A 10 kg satellite circles earth once every 2 h in an orbit having a radius of 8000 km. 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

We have
`mv_(n)r_(n)=(nh)/(2pi)`
Here `m=10 kg and r_(n)=8xx10^(6)m`. We have the time period T of the circling satellite as 2 h. That is T = 7200 s.
Thus the velocity of `v_(n)=(2pi r_(n))/(T)`
The quantum number of the orbit of satellite
`n=((2pir_(n))^(2)m)/((Txxh))`
Substituting the values
`n=((2pi xx8xx10^(6)m)^(2)xx10)/(7200xx6.64xx10^(-34)Js)`
`5.3xx10^(45Js`
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.