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Gravity on the surface of a Hollow Earth

  1. Nov 9, 2005 #1
    A friend of mine posed a question to me that I can not answer.

    He was wondering what would happen to the surface if the earth was hollow, and if all of the mass of the earth was in the crust.

    He said to ignore the fact that it would probably collapse in on itself. And assume that it stays in shape.

    My thought is that you would have a rotating focal point of gravity due to the slight ellipticity of the earths crust.

    Your thoughts?

    A question of my own. Would the effect gravity be affected on the scale of the solar system.
  2. jcsd
  3. Nov 9, 2005 #2


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    The pole's precession (the earth wobbles like a top) would probably be affected due to the change in moment of inertia, and tidal slowing of the rotation would certainly be affected, but other than that, not much would change.
  4. Nov 9, 2005 #3

    The surface gravity would change significantly. You wouldn't guess it from the basic newton's law, but that is because that treats everything as a point particle. The gravity field farther out, such as for the moon would be less affected, but at the surface, the effect should be a huge increase in surface gravity because so much matter is so much closer to you. Figure the mass on your side of the earth is roughly 1000 times closer to you (on average over the hemsiphere, more like 1000000 times closer for the stuff directly underneath you) and the stuff on the far side is only 1.5 times farther (on average over the hemisphere, more like 2 as far for the stuff directly underneath you.) (these are rough estimations, I haven't bothered to actually calculate).
  5. Nov 9, 2005 #4

    Doc Al

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    I don't think so. Taking a simplified model of a spherically symmetric earth, as long as the symmetry is maintained it doesn't matter if all the mass were contained in a thin shell: the gravitational field outside the shell will be the same.
  6. Nov 9, 2005 #5
    But what I'm saying is that model in unphysical. For a hollow earth, its simply wrong.
    Here, I'll work through the math:
    For a spherical shell of radius [tex]R[/tex], width [tex]\Delta r[/tex], and uniform density [tex]\rho[/tex] the total mass [tex]M[/tex] is:
    M = \int\int\int \rho r^2 \sin \theta dr d\theta d\phi[/tex]
    \qquad = \int^{R+\Delta r}_{R} \int^{2\pi}_0 \int^{2 \pi}_0 \rho r^2 \sin \theta dr d\theta d\phi [/tex]
    so we can stipulate that the mass of some volume slice in the shell is given by [tex]dm[/tex] as:
    dm = \rho r^2 \sin \theta dr d\theta d\phi [/tex]
    The gravitational field generate by this mass slice is given by:
    g = G \frac{dm}{s^2} \vec{\epsilon} [/tex]
    where [tex]s[/tex] is the distance to the mass slice, and [tex]\vec{\epsilon}[/tex] is the unit vector point towards the mass slice. It can be show geometrically that when calculating the gravtiational field at the point P(R, 0,0) (surface of the sphere)
    s = 2 r \sin \frac{\theta}{2} [/tex]
    The unit vector point from P to the mass slice [tex]\vec{\epsilon}[/tex] is:
    \vec{\epsilon} = \left [ \begin{array}{l} 1 \\ \theta \\ \phi \end{array} \right ] [/tex]
    Now, because this is a sphere, and as such is rotationally symmeticraly we can assume that
    g_\theta = 0 [/tex]
    g_\phi = 0 [/tex]
    so we get:
    g_r = \int \int \int G \frac{\rho r^2 \sin \theta dr d\theta d\phi}{4 r^2 \sin^2 \frac{\theta}{2}} dr d\theta d\phi [/tex]
    Integrating only over a single hemisphere, and then multiply by two (to eliminate negative r) gives
    \qquad = \frac{G \rho}{4} \int_{R}^{R+\Delta r} \int_0^{\pi} \int_0^{2\pi} \frac{\sin \theta}{\sin^2 \frac{\theta}{2}} d\phi d\theta dr [/tex]
    \qquad = \frac{G \rho}{8} \int_{R}^{R+\Delta r} \int_0^{\pi} \int_0^{2\pi} \sin \theta - \frac{\sin 2 \theta}{2} d\phi d\theta dr [/tex]
    (check your trig identities if you're not sure how I got that. i used the double angle identity)
    \qquad = \frac{ G \rho \pi \Delta r}{2}

    Now, to get a number for that, lets say our earth has a [tex]\Delta r[/tex] of 0.01 percent the normal radius of the earth (this derivation requires that [tex]\Delta r \ll R[/tex]). This gives:

    G = 6.67 \times 10^-11[/tex]
    \Delta r = 63800 \text{m} [/tex]
    \rho = 575641 \frac{kg}{m^3} [/tex]

    gives [tex] g = 3.847 \frac{m}{s^2} [/tex].

    Well that is odd. it came out smaller. Now this is gonna bug me even more. I don't suppose someone could check this for me to make sure I didn't screw something up along the way? interesting result though, if its right.
    Last edited: Nov 9, 2005
  7. Nov 9, 2005 #6

    Doc Al

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    Newton's Shell Theorem

    Don't tell Newton! The fact that the gravitational field outside of a uniform sphere or spherical shell is the same as if the mass were concentrated at the center is a famous theorem proved by Newton.

    Nowadays it's a standard exercise in calculus. Here's a site that shows how to do it: http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/MoreGravity.htm

    I haven't looked at your math, but you must have made a mistake somewhere.
  8. Nov 9, 2005 #7

    Hmm, now this is why intro physics classes shouldn't skip gravity. Hmm. Now I wonder where my math went wrong. I follow the results in the link fine, but they used a very different setup to the integral, so it doesn't help me spot my mistake.
  9. Nov 9, 2005 #8


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    so Gauss's Law as applied to a sphere should be attributed to Newton? that's an interesting piece of history. i didn't know that.
  10. Nov 9, 2005 #9
    Another thing a first year physics class should not skip.
  11. Nov 9, 2005 #10

    Doc Al

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    I wouldn't put it that way. Newton used geometrical arguments to prove his shell theorems (without even using his own calculus!) from the inverse square law. (Of course, he also developed the calculus to solve this kind of thing.) But Gauss's Law--the easy way to prove the shell theorems--didn't come until later, I believe.
  12. Nov 11, 2005 #11


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    franznietzsche: From what I see you're first triple integral for mass should just be a single integral over radius as there is no dependance on the theta or phi coordinates. Also you would not integrate over 2pi for both angles. Thus dm=(rho)4*pi*r^2dr would give M=(rho)*4/3*pi*r^3 where we can cut the minutia and just call r the thickness of the spherical shell. All this does is say that the mass of the hollow sphere is simply the density multiplied by the volume.
  13. Nov 14, 2005 #12
    Would this still hold true even though the earth isn't a sphere?
  14. Nov 14, 2005 #13
    See the divergence theorem.
  15. Nov 14, 2005 #14
    In leaving dr a variable you've left the mass of the earth non-constant. Changing dr changes the mass of the earth so you're not going to get a constant value. To get a constant value you'd have to have a variable mass density which you write as a function of dr so as to make the total mass a constant.

    I hope that helps.

  16. Nov 15, 2005 #15
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