Nonuniform density of a planet

  • Thread starter WraithM
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  • #1
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I was thinking about a situation like this: You have a planet that has a nonuniform density (mass) like this [tex]\rho = a - r[/tex] where a is just some number and r is the radius away from the center of the planet. Then you have an object falling through the planet. I had no idea how to solve this differential equation. I don't even know if there is a solution. You'd have an F=ma like this: [tex]-\frac{G(a - r)(\frac{4}{3}\pi r^3)m}{r^2} = ma[/tex]. That becomes through simplification: [tex]\frac{d^2 r}{dt^2}=\frac{4}{3}\pi a G(r^2 - r)[/tex]. I don't really know much beyond solving simple differential equations. Is this solvable? If there is a solution, how do you get it?
 

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  • #2
tiny-tim
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[tex]\frac{d^2 r}{dt^2}=\frac{4}{3}\pi a G(r^2 - r)[/tex]. I don't really know much beyond solving simple differential equations. Is this solvable? If there is a solution, how do you get it?
Hi WraithM! :smile:

Use the standard trick of writing dr/dt = v, and then

d2r/dt2

= dv/dt

= dv/dr dr/dt [chain rule]

= v dv/dr :wink:
 
  • #3
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Does that actually work? Do you just separate variables from there? I see the logic of making [tex]\frac{d^2 r}{dt^2} = v \frac{dv}{dr}[/tex], but I don't want to believe that it's that easy.

For example, take [tex]\frac{d^2 r}{dt^2} = g[/tex] (constant acceleration). Separating variables using v dv/dr: v dv = g dr, that's: v^2/2 = gr
[tex]\frac{dr}{dt} = \sqrt{2gr}[/tex]
[tex]\frac{dr}{\sqrt{2gr}} = dt[/tex]
[tex]1/g \sqrt{2gr}=t[/tex]
[tex]r = \frac{gt^2}{2}[/tex]
Wow, so that actually works?

:/ I'm feeling stupid for not knowing this.

EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?
 
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  • #4
tiny-tim
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Wow, so that actually works?
:biggrin: Woohoo! :biggrin:

(btw, that's how 1/2 mv2 gets into energy equations :wink:)
EDIT: I just did this method with my first example, and I get an integral that isn't doable... :/ Am I just stuck then?
hmm … let's see …

you didn't get vdv = 4/3 πaG (r2 - ar) dr ? :confused:
 
  • #5
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Well, yes. I did get: [tex]v \, dv = \frac{4}{3}Ga(r^2 - r)dr[/tex]
You integrate from ([tex]r_{0}[/tex] to [tex]r[/tex]) that to get [tex]v = \sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}[/tex]. That's all good, but then if you try to go another step further, it blows up. Now I'm solving for r(t). You get: [tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt[/tex] I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?
 
  • #6
tiny-tim
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… You get: [tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{8}{3}Ga(r^3/3 - r^2/2 - r_{0}^3/3 + r_{0}^2/2)}} = \int_{0}^{t} dt[/tex] I don't know how to do that integral. I don't think it's doable. Do you know how to do it? Am I perhaps doing it wrong?
ah, I didn't realise you wanted to go as far as solving for r as a function of t :redface:

Yes, that looks correct … unfortunately, not all integrals are "doable" :smile:
 
  • #7
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The equation of motion is not right, to start with.
If the density is not uniform, the force acting on the mass when is at distance r from the center will be
G*m*M(r)/r^2
where M(r) is the mass contained in the sphere of radius r.
This is not 4/3pi*r^3*density . Density is not constant. You need to integrate from 0 to r in order to find the mass.
M(r) will be 4*pi(a/3*r^3-r^4/4) - you should do it again do double check.
But before you do this, check again your density. Of course, if you are interested in a math problem only, it does not matter.
But for a physics problem...
density=a-r ?
So density is zero at r=a and then it becomes negative for r<a.
Not mentioning the units. You will have density equal to distance. (a and r are disatnces, I suppose) Not very physical...
 
  • #8
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:( You're absolutely right... I completely forgot about that... I was more interested in the math at this point. It's sort of analogous to non-uniform charge distributions and using Gauss's law. You need to integrate, but thank you! I was being stupid :/
 
  • #9
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You still end up with an integral in the same sort of form as above.
[tex]\int_{r_{0}}^{r(t)}\frac{dr}{\sqrt{\frac{2}{3} \pi G(r^3 - 2ar^2-r_{0}^3+2ar_{0}^2)}}=\int_{0}^{t}dt[/tex]
 
  • #10
3,732
414
If you expect a nice formula as a result then you'll be disappointed. Few problems will allow for such a thing. I suppose this is not one of them.
The integral that you show in the last post can be solved numerically or maybe can be expressed in term of elliptical functions, I am not sure. Try Mathematica or Maple.
 

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