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A $10\; kg$ satellite circles earth once every $2 \;hours$ 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.
$5.3 \times 10^{45}$
$6.1 \times 10^{42}$
$4.9 \times 10^{40}$
$7.1 \times 10^{48}$
Solution
$m v_{n} r_{n}=n h / 2 \pi$
Here $m=10 \,kg$ and $r_{n}=8 \times 10^{6} \,m .$
We have the time period $T$ of the circling satellite as $2 h$. That is $T=7200\, s$.
Thus the velocity $v_{n}=2 \pi r_{n} / T$
The quantum number of the orbit of satellite
$n=\left(2 \pi r_{n}\right)^{2} \times m /(T \times h)$
Substituting the values, $n=\left(2 \pi \times 8 \times 10^{6}\, m \right)^{2} \times 10 /\left(7200 s \times 6.64 \times 10^{-34}\, J s \right)$
$=5.3 \times 10^{45}$
