The Earth's Quantum Number

solved BohrModel

Q. In accordance with the Bohr's model, what is the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 × 10¹¹ m $r = 1.5 \times 10^{11}$ m with orbital speed 3 × 10⁴ m/s $v = 3 \times 10^{4}$ m/s? (Mass of earth = 6.0 × 10²⁴ kg $m = 6.0 \times 10^{24}$ kg.)

Inputs

radius

r

150000000000.00 m

speed

v

30000.0 m/s

mass

m

6e+24 kg

Outputs

number

n

How it's solved

Solved with BohrModel

Given

$r = 1.5 \times 10^{11}$ m
$v = 3 \times 10^{4}$ m/s
$m = 6.0 \times 10^{24}$ kg

Find $n$ .

Working

angular momentum comes in steps(earth) $m \cdot v \cdot r = \frac{n \cdot h}{2 \cdot π}$
$6 \times 10^{24} \cdot 30000 \cdot 1.5 \times 10^{11} = \frac{n \cdot 6.62607015 \times 10^{−34}}{2 \cdot 3.14159265}$
$n = 2.56028082 \times 10^{74}$
$n \approx 2.6 \times 10^{74}$ to 2 significant figures

Conclusion

Angular momentum comes in steps of Planck's constant over two pi, but the earth's is so vast that the steps are far too fine to notice: quantum rules fade into classical ones at large sizes.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :the_earth_s_quantum_number do4  kind :numerical5  describes "In accordance with the Bohr's model, %{number} that characterises the earth's revolution around the " \6            "sun in an orbit of radius %{radius} with orbital speed %{speed}? (Mass of earth = %{mass}.)"7  given radius: "1.5e11 m"8  given speed: "3e4 m/s"9  given mass: "6.0e24 kg"10  asks number: "what is the quantum number"11end

solution.rb

1require_relative "../questions/the_earth_s_quantum_number"2require_relative "../science/atoms/bohr_model"34solve :the_earth_s_quantum_number, using: BohrModel do5  objects :earth67  take radius: earth.orbit_radius8  take speed: earth.orbital_speed9  take mass: earth.mass1011  number = earth.quantum_number12  answer number:, figures: 21314  conclusion "Angular momentum comes in steps of Planck's constant over two pi, but the earth's is so vast that the " \15             "steps are far too fine to notice: quantum rules fade into classical ones at large sizes."16end

playground.rb

1require_relative "../solutions/the_earth_s_quantum_number"23play_with :the_earth_s_quantum_number do4  title "The Earth's Quantum Number"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	BohrModel
coulomb constant	$k$	N m² C⁻²	—	BohrModel
elementary charge	$e$	C	—	BohrModel
plancks constant	$h$	J s	—	BohrModel
mass	$m$	kg	0 … ∞	BohrModel
orbit radius	$r$	m	—	BohrModel
orbital speed	$v$	m/s	0 … ∞	BohrModel
binding energy	$E_{b}$	J	—	BohrModel
quantum number	$n$	—	—	BohrModel
atomic number	$Z$	—	—	BohrModel
potential energy	$U$	J	—	BohrModel
total energy	$E$	J	—	BohrModel
photon energy	$E_{p}$	J	—	BohrModel
frequency	$ν$	Hz	0 … ∞	BohrModel
binding energy of ground	$E_{b,g}$	J	—
orbit radius of ground	$r_{g}$	m	—
binding energy of earth	$E_{b}$	J	—
orbit radius of earth	$r$	m	—
mass of ground	$m_{g}$	kg	0 … ∞
orbital speed of ground	$v_{g}$	m/s	0 … ∞
mass of earth	$m$	kg	0 … ∞
orbital speed of earth	$v$	m/s	0 … ∞
atomic number of ground	$Z_{g}$	—	—
quantum number of ground	$n_{g}$	—	—
atomic number of earth	$Z$	—	—
quantum number of earth	$n$	—	—
potential energy of ground	$U_{g}$	J	—
potential energy of earth	$U$	J	—
total energy of ground	$E_{g}$	J	—
total energy of earth	$E$	J	—
photon energy of ground	$E_{p,g}$	J	—
frequency of ground	$ν_{g}$	Hz	0 … ∞
photon energy of earth	$E_{p}$	J	—
frequency of earth	$ν$	Hz	0 … ∞

E_{b,e} = \frac{k \cdot e^{2}}{2 \cdot r_{e}}

energy to free the electron

\frac{m_{e} \cdot {v_{e}}^{2}}{r_{e}} = \frac{k \cdot e^{2}}{{r_{e}}^{2}}

held in orbit by the proton

v_{e} = \frac{2 \cdot π \cdot k \cdot Z_{e} \cdot e^{2}}{n_{e} \cdot h}

speed set by the level

m_{e} \cdot v_{e} \cdot r_{e} = \frac{n_{e} \cdot h}{2 \cdot π}

angular momentum comes in steps

U_{e} = \frac{- k \cdot e^{2}}{r_{e}}

energy of its place

E_{e} = \frac{- k \cdot e^{2}}{2 \cdot r_{e}}

energy of the atom

E_{p,l} = h \cdot ν_{l}

one photon for one jump

E_{b,g} = \frac{k \cdot e^{2}}{2 \cdot r_{g}}

energy to free the electron(ground)

E_{b} = \frac{k \cdot e^{2}}{2 \cdot r}

energy to free the electron(earth)

\frac{m_{g} \cdot {v_{g}}^{2}}{r_{g}} = \frac{k \cdot e^{2}}{{r_{g}}^{2}}

held in orbit by the proton(ground)

\frac{m \cdot v^{2}}{r} = \frac{k \cdot e^{2}}{r^{2}}

held in orbit by the proton(earth)

v_{g} = \frac{2 \cdot π \cdot k \cdot Z_{g} \cdot e^{2}}{n_{g} \cdot h}

speed set by the level(ground)

v = \frac{2 \cdot π \cdot k \cdot Z \cdot e^{2}}{n \cdot h}

speed set by the level(earth)

m_{g} \cdot v_{g} \cdot r_{g} = \frac{n_{g} \cdot h}{2 \cdot π}

angular momentum comes in steps(ground)

m \cdot v \cdot r = \frac{n \cdot h}{2 \cdot π}

angular momentum comes in steps(earth)

U_{g} = \frac{- k \cdot e^{2}}{r_{g}}

energy of its place(ground)

U = \frac{- k \cdot e^{2}}{r}

energy of its place(earth)

E_{g} = \frac{- k \cdot e^{2}}{2 \cdot r_{g}}

energy of the atom(ground)

E = \frac{- k \cdot e^{2}}{2 \cdot r}

energy of the atom(earth)

E_{p,g} = h \cdot ν_{g}

one photon for one jump(ground)

E_{p} = h \cdot ν

one photon for one jump(earth)