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Earth gravity

  1. Mar 22, 2003 #1
    ( I'm not making a poll of this. It no opinion.) But a simple question, where would you be weighting the most?

    A - On the equator
    B - On the North pole because you are closest to the centre of the earth
    C - On the North pole because there is no centrigufal force of the spinning earth counteracting gravity
    D - Indifferent. Gravity is not depending on lattitude.

    So what is it?
  2. jcsd
  3. Mar 22, 2003 #2
    Both B&C are true, but I believe the centrifugal effect is much greater than the effect from the non-spherical Earth.
  4. Mar 22, 2003 #3
    Are you sure? Everybody agree?
  5. Mar 22, 2003 #4
    Actually B & C .

    The value of g at different latitudes is well established and well measured. At sea level, the the effective (measured) value of g:

    g (At 0* -at equator) = 9.78039 m/sec^2

    g (90* -at North pole) = 9.83217 m/sec^2

    Thus you weigh more on the pole...
    The difference in g values is about 0.05178 m/sec^2

    Of this difference it is easily calculable that a little over half is due to the centripetal acceleration provided by earth's rotation: (about 0.0336 m/sec^2).

    A 100 kg. man on the pole could loose over 5 kg. simply by moving to the equator (provided both were measured at sea level).

    P.S. We're thinking of starting a weight loss program for overweight Eskimos. We fly you to Indonesia and guaranteed a 5 % weight loss by the time you get off the plane.
    Last edited: Mar 22, 2003
  6. Mar 22, 2003 #5
    But why not "D"

    I wonder if that eskimo "g" was weighted or calculated. The textbook answer, right? But I think it's not that easy. I was intending to have "D" as the correct aswer. gravity should not be depending on lattitude

    Since the the earth is spinning the centrifugal force at the equator are the strongest it tend to make you lighter. It also makes the earth bulge outwards and it hence that bulge also put you are farther from the earth's center of gravity at the equator so gravity is weaker, according to Newtons Universal Law of Gravitation. Of course the effects of both centrifugal force and reduced gravity add together so you seem to be much lighter. right? Wrong.

    There is an error in that assumption. We assumed without saying that masses were the same as if concentrated in the centre of gravity. This is exactly true for spheres but we just deviated from the assumption that the Earth is a sphere. Hence this is not true anymore. We overlook another essential effect of the equatorial bulge. It puts more mass directly underneath you at the equator. This compensates exactly (theoretically) for that centrifugal force and the longer distance between the COG's.

    Consequently if you are on the poles, you are closer to the COG, hence a stronger gravitation force and there is no centrifugal force left, so you should be a lot heavier. Again not true (theorically), because there is a lot less mass directly underneath you and more mass to the left and the right in the equatorial bulge. This compensates again exactly for the other effects.

    Look at an extreme example and make the earth flat like a pancake. If you're on the Northpole and hence in the center of the pancake, all the mass is to the left and right. There's almost nothing underneath and you're almost weightless. When standing on the the edge of the pancake (the equator) all the mass is below you and you will weight quite a bit.

    Also consider this: the shape of the earth (geodoid) is exactly like that because of equalizing all forces. If somewhere the gravity would be less, the earth shape would correct for it automatically. (isostacy?).

    That concept is not fully operational, because the Earth is not an ideal model by all means. There are a lot of local gravity variations. You seem to weight more in Indonesia and less on the indian ocean. Tom and Jerry are trying to find out how and what
    Last edited: Mar 22, 2003
  7. Mar 22, 2003 #6


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    Can you prove any of that?

    Your pancake example is extremely pathological; assuming an actual finite mass is uniformly concentrated into a disk, an observer on the rim of the disk will feel infinite gravitational attraction, and the observer standing on the center of the disk will actually feel a gravitational force 3.414 times as strong as if his height along the axis was equal to the radius of the disk.

    The latter fact is quite familiar (think electric field lines near a large flat plate). The former is due entirely to the infinitessimal piece of the disk upon which that observer is standing.

    Clearly intuition unreliable in this matter! Also, there does not appear to be any restriction that the effects you mention would be exactly compensatory.

