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Gravitational field inside the earth

  1. Mar 7, 2006 #1
    Hi all - I broke my foot and had surgury so I'm bored (for a couple more weeks) which explains my recent posts.

    [This isn't "Homework" like some class assignment - it's just something I've been playing with since reading some thought experiments that John Wheeler expressed - and some Feynman stuff on the E field inside a charged sphere]

    I'm kind of interested in the nature of fields inside of variously shaped "sources" and would be interested to here any comments about how to best think about any of the following problems so they can be simply expressed mathematically. I could just go look up the appropriate equations but what fun is that?

    1. If the earth (mass M) were a perfect sphere with a perfectly homogenous rigid mass distribution, what would be the vector field equation for the earth's gravitational field at any location on the inside region of the sphere? Disregard all external forces and interactions. Any coordinate system is OK but let the origin be at the center of the sphere.

    2. Assume a test mass m is suspended in a cylindrical tube of negligible total volume, bored perfectly along a diameter, from one side of the planet to the other through its center. The tube contains a perfect vacuum except for the presence of the test mass, and its walls exert no influence on the test mass. The test mass is initially suspended at the surface level of the earth and is motionless relative to it. What would be the vector equation of motion for the body after it's released to free fall?

    3. Disregarding any relativistic effects, would the earth's rotation matter at all to any of the above?

    4. Would the equations remain unchanged for any diameterical path we could choose for the tube; e.g. pole to pole vs two diametrical points on the equator.

    5. What happens to G (the gravitational constant) inside the sphere?

    Any little feedback would be welcome - except maybe "hey jackiemoron - go kill yourself... slowly!!!" or "get well VERY, VERY, VERY, soon, we're all beggin' ya..." or something in that vain :wink:

    Last edited: Mar 7, 2006
  2. jcsd
  3. Mar 7, 2006 #2

    Doc Al

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    Just about all of these questions can be answered with a bit of thought, assuming you know the basics of Newtonian gravity. Since you have time on your hands :wink: , why don't you tell us what you think the answers would be and why?
  4. Mar 7, 2006 #3


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    If you know Gauss's law in electrostatics, there is a gauss's law equivalent for gravitational field. That would answer a few of your questions.


    Regardless on whether this is a "homework" question, these still qualify as homework/schoolwork-TYPE questions, and belong in that forum.

  5. Mar 7, 2006 #4


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    the gravitational field inside of the earth depends only on the mass contained inside of the radius. b/c of the geometry of a sphere the mass outside of a given radius has all of its gravitational effects cancel. also any mass dropped down the cylinder will oscillate back and forth with simple harmonic motion in a [itex] \frac{d^2r}{dt^2} = -\frac{k}{m}r [/itex] fashion
  6. Mar 8, 2006 #5
    Sorry for posting to wrong forum.

    [edit] Crap!!! Disregard all this. #1 is NO GOOD! Back to the drawing board :-(

    1. I may have messed up? Anyway, for deriving a vector field eq for inside the earth F(r) where r is position vector relative to center of the earth and |r| < re (radius of earth), I took the vector eq for the field at the surface of the earth and multiplied it by a scalar function called called "ratio" that varies as a function of r. That scaler ratio is that of the mass inside the Gaussian surface at r to that of the mass of the shell outside that surface. I did it that way since only the mass inside the surface at r contributes to the field. Since the density is constant throught the earth, I calculated the ratio in terms of volumes instead of masses, getting rid of any need for a mass density constant.

    I can't seem to get LaTeX of eqs. to preview??? Not sure if it supposed to preview OK or not? Here's a brief text version for now:

    The desired vector equation for inside the earth:
    F(r) = F_surface(re) * ratio(r)

    where F_surface(re) is the vector field eq. at the surface of the earth and re is any position vector <x,y,z> located on the surface of the earth and |re| = radius of the earth. Vector r is any position inside the sphere.

    F_surface(re) = -(mMG/|re|^3)re

    ratio(r) = (Vol inside Gaussian surface of radius |r|)/(Vol of Earth - Vol inside Gaussian surface of radius |r|)

    2. I haven't messed with this yet.

    For the other questions:

    3 & 4: I can't see any way that rotation of the earth or the choice of diametrical path for the tube through the earth can have any influence on the motion of the test mass.

    5. G is G :uhh: I couldn't tell ya what I was "thinking" there...

    Does thie reasoning in (1) look OK?
    Last edited: Mar 8, 2006
  7. Mar 8, 2006 #6


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    Here are some hints: the gravity inside a uniform spherical shell of matter is zero.

    The gravity of a sphere acts as if all the mass were concentrated into a point.

    Now, divide up the Earth into two parts - a hollow sphere for r>R, and a sphere r<=R, where R is the location of the particle.

    1) What is the net gravitational force due to the hollow sphere (see hint #1)?
    2) What is the net gravitational force due to the sphere (see hint #2).
  8. Mar 9, 2006 #7
    OK - let's try again :surprised

    F = gravitational vector field function inside of the earth
    r = position vector, <x,y,z>, with origin at center of the earth
    me = mass of earth
    m = mass of test particle
    re = radius of earth

    If I want to express the grav field vector inside the earth as a function of a position vector r (where |r| < radius of earth), the only mass that contributes to F(r) is inside a Gaussian spherical surface of radius |r|. The external "shell" of mass contributes nothing to the field according to Gauss's Law. Therefore, that quantity of mass is also a function of r, m(r). That mass m(r) is the volume of the sphere of radius |r| times the mass density. Mass density is just the volume the earth divide by the mass of the earth = ((4/3)*Pi*re^3)/me

    m(r) = ((4/3)*Pi*|r|^3) * ((4/3)*Pi*re^3)/me

    F(r) = -((G * m * m(r))/|r|^3) * r

    F(r) = -G*m*((16 * Pi^2 * re^3*)/9 * me) * r

    Is that correct? I wish I could do it in tex - it would be more readable.

    Last edited: Mar 9, 2006
  9. Mar 9, 2006 #8

    Doc Al

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    You have that inverted:
    [tex]\rho = m_e / (4/3 \pi r_e^3) [/tex]

    Latex works. It just doesn't display in preview mode (a pain in the butt, for sure).
  10. Mar 9, 2006 #9
    Crap!!! How stupid :yuck: That's what I get for hurrying and not double checking the details.
    Oh well - I just gotta work harder at not being so sloppy (in my thinking). Anyone have a spare brain I can use for a while - like 30 or 40 years? :rolleyes:

    Ohh!! That's good, though. I wasn't sure. OK - at least can go back and Edit.

    BTW - I never really understood why that mass-shell, external to the surface, canceled EVERYWHERE in the interior until studying this little problem. I like working thru problems - even if I do screw up the first five attempts at most.

  11. Mar 9, 2006 #10

    Doc Al

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    That's the only way to learn. If you aren't making mistakes, the problems are too easy. :wink:
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