Q.The time period of a satellite orbiting Earth in a circular orbit is independent of

A satellite stays in a stable circular orbit because Earth's gravitational pull supplies exactly the centripetal force needed to keep it curving around the planet rather than moving off in a straight line.

Orbital speed. For a satellite of mass $M_s$ orbiting at height $h$ (orbital radius $R_e+h$), equating the required centripetal force to the gravitational force gives

Note that $v$ depends only on the orbital height, not on the satellite's own mass -- and as $h$ increases, the required orbital speed decreases: satellites farther out move more slowly.

Time period. Since $v=$ (circumference) / (period), $v=2\pi(R_e+h)/T$, solving for $T$ gives

$T=2\pi\sqrt{\frac{(R_e+h)^3}{GM_e}},$ …