    Last edited: Mar 22, 2003
  8. Mar 23, 2003 #7
    OK Hurkyl,

    There are some differences between electrical and gravitational fields.

    Lets do more "pathological" experiments. Suppose that Earth was a water drop. It would be a perfect sphere if there were no forces or movements. Because all the molecules are subject to perfect symmetrical gravity. Now, we spin the drop and due to the centrifugal forces something happens. Normally the gravitational force on all water molecules at the surface was equal. Now, the total force of the spinning molecules decreases due to the opposite centrifugal force.

    The static not spinning molecules of the poles are pushing down stronger now than those at the equator. Hence the Earth starts to deform, the poles are coming down and the equator is moving up, so the bulge around the equator starts to form. When does this process stop? This is only possible when equilibrium is reached. Hence when the polar molecules come down, their gravity pull decreases and when the equatorial molecules move up, their gravity increases until polar and equatorial gravity is equal again.

    If that equilibrium was not reached, the deformation would have to continue until the water world was a disk again :smile: .

    Again, as I tried to point out, this equal gravity is caused by the location of the mass. The poles have less mass directly underneath compared to the equator. Consider that the poles have the gravity of a smaller earth and the equator has the gravity of a bigger earth according to their respective radius.

    Now back the pathological plate (alliteration intended :wink: ).
    Exactly in the centre of the Earth you are weighless (big difference with electrical fields). Why? Because all the mass is equally distrbuted around you and it is pulling you in all directions. The sum of all forces is zero. Now cut Earth in half and cut a slice through the middle (perhaps a few inches thick, not zero). Now you have the disk again. In the centre of that disk you are still weigtless because we still have the symmetrical forces in all directions albeit in two dimensions now (well almost). At the rim or the previous Eart surface there is still some gravity left caused by the finite mass of the disk but it certainly is not infinite.

    This was just to illustrate that textbooks can be wrong on very basic things. That differential gravity is just about the most annoying.
  9. Mar 23, 2003 #8


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    On the spinning water droplet:

    I still see no reason why the surface pressure needs to be equal... if all of the "external" force was acting on the surface of the water (such as surface tension or in hydraulics), I could see how the surface must be an equipotential surface, but since gravity and centrifugal forces act mostly on the interior of the droplet and not the surface, this argument doesn't work.

    And things, of course, can be more complicated!

    Elasticity of the earth would also seem to have a role in the equilibrium shape, one that would not extend to acting on those on the surface of the earth.

    And, of course, density variation makes things icky too. One can contrive an example where 90% of the earth's mass is concentrated underneath the equatorial observer's feet, so he'd obviously feel heavier. What effect would natural density gradients have on the solution?

    On the "more mass underneath" approach:

    That's not a very good proof! It gives a heuristic reason why it could be possible for them to be equal, but is far from actually proving it so. As a sample example why its not proof, I constructed this simple model:

    Scenario 1: The "planet" is composed of 4 point masses: (+/-1, 0) and (0, +/-1). Observer A is at (0, 2) and observer B is at (2, 0).

    Scenario 2: Flatten the planet by scaling the x axis a factor of 2 and the y axis by (1/2). Compute the forces and you'll find that the force A feels is tremendously greater than what B feels!

    Heuristically, one may argue that while the equatorial observer has more mass "beneath" him, he is on average further from that mass than the polar observer, and distance is more important!

    On the pancake:

    For an observer at a height d above the disk on its axis, if you integrate C d s^3 dA over the disk where s is the distance from the observer to a given area element and C is a constant involving masses and gravitational constant, the resulting formula is:

    F = 2 pi C (1 - d / sqrt(r^2 + d^2))

    for the magnitude of the gravitational force the observer feels, where r is the radius. While its true that there is 0 net gravitational force on the geometric point in the center of the disk, that point is a discontinuity in the field, an observer approaching the disk along its axis will find the gravitational force steadily increasing, approaching 2 pi C. Also, deviation from this is second order in (d / r), so near the disk it will very much appear as a uniform field, so for a real disk with nonzero thickness t, the gravitational force near the surface of the disk will simply be 2 pi C t. Of course, from there, it would drop nearly linearly to 0 as you go past the surface to the actual center of the disk.

