Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Earths Gravitation

  1. Jun 30, 2011 #1
    Hello I have a question.

    We all know that the center of earths gravity is ZERO, as we assume earth as 100% spherical, We also know that the earths mass is not uniformly distributed and its not 100% spherical too. Then how can the earths gravity be ZERO at the earth's center still?
     
  2. jcsd
  3. Jun 30, 2011 #2
    Hello nhemanth, welcome to Physics Forums,

    Since the Earth's shape is irregular it does not actually have a geometric centre.

    Of course any system of mass in isolation has a centre of gravity, the exact location of this is part of the subject of Geodesy.

    Since the Earth's composition is also varied, the resulting variation in gravity is also part of Geodetic work.

    However life is never that simple because the Earth does not exist in isolation, it is part of the Earth - Moon system which rotate round a common cente of gravity. This situation is responsible for the tides. In the ultimate, variation in both the Earth's fluid environments (atmosphere and ocean) contribute to (very small) variations in the position of the Earth's own centre of gravity.
    There are further weaker interactions with other heavenly bodies.

    go well
     
  4. Jul 14, 2011 #3
    Earth's gravitational force is often modeled as though the Earth were an inert sphere of uniform density.
    There is a direct relationship between gravitational acceleration and the downwards weight force experienced by objects on Earth.
     
  5. Jul 14, 2011 #4
    The centre of mass is by definition the centre of the earth.
    Everything has a centre of mass as every pair of particles has a centre of mass which can be considered a single particle of the combined mass of the particles. This can be repeated until all the particles are used up.
     
  6. Jul 18, 2011 #5

    Drakkith

    User Avatar

    Staff: Mentor

    The net force in any direction near the center of the earth (wherever that might be exactly, due to irregularities) is approximately zero from the force of gravity. The only reason we are pulled to the earths surface by gravity is because almost the entirety of its mass is underneath us. So almost all of the gravity is pulling us DOWN! If you were in the center of the earth you would have near equal amounts of mass on every side of you, so you wouldn't feel a pull in any one direction like you do on the surface.

    However, you would feel the pressure of all that mass pulling IN on itself. The matter on the surface is pulling the opposite surface towards it as is all the matter in between. The result of all this is compression.
     
  7. Jul 10, 2012 #6
    Drakkith's explanation is very good>>If you were in the center of the earth you would have near equal amounts of mass on every side of you, so you wouldn't feel a pull in any one direction like you do on the surface.>>

    To be precise, every particle of your body would be feeling an equal pull in every direction and because of this you would "experience" net 0 gravitational force.
    That is until the center of mass shifted with the shifting of the liquid core, the moon, the tides etc, then your body would shift with the center of mass. (all assuming your body could freely move through the material of the earth and yet still experience the gravitational forces created by it's mass which of course it can't)
     
  8. Jul 12, 2012 #7
    An object would always have some points where the gravity from different parts of the object add up to zero. For highly irregular objects, there could be several such points. If you want, you can define such points as gravitational "centers" of the object. However, a spheroidal shape will always have one point of zero gravitational strength at its geometric center. This was proven by Isaac Newton in Principia.
    An object with a highly irregular shape or highly irregular inhomogeneities could have several points where the total gravitational field strength is zero. There could be a finite region where the gravitational field strength is zero. However, this region would have to be very small for an object that was symmetric. In the limit of spherical or spheroidal symmetry, this region would be negligibly small. For all practical purposes, it would be a point.
    The mass distribution of the earth approximates a series of concentric shells of each with uniform density which have the shape of an oblate spheroid. There are small anomalies in the mass distribution. No doubt they shift the "center" of the earth or even broaden it by a small amount. However, this would be insignificant for most purposes.
    Gravity dominates contact forces for large objects. Gravity and centripetal force cause very large objects into oblate spheroids.
    Smaller objects are shaped as much by contact forces as by gravity. Some asteroids and comets have highly irregular shapes. Some asteroids and comets are made of clumps of material with cavities in them. It would be interesting to analyze some asteroids and comets in terms of their “centers”.
    Halley’s comet is definitely not a spheroidal object. The nucleus of Halley’s was photographed recently. It looks like a bent peanut. I doubt that Halleys’s comet would have a single center where gravity cancelled out.
    Satellite studies are the most accurate methods of determining the earth’s “shape”. Note that the measurements show that the earth is very close to being an oblate spheroid. However, it is very slightly off being an oblate spheroid. Oblate spheroids have a well defined center. You may be interested in these deviations from being an oblate spheroid.

