What Are the Limitations of the Law of Universal Gravitation?

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SUMMARY

The discussion focuses on deriving Newton's Law of Universal Gravitation using mathematical principles. The equation T² = (4π²/GM)R³ is established, linking the period of circular motion to gravitational force. The derivation utilizes circular motion equations, specifically F = mv²/R and T = 2πR/v, leading to the conclusion that gravitational force F can be expressed as F = GMm/R². This confirms the relationship between gravitational force and the masses involved, as well as the distance between them.

PREREQUISITES
  • Understanding of Newton's Laws of Motion
  • Familiarity with circular motion equations
  • Knowledge of Kepler's Laws of Planetary Motion
  • Basic algebra and manipulation of equations
NEXT STEPS
  • Study the derivation of Kepler's Laws in detail
  • Explore the implications of gravitational force in astrophysics
  • Learn about the limitations of Newtonian gravity in extreme conditions
  • Investigate the concepts of General Relativity and its differences from Newton's Law
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Students of physics, educators teaching classical mechanics, and anyone interested in the mathematical foundations of gravitational theory.

Wildcatfan
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Update: Thanks for the help! I'll work on it...
 
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I'm not sure what your equation is, but this is how I would derive Newton's Law

[tex]T^2 = \frac{4\pi^2}{GM}R^3[/tex]

In circular motion,

[tex]F = \frac{mv^2}{R}[/tex]

The period in circular motion is

[tex]T = \frac{2\pi R}{v} \Rightarrow v = \frac{2\pi R}{T}[/tex]

So, by substitution,

[tex]F = \frac{m\left(\frac{2\pi R}{T}\right)^2}{R}=\frac{4\pi^2 mR}{T^2}[/tex]

[tex]\Rightarrow T^2 = \frac{4\pi^2 mR}{F}[/tex]

Substituting into Kepler's 3rd,

[tex]\frac{4 \pi ^2 mR}{F} = \frac{4\pi^2}{GM}R^3[/tex]

[tex]\therefore F = \frac{GMm}{R^2}[/tex]
 

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