### G whiz

*The conjecture here is that g is affected by a difference in the shape of the earth's core and its sea-level surface.*

One problem with my post below is that the polar radius is wrong. Reference books are not always reliable about such things. Another problem is that, at least once, I used a wrong value for G, indicating that one's memory is also not always terribly reliable.

However, these wrong values don't tell the whole story about what is wrong.

Some have assumed that g differs from that expected of a sphere because the earth is non-spherical. But shape is unlikely to be the only issue.

The barycenter (center of mass or gravity) of an an object might be defined as the point where all internal gravitational forces cancel. That is, if we make wedges of equal mass with sides intersecting the barycenter, the gravitational force of each wedge cancels in pairs in a circle around the barycenter and also at the barycenter.

For an object of uniform mass, the barycenter is the centroid of the volume.

Hence, as long as we know the linear distance to the barycenter, we can determine g at the geoid (surface of the object) -- that is, as long as the object has no concavities in the surface or projections (i.e, as long as a tangent line at a perimeter point does not intersect another perimeter point without also intersecting the interior). A depression in the surface means that g at the bottom of the concavity will be reduced by the y component of F

_{g}coming from the higher walls and, similarly for the surface at the base of a projection.

However, the figure of the earth is very close to an ellipsoid, though reportedly it is somewhat pear-shaped, which I am guessing means that the semimajor or semiminor axis of one ellipsoid is tacked on to that of another ellipsoid; i.e., two halves of two different ellipsoids, which share one equal axis, are pasted together. If so, we would still have the situation that the perimeter is effectively a curve where a line never intersects two tangent points without intersecting the ellipsoid's interior. Hence, we would still be able to calculate g by angle (and, to be fussy, by altitude above sea level).

Anyway, to find the distance to the geoid for an ellipsoid, given the angle, we have

r = (cos

^{2}/a

^{2}+ sin

^{2}/b

^{2})

^{-0.5}

Now the value of g at latitude 45.5 degrees has been set at 9.80665.

So we plug in the following values:

Polar radius: 6357000 meters

Equatorial radius: 6378000 m

Earth mass: 5.9736 x 10

^{24}

G: 6.67259 x 10

^{-11}

Assuming confidence about the earth's mass, at 45.5 degrees, the radius of the earth's ellipsoid is 6367.29 km.

But, setting g = 9.80665, the accepted value for that latitude, r should be 6375.36, meaning that it is 8.07 km shorter than can be accounted for by g = GM

_{earth}/r

^{2}.

Yes, it may be that the earth is a bit lumpy at the geoid, but another influence may be at work: the earth's interior is not a unform body. In fact, it is believed that the mostly iron core (with a radius about the same as the moon's) is rotating separately from the remainder of the planet.

If we suppose that the core's figure has far less eccentricity than that of the geoid, we can see that the mass distribution will vary by angle, with the highest density, on average, at the poles and the least at the equator. That is, g would decrease more rapidly from pole to equator than would be so for an ellipsoid of uniform mass.

Well, I suppose we might also like to calculate the moon's influence, which should cause the effective g to be less when the moon is in the sky and more when it is behind the earth from the observer. However, I am sure that the official guardians of g have taken the moon's effect into account. In general, I would expect that the effect would on average cancel out, but then again, its effect on local g may well depend on the moon's orbit with respect to latitude.

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