Question about the Derivation of the Gravitational Law

  • #1
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Main Question or Discussion Point

The derivation of the law have been put up in the forums but I have a question regarding its derivation.

I understood everything from the assumptions to the application of Newton's Third Law, but I got stocked at this step:

[tex] \frac{m}{k} = \frac{M}{k'} [/tex].

This is similar to

[tex] \frac{C}{M} = \frac{c}{m} = \frac{k}{4 \pi^2} [/tex]

at this site, http://www.relativitycalculator.com/Newton_Universal_Gravity_Law.shtml.

According to the same site, the next step requires the force to be squared. Why is this so? Is it merely to acquire the force ##F## between the two bodies? Aren't there any other ways to calculate the force other than multiplication?
 

Answers and Replies

  • #2
Nugatory
Mentor
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According to the same site, the next step requires the force to be squared. Why is this so? Is it merely to acquire the force ##F## between the two bodies? Aren't there any other ways to calculate the force other than multiplication?

It's just a convenient algebra trick to get both ##m## and ##M## into the equation for ##f##. We have ##f=f'## so we can multiply both sides of that equation by ##f## to get one equation that can be solved for ##f## in terms of ##k##, ##m##, and ##M##.
 
  • #3
251
8
Alright. Thank you for your help. :D
 
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