A 20kg satellite circles the earth 2 hour in an orbit having a radius of `8000km`. If Bohr's angular momentum postulate is applied to the satallite. Find the quantum number of the orbit of the satallite. Take `h=6.6xx10^(-34)J-s`.
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Here, m=20kg, `T=2h=2xx60xx60s` `r=8000km=8xx10^6m` Total angular momentum of the satellite `=mvr=m(r omega)r=mr^2 omega=mr^2xx(2pi)/T` According to Bohr's concept, `mr^2((2pi)/T)=nh/(2pi)` `n=(4pi^2mr^2)/(Th)` `=(4xx(22//7)^2xx20(8xx10^6)^2)/(2xx60xx60xx6.6xx10^(-34))` `n=1.064xx10^(46)`
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