Kepler's Third Law

using harmonic law, ${ P }^{ 2 }=\frac { 4{ \pi }^{ 2 } }{ G({ M }_{ 1 }+{ m }_{ 2 }) } { R }^{ 3 }$ , if the mass of the star is 4 times the mass of the planet, then ${ P }^{ 2 }=\frac { 4{ \pi }^{ 2 } }{ G({ 4m }_{ 1 }+{ m }_{ 1 }) } { R }^{ 3 }$ . substitute all the given values, we arrived at

$3.13\times { 10 }^{ x }\quad =\quad \frac { { 4\pi }^{ 2 }\left( { 10 }^{ 3 } \right) }{ { \left( 6.67\times 10 \right) }^{ -11 }\left( 5 \right) ({ 2 })^{ 2 } }$

to get the value of x, apply the logarithms on both sides, and we get:

$x\quad =\quad \log { \frac { { 2.959\times 10 }^{ 13 } }{ 3 } } =\quad 12.99\approx 13$

13, because Tau squared is equal to 4 pi^2 a^3 all divided by G(Star mass + Planet mass) according to Kepler's third law.