Question

An astronaut on the surface of the Moon fires a cannon to launch an experiment package, which leaves the barrel moving horizontally. Assume the free-fall acceleration on the Moon is one-sixth of that on the Earth.

(a) What must the muzzle speed of the package be so that it travels completely around the Moon and returns to its original location?

(b) What time interval does this trip around the Moon require?

Short answer

(a) The required speed of the package is $1.69\text{km}$.

(b) The required time interval for the trip around the moon is $1.8\text{h}$.

Step-by-step solution

Understanding the concept:

Write the formula to calculate the acceleration due to gravity on the moon

${\mathit{g}}_{\mathbf{m}}\mathbf{=}\frac{{\mathbf{v}}^{\mathbf{2}}}{{\mathbf{R}}_{\mathbf{m}}}$

Here,${\mathit{g}}_{\mathbf{m}}$is the acceleration due to gravity on the moon,${\mathbf{R}}_{\mathbf{m}}$is the radius of the moon and$\mathbf{V}$is the velocity of the package travelled.

The expression for the time interval for the trip around the moon is given by,

$\mathit{T}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }{\mathbf{R}}_{\mathbf{m}}}{\mathbf{v}}$

Here, $\mathit{T}$is time interval for trip around the moon, ${\mathbf{R}}_{\mathbf{m}}$is the radius of the moon, $\mathbf{v}$is the muzzle speed of the package.

Determine the muzzle speed of the package, so that it travels completely around the moon and returns to its original location

The free fall acceleration on the moon is one sixth of that on the earth.

Substitute $1.737\times {10}^6\mathrm{m}$for $R_m$and $\frac{9.81}{6}\mathrm{m}/{\mathrm{s}}^2$for $g_m$to find the $v$.

$\begin{array}{c}\frac{9.81}{6}\mathrm{m}/{\mathrm{s}}^2=\frac{v^2}{1.737\times {10}^6\mathrm{m}}\\ v=\sqrt{2839995}\mathrm{m}/\mathrm{s}=(1685.22\mathrm{m}/\mathrm{s})\left(\frac{1\mathrm{km}}{{10}^3\mathrm{m}}\right)\\ =1.69\mathrm{km}/\mathrm{h}\end{array}$

Therefore, the required speed of the package be travelled completely around the moon and return its original location is $1.69\mathrm{km}/\mathrm{h}$.

The required time interval for the trip around the moon

(b)

Write the formula to calculate the time interval for the trip around the moon

$T=\frac{2\pi R_{\mathrm{m}}}{v}$

Here, $T$is time interval for trip around the moon.

Substitute $1.737\times {10}^6\mathrm{m}$for $R_m$and $1685.22\mathrm{m}/\mathrm{s}$for $v$to find $v$.

$\begin{array}{c}T=\frac{2\pi \times 1.737\times {10}^6\mathrm{m}}{1685.22\mathrm{m}/\mathrm{s}}\\ =6477\mathrm{s}\times \frac{1\mathrm{h}}{3600\mathrm{s}}\\ =1.8\mathrm{h}\end{array}$

Therefore, the required time interval for the trip around the moon is $1.8\mathrm{h}$.