A 10; kg satellite circles Earth once every 2 | vedclass

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(A) According to Bohr's postulate, the angular momentum $L$ is given by $m v_n r_n = n h / 2 \pi$.Given: mass $m = 10\; kg$, radius $r_n = 8000\; km =

Vedclass · Accepted Answer

$5.3 	imes 10^{45}$

Vedclass · Answer

(A) According to Bohr's postulate, the angular momentum $L$ is given by $m v_n r_n = n h / 2 \pi$.Given: mass $m = 10\; kg$, radius $r_n = 8000\; km = 8 	imes 10^6\; m$, and time period $T = 2\; hours = 7200\; s$.The orbital velocity $v_n$ is given by $v_n = 2 \pi r_n / T$.Substituting $v_n$ into the angular momentum equation: $m (2 \pi r_n / T) r_n = n h / 2 \pi$.Rearranging for the quantum number $n$: $n = (2 \pi r_n)^2 	imes m / (T 	imes h)$.Substituting the values: $n = (2 \pi 	imes 8 	imes 10^6)^2 	imes 10 / (7200 	imes 6.63 	imes 10^{-34})$.$n = (25.13 	imes 10^6)^2 	imes 10 / (4.77 	imes 10^{-30})$.$n = 6.315 	imes 10^{14} 	imes 10 / 4.77 	imes 10^{-30} \approx 5.3 	imes 10^{45}$.

$A$ $10\; kg$ 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.

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