Just wondering. All the books say that at the equator the field is weaker than the pole because of the distance. hmm.. But this is what I have in mind initially but need a stronger explanation from you guys.. haha.. From the centre of the earth, the field is zero. as it goes towards the the surface, the field increase with radius and mass. So the equator has greater radius, wouldn't the field get stronger? I know that g is proportional to 1/r^2 but that applies to the field beyond the surface of the earth right?
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It requires an extensive study of the superposition principle, adding up the infinitesimal gravity contributions of each local unit of the Earth's mass to actually understand the situation. The entire body of Earth acts on the target object, so you must consider gravity due to every unit of mass of Earth, and add them all up. This can make a calculus nightmare for bodies that aren't uniform.
In any case of the poles or the equator, the location is EXTERNAL to ALL of Earth's mass, and thus in both cases, all of Earth's mass applies as the numerator of the formula. The only distinguishing term is the radius in the 1/r^2 relationship, which would indeed imply that TRUE GRAVITY at the equator is less than TRUE GRAVITY at the poles. The values for TRUE GRAVITY are 9.81 N/kg at the equator and 9.83 N/kg at the poles, for sea level for both cases.
"True gravity"? You might as well ask what would be "false" gravity?
Well, because we rotate along with the Earth, it is very hard to make a measurement of the true gravity. We are subject to the illusion of the rotation of the Earth. That will make the gravity seem to be a little less, since some of gravity is already "used up" in providing the force to cause centripetal acceleration.
Only the gravity that needs to be opposed by another constraint force ever gets noticed or measured. So at the equator, the measured gravity is slightly less than the true gravity. 9.78 N/kg instead of 9.81 N/kg. At the poles, true gravity is the measured gravity.
In any case of the poles or the equator, the location is EXTERNAL to ALL of Earth's mass, and thus in both cases, all of Earth's mass applies as the numerator of the formula. The only distinguishing term is the radius in the 1/r^2 relationship, which would indeed imply that TRUE GRAVITY at the equator is less than TRUE GRAVITY at the poles. The values for TRUE GRAVITY are 9.81 N/kg at the equator and 9.83 N/kg at the poles, for sea level for both cases.
"True gravity"? You might as well ask what would be "false" gravity?
Well, because we rotate along with the Earth, it is very hard to make a measurement of the true gravity. We are subject to the illusion of the rotation of the Earth. That will make the gravity seem to be a little less, since some of gravity is already "used up" in providing the force to cause centripetal acceleration.
Only the gravity that needs to be opposed by another constraint force ever gets noticed or measured. So at the equator, the measured gravity is slightly less than the true gravity. 9.78 N/kg instead of 9.81 N/kg. At the poles, true gravity is the measured gravity.
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