The orbit of a geostationary satellite is cir | vedclass

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(D) The gravitational force provides the necessary centripetal force for the circular orbit of the satellite: $\frac{GMm}{(R+h)^{2}} = m \omega^{2}(R+

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$(ii), (iii)$ and $(iv)$

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(D) The gravitational force provides the necessary centripetal force for the circular orbit of the satellite: $\frac{GMm}{(R+h)^{2}} = m \omega^{2}(R+h)$.Here,$G$ is the gravitational constant,$M$ is the mass of the Earth,$m$ is the mass of the satellite,$R$ is the radius of the Earth,and $h$ is the height of the satellite above the surface.Simplifying for angular velocity $\omega$: $\omega = \sqrt{\frac{GM}{(R+h)^{3}}}$.Since $\omega = \frac{2\pi}{T}$,the time period $T$ is given by $T = 2\pi \sqrt{\frac{(R+h)^{3}}{GM}}$.The radius of the orbit is $r = R+h$.Thus,$T = 2\pi \sqrt{\frac{r^{3}}{GM}}$.From this expression,$T$ depends on the mass of the Earth $(M)$,the radius of the orbit $(r)$,and the height $(h)$ (since $r = R+h$). It does not depend on the mass of the satellite $(m)$.Therefore,the correct factors are $(ii), (iii),$ and $(iv)$.

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

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