    Here are some links to articles concerning the shape of the earth.

    http://en.wikipedia.org/wiki/Figure_of_the_Earth
    Figure of the Earth
    “Closer approximations range from modelling the shape of the entire Earth as an oblate spheroid or an oblate ellipsoid, to the use of spherical harmonics or local approximations in terms of local reference ellipsoids. The idea of a planar or flat surface for Earth, however, is still acceptable for surveys of small areas, as local topography is more important than the curvature.”

    http://adsabs.harvard.edu/full/1966SSRv....5..818K
    "Recent works on the earths gravitational potential by its dynamical effects on the motion of a satellite are reviewed."


    Please note that at the center of the earth, the gravitational field strength won't be nearly as important as the pressure. The pressure at the center of the earth would be caused by contact forces, not gravity.
     
  9. Aug 18, 2012 #8

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Sorry about reviving a month-old thread. I don't visit the Earth Sciences forum very often, and this thread interested me.
    That is an all-too-common error. It's actually about 4 times denser at the center than near the surface.

    http://www.cas.umt.edu/geosciences//faculty/sheriff/438-Gravity_Electromagnetics/images/Density%20Depth.gif [Broken]​

    It seems to me that must be true, but I am wondering if there is a straightforward proof of this. Do you happen to know?
    That statement threw me at first, until I thought -- of course -- a hollow spherical shell has more than several such points :wink:

    I would consider a hollow spherical shell to have spherical symmetry, yet the region of zero field strength is nearly as large as the object itself.
     
    Last edited by a moderator: May 6, 2017
  10. Aug 19, 2012 #9
    Yes, that is true. However, Newton proved that the gravitational field outside a sphere is equivalent to the gravitational field generated by a point particle with the same mass as the entire sphere. This is true whether or not the density of the sphere is uniform. The density of the sphere in this model just has to have a spherical symmetry and go to zero for radii above a certain maximum radius. The maximum radius defines the outer surface of the sphere. The same applies to spheroidal symmetry.
    The nongaseous part of the earth earth has a density with a nearly spheroidal symmetry. Therefore, it has at least one point of zero symmetry at the center of the earth.
    As I wrote this, I realized that the atmosphere very likely has some deviation from spheroidal symmetry due to the differences in temperature, humidity and centripetal force as one moves from pole to pole. The atmosphere is a compressible fluid, so these things would effect its density. Therefore, what we have been discussing may not be valid for the atmosphere of a planet. On earth, the atmosphere probably has a negligible contribution to gravity. However, Jupiter and Saturn are mostly atmosphere. I wonder whether the asymmetry in their atmospheres affect their gravitational field.
    I had stated in a previous post that even an irregular object has to have at least one point of zero gravity. However, I was speaking out of my physical intuition. I may be wrong about that. Sorry.

    No. I can think of lots of examples, but I don't know if what I said applies to all irregular shapes.

    Thank you very much! You are correct! I was wrong!
    Your example threw me. I didn't occur to me that a shape with spheroidal symmetry can have more than one point of zero gravity. I actually thought that a spheroidal shape implied a "unique" point of zero gravity. Obviously, I was wrong.

