1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Nonuniform density of a planet

  1. Apr 29, 2009 #1
    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?
     
  2. jcsd
  3. Apr 29, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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:
     
  4. Apr 29, 2009 #3
    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?
     
    Last edited: Apr 29, 2009
  5. Apr 30, 2009 #4

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:

    (btw, that's how 1/2 mv2 gets into energy equations :wink:)
    hmm … let's see …

    you didn't get vdv = 4/3 πaG (r2 - ar) dr ? :confused:
     
  6. Apr 30, 2009 #5
    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?
     
  7. Apr 30, 2009 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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:
     
  8. Apr 30, 2009 #7
    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...
     
  9. Apr 30, 2009 #8
    :( 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 :/
     
  10. Apr 30, 2009 #9
    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]
     
  11. May 1, 2009 #10
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Nonuniform density of a planet
  1. Bulding Planets (Replies: 5)

  2. Revolution of planets (Replies: 3)

Loading...