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

Explanation

From Eq. (12.13), we have
`upsilon_(n)r_(n)=nh//2pi`
Here m = 10 kg and `r_(n) = 8 × 10^(6)` m. We have the time period T of the
circling satellite as 2 h. That is T = 7200 s.
Thus the velocity `upsilon_(n) = 2pi r_(n)/T`.
The quantum number of the orbit of satellite
`n=(2pin_(n))^(2)xxm//(Txxh)`
Substituting the values,
`n=(2pixx8xx10^(6)m)^(2)xx10//(7200 s 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
quantisation conditions tend to those of classical physics.