    If the density increases with distance from the center, then there will be only one point of zero gravity in the sphere or spheroid. Obviously, the gravitational field at any point is the sum of gravitational fields from the series of shells between the point and the center. Thus, the gravitational field in this case increases. This is the case with the earth.
    The density of any planet would tend to decrease with distance from the center due to pressure. Therefore, I hypothesize that in all the rocky planets there is only one point of precisely zero gravity.
    Thanks to your suggestion about the hollow shell, I now have doubts about the gas giants. Maybe they have more than one point of zero gravity.
    I don't think what we are saying is of any practical interest. The pressure and viscosity at the center of a planet would be so great that a small nonzero gravitational field itself would not be noticeable. This is more a mathematical exercise than a physical one. However, it is interesting from a mathematical point of view.
    The points of zero gravity may be of importance in understanding the dynamics of asteroids and comets. Asteroids and comets can have highly irregular shapes. After a collision, when the asteroid or comet has shattered and is coalescing, there may be a period of time where the density at the "center" is very small compared to the density at the edges. Thus, there may be "hollow asteroids" or "hollow comets". This is just my conjecture, based on what I learned in this conversation. The gravity on the surface of these hypothetical objects would be very small, anyway.
    I have noticed among some physics newbies a tendency to confuse pressure with gravitational strength. Some make the mistake believing that the pressure at the center of the earth should be small. They should rest assured: the pressure at the center of the earth is huge. There is no "hollow earth" consistent with the known laws of physics. There probably is no "hollow planet" of any type.
     
    Last edited by a moderator: May 6, 2017
  11. Aug 19, 2012 #10
    Looking at your picture, I note that there is no discontinuity in density between the outer core, which is supposed to be liquid, and the inner core, which is supposed to be solid. Yet, there is a discontinuity in density between the mantle and the core. That discontinuity may be due to a sudden change in composition. Maybe your reference decided that the inner and outer core had the same composition. Yet, I that still doesn't explain why the transition from inner to outer core is so smooth.
    This seems odd to me, since the liquid and solid states of a substance usually have noticeable differences in pressure. For instance, there would be a finite difference in density between the ice and liquid water phases at atmospheric pressure.
    I am wondering if there is a significant difference between the liquid and solid phases at the core of the earth. Maybe the temperatures and pressures in the core of the earth are above the critical points for the transition from liquid to solid.
     
    Last edited by a moderator: May 6, 2017
  12. Aug 19, 2012 #11
    If you google around a bit, you will find several models about the density depth assumption, (could not find a valid ref) but most with a discontinuity at the inner -outer core boundary. And yes the inner core is assumed to be solid due the immense pressure Clausius Clapeyron relation
     
