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Strange: Earth's center of gravity

  1. Jun 22, 2011 #1
    Hi there, so I'm a bit confused...I'm 14 years old and I'm living in Croatia.I heard for Hallow Earth theory one year ago so I wanted to research it, but I had no material i needed until today...data(material9 i needed is what is radius of Earth on poles and Equator and what g is there (gravitational constant)...
    r(equator)=6378137 m
    r(poles)=6356752.3 m
    smanjenje (croatian for getting smaller)=
    =delta g/delta r=
    =-0.002476 (N/kg)/km
    delta r1=
    =3178 km
    g(when we are 3200 km from earth's center)=g1+smanjenje*delta r1
    =1.912 N/kg
    bla, bla, bla...so I've repeated it with bigger and bigger depth, until g turned negative:
    delta r4=6378-2000=
    =4378 km
    g(2000km from center)=g1+smanjenje*delta r4
    =-1.06 N/kg
    But that is Impossible! That would mean that there would be anti gravity, but since w are still in one peace, shouldn't it mean that a real center of gravity is where g=0, that would mean somewhere between 2500 and 2000 km deep according to my calculations...anyone explanation? wouldn't it mean that there is highest pressure on sphere somewhere between 2000 and 2500km from center? And doesn't it mean that calculations for objects close to earth (moon and so) are false? wouldn't it change our calculations how earth is heavy? answer please!
    PS don't use miles and so cuz I don't know how much it is, use standard units (kg, m and so)
    PSS i told you where from am I only to promote a bit my country, we have gorgeous coast, sea, and we are christian country aldo we really don't care for it, and there isn't war at our country for decades, and we are kinda modern country...sorry for oftopic, I got carried away...
  2. jcsd
  3. Jun 22, 2011 #2
    I think you are implicitly assuming that the mass of the Earth is constant as you move into the planet, it can be shown that for a spherically symmetric object which Earth could be approximated as, that only the mass inside the radius of the sphere is used when calculating the gravitational influence of the object. You also appear to assume that the Earth would be of constant density which is not the case. As it happens there was a thread a while back on here which discussed this in more detail and showed that g does actually increase slightly as you move towards the Earth's core through the crust.
  4. Jun 22, 2011 #3
    Welcome to Physics Forums.

    I have not carefully worked through your calculations, but I assure you there is lots of evidence that the Earth is not hollow. Perhaps the most convincing arguments come from earthquake waves. When an earthquake occurs it sends out various kinds of waves that seismographs can pick up all over the world. Some of the waves go through the Earth and some move along the surface. They move at different speeds. One kind will not travel through a liquid. We can use all that information, the distances they travel, how long they travel and the physics of reflection and refraction of waves to figure out what is going on in side the Earth.

    When I first heard about this, geophysicists had figured out there is a nickel-iron core in the center of the Earth, a liquid core around that, a mantle around that and then the crust we live on. When I studied it in college, they had measured so many earthquakes that they had identified seven or eight layers within the mantle. More earthquakes (there are hundreds of thousands each year, though many are too small to be much use for this) let them discover that there are mountains and canyons on the surface of the inner core. Then they measured the rotational speed of the inner core by tracking those features. More recently they identified the location of parts of tectonic plates that break off in subduction zones and sink into the upper mantle.

    It is pretty spectacular and very detailed. The evidence makes awfully clear that the earth is not hollow. There are several independent lines of reasoning (that don't rely on seismic waves) that lead to the same conclusion.
  5. Jun 22, 2011 #4
    Inside of a hollow, spherical shell, g=0 everywhere.

    Inside of a sphere of uniform density, g=kr, where:

    k = 4/3 πGρ
    G = Newton's gravitational constant
    ρ = mass density
    r = distance from the center.
  6. Jun 22, 2011 #5
    You calculate the derivative of g with respect to r: dg/dr by dividing the differences of g and r at the poles and the equator.
    and then you use that to calculate g at various depths by using

    [tex] g = g_{surface} - d \frac {dg}{dr} [/tex]

    (where d is the depth)

    The first problem with this, is that the difference in g comes from both
    the distance to the center of the earth, and from the centripetal acceleration that
    is necessary to keep someone on the equator from flying off into space in a straight
    line. This will reduce gravity on the equator. If you climbed a 21 km mast at the poles,
    gravity would still be bigger than at the equator.

    The second problem is that there's no reason for dg/dr to be constant, or for a graph of
    g vs. r to be a straight line.
  7. Jun 23, 2011 #6
    I had thought it was linear, but the posts by superstring and willem2 made me look further into it. I found http://en.wikipedia.org/wiki/File:EarthGravityPREM.jpg" [Broken] showing g at depth for various assumptions about the density of the interior. The key is a little unclear. The PREM in it is the Preliminary Reference Earth Model, which is a synthesis of of geophysical data done about 30 years ago.

    I am surprised by the increase in g when the density is taken into account. I thought Airy's pendulum experiments in coal mines gave a larger period in the mine, indicating a reduction in the acceleration due to gravity. Am I missing something, or is there an inconsistency there?
    Last edited by a moderator: May 5, 2017
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