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  • Help me understand! The time period of an earth satellite in circular orbit is independent of :

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

 

Answers (3)

best_answer

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

\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.
But it depends on the radius of the orbit.

Posted by

divya.saini

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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

\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.
But it depends on the radius of the orbit.

 

Posted by

Deependra Verma

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