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A \(10\;\mathrm{km}\) satellite circles earth once every \(2\;h\) in an orbit, having a radius of \(8000\;\mathrm{km}\) . 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.

Answer

We have
\(mv_nr_n=\frac{\mathit{nh}}{2}\)
Here \(m=10\;\mathrm{kg}\) and \(r_n=8\times 10^6\;\mathrm{m.}\)  We have the time period \(T\) of the circling satellite as \(2\;h\). That is \(T=7200\;\text{s}\).
Thus the velocity \(v_n=\frac{2\pi r_n}{T}\)  The quantum number of the orbit of satellite \(n=\left(2\pi r_n\right)^2\times \frac{m}{\left(T\times h\right)}\)
Substituting the values, \(n=\left(2\pi \times 8\times 10^6m\right)^2\times \frac{10}{(7200\;\text{s}\times 6.64\times 10^{-34}\;\mathit{Js})}\)
\(=5.3\times 10^{45}\)
Note that the quantum number for the satellite motion is extremely large! In fact for such large quantum numbers the results of quaritisation conditions tend to those of classical physics.
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