Newton showed that F=G(m1m2)/r2
When calculating the force of gravity on the surface of the Earth, why do we use the radius of the Earth for our r value? Of course, the above Newtonian equation suggests we should, but it seems rather counter-intuitive.
I must admit to not having taken a relevant course since an Honors Physics course in high school. If we were ever given a logical explanation (with minimal calculus required) for this, I cannot recall it. I have also posed this question to a friend of mine who, while not a physics major, attended a small, but well respected, science and engineering school in Southern California. He could not recall an answer to this question, though he did remember learning about Shell Theorem. Despite my limited understanding of Calculus, Shell Theorem does make some degree of logical sense.
In any case, my question is this: Since the mass directly under my feet contributes significantly more gravitational force than the equally dense mass on the opposite side of the Earth (in a spherically symmetrical Earth where density only varies by depth), shouldn't the center of gravity be somewhat closer to me than the center of the Earth?
If I were to compose a fictional earth consisting of tens of thousands of spheres, of say 100km radius, would I not find that the sphere directly below my feet with radius=100km contributes many magnitudes more force than the sphere opposite it with radius= approx 12,500km? By my calculations the nearer 100km radius sphere would exert over 15,000 times the force of the sphere on the opposite side of Earth.
I have been unable to find a satisfactory answer to this quandary (based on my limited knowledge of both physics and calculus).
I have tried a variety of sources, but keep getting answers similar to this one from, groan, Wikipedia:
"If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become 'infinitely small', this entails integrating the force (in vector form, see below) over the extents of the two bodies.
In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object's mass were concentrated at a point at its centre.[3] (This is not generally true for non-spherically-symmetrical bodies.)"
It seems logical that if the two masses are sufficiently far apart, r would approach the distance from their centers, and yet on the surface of the Earth where the distance is relatively short, my brain cannot comprehend how this is so.
Is someone able to explain the above in a manner that makes logical sense to someone with my limited math and science skills and is consistent with Newton's Law of Gravitation? How would you demonstrate this to an intelligent Undergraduate who is unable or unwilling to do the vector math that appears to be required.
Thanks and I have already learned a great many things from the superb minds on this forum.
When calculating the force of gravity on the surface of the Earth, why do we use the radius of the Earth for our r value? Of course, the above Newtonian equation suggests we should, but it seems rather counter-intuitive.
I must admit to not having taken a relevant course since an Honors Physics course in high school. If we were ever given a logical explanation (with minimal calculus required) for this, I cannot recall it. I have also posed this question to a friend of mine who, while not a physics major, attended a small, but well respected, science and engineering school in Southern California. He could not recall an answer to this question, though he did remember learning about Shell Theorem. Despite my limited understanding of Calculus, Shell Theorem does make some degree of logical sense.
In any case, my question is this: Since the mass directly under my feet contributes significantly more gravitational force than the equally dense mass on the opposite side of the Earth (in a spherically symmetrical Earth where density only varies by depth), shouldn't the center of gravity be somewhat closer to me than the center of the Earth?
If I were to compose a fictional earth consisting of tens of thousands of spheres, of say 100km radius, would I not find that the sphere directly below my feet with radius=100km contributes many magnitudes more force than the sphere opposite it with radius= approx 12,500km? By my calculations the nearer 100km radius sphere would exert over 15,000 times the force of the sphere on the opposite side of Earth.
I have been unable to find a satisfactory answer to this quandary (based on my limited knowledge of both physics and calculus).
I have tried a variety of sources, but keep getting answers similar to this one from, groan, Wikipedia:
"If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become 'infinitely small', this entails integrating the force (in vector form, see below) over the extents of the two bodies.
In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object's mass were concentrated at a point at its centre.[3] (This is not generally true for non-spherically-symmetrical bodies.)"
It seems logical that if the two masses are sufficiently far apart, r would approach the distance from their centers, and yet on the surface of the Earth where the distance is relatively short, my brain cannot comprehend how this is so.
Is someone able to explain the above in a manner that makes logical sense to someone with my limited math and science skills and is consistent with Newton's Law of Gravitation? How would you demonstrate this to an intelligent Undergraduate who is unable or unwilling to do the vector math that appears to be required.
Thanks and I have already learned a great many things from the superb minds on this forum.
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