Circular Motion: Kepler's Third Law

Question	Answer
Examples of planets undergoing circular motion	Moon orbiting the earth; Earth orbiting the sun
Equating the equation for gravity and the equation for centripetal force	v² ∝ 1/R
Equating the equation for gravity and the equation for centripetal force: Derivation	GMm/d² = mv²/r --> d = r = R --> GMm/R = mv² --> v² = GM/R --> v² ∝ 1/R (as G & M are constants)
Kepler's Third Law	The square of the periodic time of a satellite's orbit around a planet is proportional to the cube of the radius of the orbit (including the radius of the planet) and inversely proportional to the mass of the planet
Formula for Kepler's Third Law	T² = 4π²R³/GM
Derivation of formula of Kepler's Third Law	Gravitational force = centripetal force --> GMm/R² = mω²R --> GM/R² = ω²R --> GM/R² = (2π/T)²R --> GM/R² = 4π²R/T² --> T² = 4π²R³/GM

Resource summary

	Created by alex.examtime9373 over 10 years ago

	Copied by tatemae.honne over 10 years ago