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Finding the force on a particle inside the radius of the Earth

  1. Aug 15, 2009 #1
    1. The problem statement, all variables and given/known data
    The mass of Earth is M, the radius of earth is R. For a particle of mass m within the earth at a distance r from the center of the earth, the gravitational force attracting m toward the center of the Earth is [tex]F_r = \frac{-GM_rm}{r^2}[/tex], where Mr is the mass of the earth within a sphere of radius r. Show that [tex]F_r = \frac{-GMmr}{R^3}.[/tex]


    3. The attempt at a solution

    I am completely blanking on how to go about this - I could really use a hint to get me started. :confused:
     
  2. jcsd
  3. Aug 15, 2009 #2
    If the question is "what is the force as a function of r?", you will need Mr as a function of the radius.
     
  4. Aug 15, 2009 #3
    Shell Theorem may help you simplify the problem a lot.
     
  5. Aug 15, 2009 #4

    Redbelly98

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    They are assuming a uniform density for the Earth here. What is the Earth's density, given M and R?
     
  6. Aug 15, 2009 #5
    So if the density of the Earth is taken to be a constant, then we have [tex]\rho = \frac{M}{\frac{4}{3}\pi R^3}[/tex], and the mass of a thin shell is [tex] dm = 4\pi r^2 \rho[/tex]. So the total force towards the center experienced by a particle inside the earth at a radius r would be:

    [tex] Fr = - \frac{M}{\frac{4}{3}\pi R^3}\frac{4\pi Gm}{r^2}\int_{0}^{r}r^2 dr[/tex]. That gets me pretty close to the equation I'm going after! :approve:
     
    Last edited: Aug 15, 2009
  7. Aug 15, 2009 #6
    You are very close. Actually I think in it's current form it will give you the correct answer, but for the wrong reasons.
    But in any case, consider just the portion of your equation:
    [tex]4\pi \int^{r}_{0}r^{2} dr[/tex]

    what is this equal to?
     
  8. Aug 15, 2009 #7

    Redbelly98

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    [/URL]

    Looks good to me :smile: , but why stop there?
     
    Last edited by a moderator: May 4, 2017
  9. Aug 15, 2009 #8
    Sorry for the interjection.
    But is this really that difficult?

    If r is the radius at some place inside the earth where the mass m lies, then all you need to know is the mass of a planet with the same density as the earth, only with a radius r. Doesnt all the other mass outside of r cancel itself out as far as force on the mass m? or what am I missing?
     
  10. Aug 15, 2009 #9
    I believe I see what you're getting at: That integral is just the volume of the sphere with radius r, and the gravitational force exerted by a sphere with density [tex]\rho[/tex] and volume [tex]\frac{4}{3}\pi r^3[/tex] is just the volume times the density, plugged into [tex]\frac{GMm}{r^2}[/tex] with rho*v substituted for litte m. The integration wasn't really necessary.

    This question did get me thinking about the shell theorem, though, and I have a further question about the explanation at http://en.wikipedia.org/wiki/Shell_theorem . Right at the top of the page it says "A solid, spherically symmetric body can be modeled as an infinite number of concentric, infinitesimally thin spherical shells." However, in the diagram directly below, the shaded area represented as dM doesn't look at all like a spherical shell to me - it looks more like a thin "slice" cut through the axis of the sphere, like a slice taken out of a sphere-shaped bread loaf. What's going on here?
     
  11. Aug 15, 2009 #10
    Yeah, you're right. I was making the problem a lot more difficult than it really was. :yuck:
     
  12. Aug 15, 2009 #11

    Doc Al

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    The shaded portion is not the shell, but one thin infinitesimal slice from the shell. The outline of the shell itself is the black circle.
     
  13. Aug 15, 2009 #12
    Well you know what. I was looking at it from a symmetry point of view from just drawing on a 2-d piece of paper. But now that I think about it in 3-d its not that intuitive. I think the math gets more complex. So if you have to prove that before you do the problem the easy way... (ie just assuming all the mass cancels out based on 2-d symmetry)... yikes. It may be more difficult. I am no expert. just learning along with you and everyone else.
     
  14. Aug 15, 2009 #13
    You don't actually need the shell method to be incorporated into the problem.
    I'm sure Fightfish brought it up so that you could see that the outer shells cancel out and all you are left with is what is under you.
    But instead of doing concentric shells, just consider a single shell of arbitrary radius that is surrounding you. Then calculate the force of gravity that each point on the shell applies to you and you should find that the force is zero no matter where you are inside the shell, just as long as you are not on it or outside it. Then incorporating the shell method you can just say that ALL of the outside matter could just be thought of as individual shells which we now know each contributes a force of 0N.
     
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