Question

Whenever two Apollo astronauts were on the surface of the Moon, a third astronaut orbited the Moon. Assume the orbit to be circular and $\mathbf{100}\mathbf{}\mathbf{\text{km}}$above the surface of the Moon, where the acceleration due to gravity is role="math" localid="1663698425984" $\mathbf{1}\mathbf{.}\mathbf{52}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$. The radius of the Moon is $\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{m}$. Determine (a) the astronaut’s orbital speed and (b) the period of the orbit.

Short answer

The solution is

1. The astronaut’s orbital speed $v=1.65\times {10}^3\mathrm{m}/\mathrm{s}$
2. Period of the orbit $T=1.90\text{h}$

Step-by-step solution

Given data

The radius of the circular orbit above the surface of the moon is 100 km

The acceleration due to gravity is $1.52{\text{m/s}}^2$.

The radius of the Moon is $1.70\times {10}^6\mathrm{m}$

Concept Introduction

The centripetal force will act inward, causing centripetal acceleration.

The expression for the centripetal force is,

$\mathbf{F}\mathbf{=}\frac{{\mathbf{mv}}^{\mathbf{2}}}{\mathbf{r}}$

Here, $\mathit{m}$is the mass of the electron, $\mathit{v}$ is the velocity, $\mathit{r}$ is the radius.

Calculation of velocity.

(a)

Newton’s second law is used to calculate the astronaut's orbital speed.

$\sum F_y=m{a}_y:m{g}_{moon}down=\frac{m{v}^2}{r}down$

Solving for the velocity gives,

$\begin{array}{c}v=\sqrt{g_{moon}r}\\ =\sqrt{\left(1.52\mathrm{m}/{\mathrm{s}}^2\right)\left(1.7\times {10}^{6}m+100\times {10}^{3}m\right)}\\ =1.65\times {10}^3\mathrm{m}/\mathrm{s}\end{array}$

Calculation of time period.

(b)

To find the period, we use$v=-\frac{2\pi r}{T}$and

$\begin{array}{c}T=\frac{2\pi \left(1.8\times {10}^{6}m\right)}{1.65\times {10}^3\mathrm{m}/\mathrm{s}}\\ T=6.84\times {10}^{3}s\left(\frac{1h}{86400s}\right)\\ T=1.90\text{h}\end{array}$