    As for the observer on the rim, there's a significant difference between zero thickness and small thickness! Taking all of that material on the outer edge that was originally a small nonzero distance away and compressing it into that point 0 distance away from the observer is sufficient to introduce an infinity. (Again computed by integrating Newton's law). I can't manage to produce a closed form solution to this integral and I can't get the TI-89 to give me a numeric answer either, so I can't give any actual figures for nonzero thickness.
  10. Mar 23, 2003 #9
    Hmmm, this is actually a really interesting question.... I'm willing to buy that under certain idealized conditions our spinning drop would be an equipotential surface -- but that doesn't necessarily mean gravitational force, dU/dh, would be equal at all points. Hmm.
  11. Mar 23, 2003 #10
    -> Hurkyl

    Your "More mass underneath" analysis is incorrect. Distance is not more important, distribution is ! An observer that is very close to the center of mass does not necessarily experience more attraction. In the case with the disc, the mass is distributed fairly uniformed around the observer at the axis so he would barely experience any attraction at all.


    I'm with Andre on all points.
  12. Mar 27, 2003 #11
    Final blow.

    Have a look at this exagarated gravity map Not much evidence of lattitude related gravity. Just strong local anomalys. You'd be as heavy in Indonesia as in Norway.
  13. Mar 30, 2003 #12


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    That's just not so.

    Assuming Newtonian style gravity:

    Suppose the gravitational source is a disk in the x-y plane centered on the origin, with a uniform density &rho and radius a.

    Suppose we have a point observer on the positive z axis measuring gravity. Let z represent his z-coordinate, and let his mass be m.

    For an area element of the disk dA located a distance r from the origin, its mass is &rho r dr d&theta and its distance from the observer is sqrt(z^2 + r^2), so the z-component of the gravitational force on the observer by this area element is:

    dF = -G m z / (z^2 + r^2)^(3/2) &rho r dr d&theta

    Integrating over the disk:

    &int dF = &int -G m z &rho r / (z^2 + r^2)^(3/2) dr d&theta
    = (-G m z &rho) &int d&theta &int r / (z^2 + r^2)^(3/2) dr
    = (-G m z &rho) (2&pi) (-1) (r^2 + z^2)^(-1/2) | r = 0..a
    = (-2&pi G m z &rho) (1 / z - 1 / sqrt(z^2 + a^2))
    = -2&pi G m &rho (1 - z / sqrt(z^2 + a^2))

    Due to symmetry, the x and y components of the gravitational force are 0, so the magnitude of the gravitational force experienced by our axial observer is:

    F = 2&pi G m &rho (1 - z / sqrt(z^2 + a^2))

    Notice that:

    F -> 2&pi G m &rho as z -> 0

    so that it is not true that F -> 0 as z -> 0.

    Plugging in some numbers for comparison:

    F(z = a) = 2&pi G m &rho (1 - 1 / sqrt(2)) = 2&pi G m &rho * 0.2929
    F(z -> 0) = 2&pi G m &rho

    So we see that as our observer approaches the disk's surface, the gravitational force he feels increases up to about 3.414 times as strong as the force he felt when his distance from the disk was equal to its radius.

    Taylor expanding around z = 0 shows that near the surface, deviation from the uniform field F = 2&pi G m &rho is second order in z, so when you're sufficiently close to the surface, you appear to be in a uniform gravitational field.

    This is all exactly the same as the case for a uniformly charged plate in electrostatics.

    For a disk with thickness t, because of the aforementioned near-uniformity, an observer near the surface of the cylinder (or on the surface of the cylinder, as long as he's not sinking into the cylinder) will experience a gravitational force:

    F = 2&pi G m &rho t

    For the exact solution for an observer who is a height h above the surface of the cylinder, integrate F(z) for z in h .. h + t yielding

    F(h) = 2&pi G m &rho (t + sqrt(h^2 + a^2) - sqrt(h^2 + (t + a)^2))

    Plugging in h = 0 yields:

    F(h = 0) = 2&pi G m &rho (t + a - sqrt(t^2 + a^2))

