As per this article:
(My bolds.) http://www.webspawner.com/users/gravibiasGravitational Bias and Escape Velocity
Author: Greg Alexander
Date: 7th May 2003.
When using the equation describing the acceleration due to gravity, g = GM/r², it is always assumed for a body like the Earth that the gravitational field acts from a point at the exact centre of the mass such that when stood at its surface the distance between you and that point is always exactly equal to the radius of the Earth. However knowing that the inverse square law applies to any gravitational field, wouldn’t it be the case that the mass directly beneath your feet would have a disproportionately greater effect that that on the other side of the planet? As the relation between the field intensity and distance isn’t linear, surely it can’t be valid to assume that one can summate all the gravitational forces produced by all the particles within the Earth’s mass into a single force acting from an isolated, central point?
The following example would appear to demonstrate quite clearly that this indeed is not a valid assumption :
If we divide the Earth’s mass into 100 smaller units taking the form of spheres (these spheres needn’t necessarily have to fit alongside each other without overlapping, and assuming, for the sake of the argument, that the density of the Earth is uniform throughout), we obtain the following proportions for such a sphere; a mass of 5.98 x 10exp22 kg, and a radius of 1,372.6 km. Stood at the surface of the Earth directly over one such sphere the observer would be exactly 1,372.6 km from its mathematical centre and experience a ‘g’ of 2.13 m/sec² as a consequence. Placing a second theoretical sphere at the Earth’s exact centre a further ‘g’ is produced of 0.098 m/sec² where ‘r’ is the radius of the Earth. (It can be noted that 0.098 is exactly 100 times less than 9.8 m/sec²). A third theoretical sphere is placed on exactly the opposite side of the Earth to the first such that the distance between its centre and the observer is 11,383.4 km. This would produce a ‘g’ of 0.0309 m/sec².
From this example it is apparent that the sum of the g’s produced by the first and third spheres, 2.13 + 0.0309 = 2.16 m/sec², is not at all equal to twice the ‘g’ produced by the second centrally located sphere of 0.196 m/sec². In fact there is a definite bias of over a 1,000%!
If such a bias does occur on the Earth’s surface as a result of the non-finite proportions of its mass which is spread out within a substantial volume of space, then this must reflect upon the relevant accuracy of the equation g = GM/r² (when applied at the surface). When the value of the universal gravitational constant G (6.7 x 10exp-11 N m² kg-1), the mass of the Earth in kilograms (5.98 x 10exp24 kg) and the Earth’s radius in metres (6,378,000 m) are put into the equation in the relevant places you always end up with g = 9.8 m/sec². But how much of this value (which is measurable experimentally) is the result of such a gravitational bias?
If the 9.8 m/sec² acceleration at the Earth’s surface is, to a significant part, the result of such a bias, one consequence of this would be that the calculated value of ‘g’ would decrease far more rapidly as you left the Earth’s surface, say, in a rocket. This in turn would inevitably effect the Earth’s escape velocity which would inevitably decrease. However currently it is calculated as being 11.18 km/sec.
Is it possible that this value for the Earth’s escape velocity is wrong? Although such a suggestion may sound slightly outrageous there is at least some evidence that such may be the case. On the 10th of August 1972 a meteorite was seen to enter the Earth’s atmosphere and travel in a horizontal direction over Utah and Montana, eventually leaving the atmosphere somewhere over Alberta. It apparently approached the Earth at some 10 km/sec and was accelerated to 15 km/sec by the Earth’s gravity.
Straight forward enough however one can’t help making the following observation: it is an indisputable fact that just as any object leaving the Earth’s gravitational field needs to travel at the escape velocity, any object entering it would be accelerated by the same amount. In the case of the meteorite described above, whose trajectory was apparently well documented, it had accelerated by only some 5 km/sec, far less than 11.18 km/sec one might have expected.
A further consequence of such a gravitational bias is that the calculated g’s on the surface of other bodies such as the Moon or Mars, may actually be different. Because the mass immediately beneath the observer’s feet has a far greater effect, this would mean that as the body in question became smaller and the mass less, the acceleration due to gravity at its surface would not decrease in the same proportion but significantly less than this. As a result one might expect that the g’s on the surface of the Moon and Mars would be greater than the currently quoted values.
Also apparently affected by this bias are the celestial mechanics of the Solar System. Because the value of ‘g’ tails off quicker as you leave the surface of the gravitating body in question, its gravitational profile is also changed completely and tends to be weaker at a distance than previously calculated. Any body in orbit about it would by necessity have to move in a little closer to retain the same period of orbit. This would equally affect the orbits of the Moon about the Earth as well as each of the planets around the Sun.
Of course we have all been told that the distance between the Earth, the Moon and the Sun, and all the planets have already been calculated precisely but how are we to reconcile this with the above observations?
This is by the same writer who claims to have spotted the ISS in a telescope and seen it is physically little more than a round ball of some kind sitting in place for the fake space station. (See: ISS thread)