The orbit of geostationary satellite is circular, the time period of satellite depends on (i) mass of the satellite, (ii) mass of the earth, (iii) radius of the orbit and (iv) height ofthe satellite from the surface of the earth

Orbital velocity, ${\mathrm{v}}_{\mathrm{o}} = \sqrt{\frac{{\mathrm{GM}}_{\mathrm{E}}}{{\mathrm{R}}_{\mathrm{E}}+\mathrm{h}}}$

Time period, T = $\frac{2\mathrm{\pi }({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}{{\mathrm{v}}_{\mathrm{o}}} = \frac{2\mathrm{\pi }{({\mathrm{R}}_{\mathrm{E}}+\mathrm{h})}^{{\displaystyle \frac{1}{2}}}}{{\left({\mathrm{GM}}_{\mathrm{E}}\right)}^{{\displaystyle \frac{1}{2}}}}$

Thus, the time period of satellite in independent of mass of satellite but depends on mass of the earth, radius of the orbit (${\mathrm{R}}_{\mathrm{E}}+\mathrm{h}$), height of the satellite from the surface of the earth.