    So again we see that an observer near (or on) the surface experiences a nonvanishing gravitational force. The total mass of the cylinder is &pi a^2 t &rho, so we can rewrite the above equation using M as the total mass of the cylinder:

    M = &pi a^2 t &rho
    &pi &rho = M / (a^2 t)

    F(h = 0) = 2 G m M / (a^2 t) * (t + a - sqrt(t^2 + a^2))

    Plugging in some numbers for comparison, suppose the thickness of the disk is 1 unit and the radius is 1000 units. Then, the gravitational force is approximately:

    F(h = 0) = G m M (2 * 10^-6)

    Which is the same gravitational force felt by an observer 707 units from a point gravitational source with the same mass as the cylinder. Hardly "barely any attraction at all".

    For completeness, another example would be a thickness of 1 unit and a radius of 10^6 units. In this case, the equivalent observer would be 7*10^6 units from the point mass. Clearly the ratio between thickness and radius has a dramatic effect on things, but as pointed out before the force will never vanish, and an observer standing on (not in) the disk will always experience a greater gravitational attraction than an observer who is some distance away from the disk along the axis.

    When a is much larger than t, the equation reduces to:

    F = 2 G m M / a^2

    An observer standing on a disk of negligable thickness and radius a will experience a gravitational force twice as much as an observer standing on a sphere with the same mass and radius.

    This all assumes uniform thickness; in reality we would expect the disk (or cylinder) to be more dense in the center, this increasing the force felt by our axial observer.

    Last edited: Mar 30, 2003
  14. Apr 12, 2003 #13
    Very interesting! A lecturer once told me that first people thought the earth to be flat, then like a circular plate, then spherical.

    So far so good. But they Newton came up with a calculation showing that its a flattened sphere or "oblate spheroid". Next we were told that more accurate determinations show it to be "pear shaped". I can't remember whether the big bulge is supposed to be near the north or the south pole, or whether it is big enough to have a significant effect.

    The problem is, when you look at the NASA pictures of the earth from space, they show a very spherical shape. It may be true that a minor distortion can have a measurable effect.
  15. Apr 12, 2003 #14
    Re: Re: Earth gravity

    Wait, are you saying that the centripetal accelration makes you heavier or lighter at the equator?
  16. Apr 13, 2003 #15
    Re: Re: Re: Earth gravity

    Lighter on the surface of earth (at sea level) at the equator relative to sea level at the poles. The figures I gave are MEASURED VALUES of surface gravity acceleration (g) at the equator vs. that of g at the poles. Half of that reduction is due to centripetal accel.

    Last edited: Apr 13, 2003
  17. Apr 13, 2003 #16
    I really must be missing something, wouldn't the increased centripetal acceleration at the equator make you heavier?
  18. Apr 13, 2003 #17
    Sorry, climbhi; I should have said 'centrifugal' accel., meaning in a direction opposite to that of gravity.
    So over 1/2 of the reduction in gravity at the equator is due to earth rotation.

    The other part of the reduction in weight at the equator is due to being at a greater distance from the earth's center.

    Last edited: Apr 13, 2003
  19. Apr 13, 2003 #18
    Quick question... Is centrifugal force even factored into the accepted 9.8m/s^2 ? Now that I think about it, I'm not so sure.

    When an object is outside of the earths atmosphere and 'falling' twords earth (not orbiting, but just going right 2 the center) it is not affected by centrifugal force. For objects that are in the earths atmosphere and are falling twords earth, they would be spinning with the earth and would be affected by centrifugal force.

    So back my question, Is the 9.8m/s^2 for objects that are affected by centrifugal force or for objects that are not?
  20. Apr 13, 2003 #19


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    Didn't you just answer your own question? ;)
  21. Apr 14, 2003 #20

    I'm puzzled. Reading your post you seem to persist in being lighter at the Equator, while I'm trying to promote that the total force (gravity and centrifugal) is not depending on lattitude. So we actually seem to have a disagreement. May I assume that your calculations are based on a simplification that all the mass of the Earth is working from it's centre. But again that is only valid for spheres and the Earth is no sphere. This link shows a neat real gravitation model of the earth.

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