Question about the Derivation of the Gravitational Law

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SUMMARY

The discussion centers on the derivation of Newton's Law of Universal Gravitation, specifically the equation \(\frac{m}{k} = \frac{M}{k'}\) and its relation to force calculations. Participants clarify that squaring the force is necessary to incorporate both masses \(m\) and \(M\) into the equation for force \(F\). This algebraic manipulation simplifies the derivation process, allowing for a clearer solution in terms of the constants \(k\), \(m\), and \(M\). The conversation emphasizes the importance of understanding these mathematical relationships in gravitational physics.

PREREQUISITES
  • Understanding of Newton's Third Law of Motion
  • Familiarity with algebraic manipulation in physics
  • Basic knowledge of gravitational force equations
  • Experience with mathematical relationships in physics
NEXT STEPS
  • Study the derivation of Newton's Law of Universal Gravitation in detail
  • Learn about the implications of squaring forces in physics
  • Explore algebraic techniques for solving equations in gravitational contexts
  • Investigate the role of constants like \(k\) and \(k'\) in gravitational calculations
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Students of physics, educators teaching gravitational concepts, and anyone interested in the mathematical foundations of gravitational laws will benefit from this discussion.

ecastro
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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:

\frac{m}{k} = \frac{M}{k'}.

This is similar to

\frac{C}{M} = \frac{c}{m} = \frac{k}{4 \pi^2}

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?
 
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ecastro said:
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##.
 
Alright. Thank you for your help. :D
 

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