# Does "g =GM/d^2" best in situ in mathematical equation of "F = GMm/d^2 = mg"?

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#### E E K

• 486 ##### Does "g =GM/d^2" best in situ in mathematical equation of "F = GMm/d^2 = mg"?
« on: February 09, 2018, 03:38:41 PM »
Both earth and an apple (smaller objects) accelerate toward each other due to the force of gravitation but an apple appears a lot to the earth due to its greater acceleration as compared to the earth toward an apple, which is so minuscule to be distinguished.

Since the difference in masses is mammoth therefore it seems that not only the earth is stationary as compared to an apple but also the reduction in on-center distance "d" occurs due to falling of an apple ONLY (in its acceleration mode), but verily, both masses are changing their positions and hence on-center distance decreases due to the falling of both masses in their higher derivatives of motion (complex motion) before they strike each other.

This can easily be observed if the difference in masses is not so huge or if we consider the following two identical spherical masses (from point to celestial), which are separated by on-center distance d.

First Mass = M1, Second Mass = M2, M1 = M2 = Identical, Centre-to-Centre distance b/w M1 and M2 = d, d1 = d2, d1 + d2 = d, Gravitational acceleration of M1 = g1, Gravitational acceleration of M2 = g2, g1=g2 and c be the mid point of d".

Although, both M1 and M2 strike each other at c as per universal law of gravitation but since none of the M1 or M2 is stationary at c therefore neither M1 covers a distance d1 with g1=g2 nor M2 covers a distance d2 with g1=g2 on their road to c.  Acceleration g1=g2 is only possible if either M1 or M2 is stationary at c.

The earliest imaginable motions of M1 and M2 towards c might be due to the generation of g1 & g2 and the reduction in d (reduction in "d1" and "d2" equally on both sides of c) as well but after that both M1 and M2 start moving toward c at higher types of motion (such as gravitational jerk, jounce, crackle, pop, lock, drop etcetera or complex motion) as d1 and d2 decreases equally on both sides due to the change in positions of both M1 and M2 while on their ways to "c".

Both M1 and M2 move at much faster rate due to the formation of the complex motion instead of simply with accelerations before they hit each other at c. The actual striking time of M1 and M2 at "c" is much less than estimated by g1 and g2 combined.

So is F=GMm/d^2 well formulated?

Addendum: As gravity is universal therefore physical objects orbits other physical objects and hence causes the motions of planets, stars, and galaxies in the universe.
« Last Edit: February 09, 2018, 06:28:00 PM by E E K » #### rabinoz

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• Real Earth Believer ##### Re: Does "g =GM/d^2" best in situ in mathematical equation of "F = GMm/d^2 = mg"?
« Reply #1 on: February 09, 2018, 05:54:49 PM »
Both earth and an apple (smaller objects) accelerate toward each other due to the force of gravitation but an apple appears a lot to the earth due to its greater acceleration as compared to the earth toward an apple, which is so minuscule to be distinguished.

Since the difference in masses is mammoth therefore it seems that not only the earth is stationary as compared to an apple but also the reduction in on-center distance "d" occurs due to falling of an apple ONLY (in its acceleration mode), but verily, both masses are changing their positions and hence on-center distance decreases due to the falling of both masses in their higher derivatives of motion (complex motion) before they strike each other.

This can easily be observed if the difference in masses is not so huge or if we consider the following two identical spherical masses (from point to celestial), which are separated by on-center distance d.

First Mass = M1, Second Mass = M2, M1 = M2 = Identical, Centre-to-Centre distance b/w M1 and M2 = d, d1 = d2, d1 + d2 = d, Gravitational acceleration of M1 = g1, Gravitational acceleration of M2 = g2, g1=g2 and c be the mid point of d".

Although, both M1 and M2 strike each other at c as per universal law of gravitation but since none of the M1 or M2 is stationary at c therefore neither M1 covers a distance d1 with g1=g2 nor M2 covers a distance d2 with g1=g2 on their road to c.  Acceleration g1=g2 is only possible if either M1 or M2 is stationary at c.

The earliest imaginable motions of M1 and M2 towards c might be due to the generation of g1 & g2 and the reduction in d (reduction in "d1" and "d2" equally on both sides of c) as well but after that both M1 and M2 start moving toward c at higher types of motion (such as gravitational jerk, jounce, crackle, pop, lock, drop etcetera or complex motion) as d1 and d2 decreases equally on both sides due to the change in positions of both M1 and M2 while on their ways to "c".

Both M1 and M2 move at much faster rate due to the formation of the complex motion instead of simply with accelerations before they hit each other at c. The actual striking time of M1 and M2 at "c" is much less than estimated by g1 and g2 combined.

So is F=GMm/d^2 well formulated?
The general expression, F = GMm/d2,  is fine so long as the objects either have spherical symmetry or are so far apart that the uncertainties in the locations of their centres-of-gravity are insignificant.

But F = mg, where g = GM/d2, is meaningful only when M >> m, so that the acceleration of mass, M, can be neglected.

This is well satisfied on earth, with a mass of almost 6 x 1021 tonnes and the largest ship with a mass of about 400,000 tonnes.

And, of course, g = GM/d2 depends on d and hence altitude, so even on earth g is not quite constant.

Addendum: If you want the long version, here it is: The Free Fall Law of Galileo is Only Approximately Correct, by C P Viazminsky and P C Vizminiska
« Last Edit: February 10, 2018, 03:43:18 AM by rabinoz »