r)}{r}}$$

Re-express as follows:

$$\sqrt{\frac{r}{2R-r}}dr=-\sqrt{gR}dt$$

Substitute $2R-r = y^2$: This gives:

$$2\sqrt{2R-y^2}dy=\sqrt{gR}dt$$

The limits of integration on this equation are from 0 to t and from y = 0 to y = √R

Now substitute:

$$y=\sqrt{2R}\sin{\theta}$$

This gives:

$$4\cos^2{\theta}d\theta=\sqrt{\frac{g}{R}}dt$$

where θ goes from 0 to π/4

This equation can be rewritten as:

$$(\cos2\theta+1)d(2\theta)=\sqrt{\frac{g}{R}}dt$$

Carrying out the integration between the indicated limits gives:

$$1+\frac{\pi}{2}=\sqrt{\frac{g}{R}}t$$

So,

$$t=(1+\frac{\pi}{2})\sqrt{\frac{R}{g}}$$

I think I did the "arithmetic" correctly.[/QUOTE]