What is the Newton's version of Kepler's third law?

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Answer 1 · 2018-06-26T08:45:33

Newton's law $F_{g} = G \cdot \frac{M_{s} \cdot M_{p}}{R^{2}}$ $F_g=G·(M_s·M_p)/R^2$ where $M_{s}, M_{p}$ $M_s, M_p$ are the

mass of Sun and a planet, $G$ $G$ is a constant value and $R$ $R$ is the distance between Sun and Planet.

Kepler's Law is $\frac{T^{2}}{R^{3}} = K$ $T^2/R^3=K$ constant and T is period of traslation in orbit and R again, distance between Sun and Planet.

We know that centrifuge force is given by

$F_{c} = M_{p} \cdot a = M_{p} {(2 \frac{π}{T})}^{2} \cdot R$ $F_c=M_p·a=M_p(2pi/T)^2·R$ where a is acceleration in orbit

Then combining both expresions

$\frac{T^{2}}{R^{3}} = \frac{4 π^{2}}{G M_{s}}$ $T^2/R^3=(4pi^2)/(GM_s)$