Kepler’s third law, obtained by equating the gravitational force to the centripetal force, relates the orbital period and radius to the mass of the central body:
T2=GM4π2r3
Isolating the mass M:
M=GT24π2r3
Calculating the factors: r3=(108)3=1024 m3; T2=(86400)2=7,46⋅109 s2; 4π2=39,5. So:
M=6,67⋅10−11⋅7,46⋅10939,5⋅1024=4,98⋅10−13,95⋅1025≈7,94⋅1025 kg
Comparing with the Earth’s mass MT≈5,97⋅1024 kg, the planet has about 13 Earth masses.
M≈7,94⋅1025 kg (∼13 MEarth)