Question

The Moon orbits Earth at a radius of ${\mathbf{\text{3.84×10}}}^{\mathbf{\text{8}}}\mathbf{\text{m}}$. To do so as a classical particle. Its wavelength should be small. But small relative to what? Being a rough measure of the region where it is confined, the orbit radius is certainly a relevant dimension against which to compare the wavelength. Compare the two. Does the Moon indeed orbit as a classical particle? (localid="1659095974931" ${\mathbf{\text{m}}}_{\mathbf{\text{Earth}}}{\mathbf{\text{=5.98×10}}}^{\mathbf{\text{24}}}\mathbf{\text{kg}}$and ${\mathbf{\text{m}}}_{\mathbf{\text{moon}}}{\mathbf{\text{=7.35×10}}}^{\mathbf{\text{22}}}\mathbf{\text{kg}}$)

Short answer

The moon is unmistakably orbiting in the manner of a classical particle.

${\text{λ}}_{\text{moon}}=\text{8.84}\times {\text{10}}^{\text{-60}}\text{\hspace{0.17em}m}$

Step-by-step solution

Step 1:Concept Introduction

To be treated as a classical particle, the wavelength of the orbiting moon must be modest in comparison to the orbiting radius, according to the problem statement. Will find the appropriate velocity of the moon using Newton's second law, and then apply the de-Broglie equation to find the associated wavelength.

$\begin{array}{l}{\mathbf{\text{F}}}_{\mathbf{\text{c}}}\mathbf{=}\frac{{\mathbf{\text{Gm}}}_{\mathbf{\text{e}}}{\mathbf{\text{m}}}_{\mathbf{\text{m}}}}{{\mathbf{\text{r}}}^{\mathbf{\text{2}}}}\mathbf{=}\frac{{\mathbf{\text{m}}}_{\mathbf{\text{m}}}{\mathbf{\text{v}}}^{\mathbf{\text{2}}}}{\mathbf{\text{r}}}\\ \mathbf{\text{v}}\mathbf{=}\sqrt{\frac{{\mathbf{\text{Gm}}}_{\mathbf{\text{e}}}}{\mathbf{\text{r}}}}\end{array}$

Substituting formula.

$\begin{array}{c}\text{v}=\sqrt{\frac{(\text{6.67}\times {\text{10}}^{\text{-11}}{m}^3\cdot k{g}^{-1}\cdot s^{-2})\times (\text{5.98}\times {\text{10}}^{\text{24}}\text{\hspace{0.17em}kg})}{\text{3.84}\times {\text{10}}^{\text{8}}\text{m}}}\\ =\text{1.02}\times {\text{10}}^{\text{3}}\text{m/s}\end{array}$

$\begin{array}{c}{\text{λ}}_{\text{m}}=\frac{\text{h}}{{\text{m}}_{\text{m}}\text{v}}\\ {\text{λ}}_{\text{m}}=\frac{\text{6.63}\times {\text{10}}^{\text{-34}}J\cdot s}{(\text{7.35}\times {\text{10}}^{\text{22}}\text{kg})\times (\text{1.02}\times {\text{10}}^{\text{3}}\text{\hspace{0.17em}m/s})}\\ {\text{λ}}_{\text{m}}=\text{8.84}\times {\text{10}}^{\text{-60}}\text{m}\end{array}$

Because this is such a small value in comparison to the orbital radius, the moon is orbiting as a classical particle.

${\text{λ}}_{\text{moon}}=\text{8.84}\times {\text{10}}^{\text{-60}}\text{\hspace{0.17em}m}$

Therefore, the moon is unmistakably orbiting in the manner of a classical particle.