  13. Aug 19, 2012 #12
    The earths mass closer to spheroidal than spherical due to centrifugal force. However, this doesn't change the fact.
    The geodesic surfaces of the earth are still concentric. Mass density increases with distance from the center. However, the symmetry of this distribution is still spheroidal.
    Wait, lets look at the symmetry of the problem rather than specifics. If the symmetry of the mass density just has to be invariant to inversion through a point, then the gravitational field vector at that point has to be a zero vector.
    It doesn't matter that the earth is shaped like an ellipsoidal solid. An ellipsoidal solid is invariant to inversions through their center. Therefore, the gravitational field of the earth is zero at the center.
    The mass density of the earth is invariant with respect to inversion transformations through the center. By inversion, I mean that if one calculates the density at a point (x,y,z), then has the same value of density at the point (-x, -y, -z). If you do the necessary integrals, you will find that the gravitational field vector is anti-invariant at regards to inversion transformations. If you calculate the gravitational field vector at a point (x,y,z), then it is equal to the negative of the gravitational field vector at (-x,-y,-z).
    If the mass density is invariant to inversions at a point, then the gravitational field has to be zero at that point. This is because the only vector that is invariant to inversion is the zero vector.
    The processes and objects that break the inversion symmetry have to cause a small gravitational field at the center of the earth. These things that break the inversion symmetry probably are small perturbations. However, let us discuss two of them: topology and tides.
    Although the topological features at the earths surface may break the inversion symmetry by a small amount, they are probably insignificant. The concentration of continents in the North hemisphere probably breaks the symmetry at the center by a very small amount. However, continents are only a very small fraction of the earths mass. So I suppose this is negligible for practical purposes.
    The tidal bulge of the earth probably breaks the inversion symmetry at the center by a very small amount. It wouldn't break the inversion symmetry if the moons orbital plane coincided with the earth's equator plane. However, it is a little offset. Given this offset, the rock tides must be causing some deviation from inversion symmetry. The same goes for solar tides.
    I conjecture that the tidal forces of both the moon and the sun break the inversion symmetry of the center of the earth. The tidal forces cause the mass density to vary just slightly from a state of inversion symmetry. The tidal force probably disturb the inversion symmetry more than the imbalance in topography. However, the total effect at the center of the earth must be very small.
    Now this I can't prove mathematically. However, it appears to me that the things that break inversion symmetry may be shifting the point of zero gravity just a small distance off center. Therefore, there may be a point of zero gravity just a small distance from the geometric center of the earth. If this is true, then one could redefine the center of the earth as this hypothetical point of zero gravity. It won't make a big difference in the physics, but it may keep some purists happy.
    I don't see how this shift can make a significant difference in the physical dynamics of the earth. The things near the center of the earth are subject to large forces that are not gravitational. The pressure at the center of the earth is enormous, whatever the gravitational field vector may be. The pressure is caused by the gravity from mass contributions far from the center of the earth.
    This is a good mathematics exercise, but not a real physics problem. I don't know how we can experimentally prove that the gravitational field vector at the center of the earth is not zero. The pressure and temperature at the center of the earth may be important in experiments, but not the gravitational field vector.
    There is probably a nonzero gravitational field vector at the center of the earth due to tidal forces distorting the earths mass density. There may also be a smaller contribution to the gravitational field at the center due to the asymmetry of plate tectonics. The tides are the most important contributions to asymmetry.
    Therefore, the statement that there is no gravity at the center of the earth is an approximation. The statement is slightly wrong and you are barely right. However, the gravitational vector at the surface of the earth is only a small fraction of the gravity that is on the surface of the earth. The nonzero gravitational field vector at the center of the earth probably has no significant effect.
     
  14. Aug 19, 2012 #13
    The inner core is observed to be solid -- from seismology.

    In fact that observation, coupled with phase relations, give us some constraints on the pressure and temperature conditions in the core.
     
  15. Aug 19, 2012 #14

    Redbelly98

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Agreed. FYI my bringing up the hollow shell was strictly for mathematical interest. I presume that any real, approximately spherical, body would not be hollow.

    FYI, you might be interested in a couple of other past discussions on Earth's varying density and its effect on gravitational field strength:

    https://www.physicsforums.com/showthread.php?p=1560428#post1560428 (See Andre's Post #8)
    https://www.physicsforums.com/showthread.php?t=463706 (This is where I first learned that g was not a maximum at Earth's surface!)

    Good points about the discontinuity. I had done a google image search on Earth density depth, and simply used the first decent looking graph to post here. Looking at other graphs in that same search, they show a 5-10% change in density at the solid-liquid interface. Apparently the source of the image I posted simplified things by ignoring this amount of change.
     
  16. Aug 19, 2012 #15

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Think about the shape of the gravitational potential function, in a sphere enclosing the body that is big enough so that at the surface of the sphere the potenntial is approximately the same as replacing the body with a point mass.

    The potential is a smooth function and must have a point with zero gradient somwhere inside the sphere. At that point (or points) the gravitational force = the gradient of the potential function = 0.

    (Making the argument mathematically rigorous is left as an exercise!)
     
  17. Aug 20, 2012 #16
  18. Aug 20, 2012 #17
    Several investigators, from Lehmann to Deuss, have concluded that the inner core is solid, based on their observations. The quality of the data, and the confidence of the conclusion has increased over time. Nowadays I think it is fair and uncontroversial to say that the inner core has been observed to be solid.

    ps. I don't understand the significance of the link you posted.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Earths Gravitation
  1. Pictures of the Earth (Replies: 6)

  2. Death of earth (Replies: 56)

  3. Earth nucleus (Replies: 4)

Loading...