A 10 kg satellite circles earth once every 2h in an orbit having a radius of 8000km. Assuming that Bohr's angular momentum postulate applies to a satellite just as it does to an electron in the hydrogen atom, then the quantum number of the orbit of satellite is

Explanation

`mv_(n) r_(n) `= nh /`2pi` . Here m = 10 kg and `r_(n) 8 xx 10^(6)`m . we have the time period T of the circling satellite as 2h. That is T = 7200 s.
Thus the velocity `v_(n) = 2pi r_(n)`/T
The quantum number of the orbit of satellite
n = `(2pi xx 8 xx 10^(6) m)^(2) xx 10 `/(7200s `xx 6.64 xx 10^(-34)`JS) = 5.3 `xx 10^(45)`
Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quantization condition tend to those of classical physics.