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A 10kg satellite circles earth once ever...

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.

Text Solution

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From Eq., we have
`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 2 h. That is `T = 7200 s`.
Thus the velocity `v_n = 2pi r_n//T`.
The quantum number of the orbit of satellite
`n = (2pi r_n)^2 xx m//(T xx h)`.
Substituting the values, `n = (2pi xx 8 xx 10^6 m)2 × 10//(7200 s xx 6.64 xx 10^(-34) J s)`
`= 5.3 xx 10^(45)`.
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