Question

The time period of an earth satellite in circular orbit is independent of(a) both the mass and radius of the orbit(b) radius of its orbit(c) the mass of the satellite(d) neither the mass of the satellite nor the radius of its orbit.

Answer 1

As we learnt in  

For a satellite , Centripetal force = Gravitational force

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

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

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

\therefore \; \; T\; is \; independent \; of \; mass\; (m)\; of\; satellite.

mw^2R = \frac{G_{e}m}{R^3} = \sqrt{\frac{Gm_{e}}{\left ( re+h \right )^3}}mw^2R = \frac{G_{e}m}{R^3} where R = R_{r} h

w = \sqrt{\frac{Gm_{e}}{r^3}} = \sqrt{\frac{Gm_{e}}{\left ( re+h \right )^3}}

T = \frac{2_{x}}{w} = 2\times \frac{\sqrt{\left ( r_{e} +h \right )}}{Gm_{e}}

\therefore T\ is\ independent\ of\ mass\ of\ Satelite