#### The time period of an earth satellite in circular orbit is independent of : Option 1) the mass of the satellite Option 2) radius of its orbit Option 3) both the mass and radius of the orbit Option 4) neither the mass of the satellite nor the radius of its orbit

Option  - 1 is correct,
For a satellite , Centripetal force = Gravitational force

Let,

m = mass of the satelite, Me = Mass of the earth, $\omega$ = Angular velocity of the satelite,
R = Distance of the satelite from the center of the earth, re = Radius of the earth,
h = Distance of the satelite from the surface of the earth

$\dpi{100} \therefore \; \; mR\omega ^{2}=\frac{GmM_{e}}{R^{2}}\; \; \; \; \; \; where\; R=r_{e}+h$

$\dpi{100} or\; \; \; \omega =\sqrt{\frac{GM_{e}}{R^{3}}}=\sqrt{\frac{GM_{e}}{(r_{e}+h)^{3}}}$

$\dpi{100} \therefore \; \; \; T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{(r_{e}+h)^{3}}{GM_{e}}}$

$\dpi{100} \therefore \; \; T\; is \; independent \; of \; mass\; (m)\; of\; satellite.$
But it depends on the radius of the orbit.

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For a satellite , Centripetal force = Gravitational force

Let,

m = mass of the satelite, Me = Mass of the earth, $\omega$ = Angular velocity of the satelite,
R = Distance of the satelite from the center of the earth, re = Radius of the earth,
h = Distance of the satelite from the surface of the earth

$\dpi{100} \therefore \; \; mR\omega ^{2}=\frac{GmM_{e}}{R^{2}}\; \; \; \; \; \; where\; R=r_{e}+h$

$\dpi{100} or\; \; \; \omega =\sqrt{\frac{GM_{e}}{R^{3}}}=\sqrt{\frac{GM_{e}}{(r_{e}+h)^{3}}}$

$\dpi{100} \therefore \; \; \; T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{(r_{e}+h)^{3}}{GM_{e}}}$

$\dpi{100} \therefore \; \; T\; is \; independent \; of \; mass\; (m)\; of\; satellite.$
But it depends on the radius of the orbit.