Chapter 8.5, Problem 43E

Solution

a)

To find:the velocity of the lunar module which is in a circular orbit $250$ km above the surface of the moon, assuming the moon’s radius is $1740$ km and that $k=0.012$ .

The velocity of the lunar module is $1551$ m/sec.

Given information:

  $v=\sqrt{\frac{3.99\times {10}^{14}k}{r}}$

Calculation:

The lunar excursion module is in a circular orbit $250$ km above the surface of the moon. Therefore,

  $\begin{array}{c}r=1740+250\\ =1990\end{array}$

Radius is in km, but the velocity is given in m/sec, therefore convert the radius in meters.

  $1$ km = $1000$ m

So, $1990$ km = $1990\times {10}^3$ m

Substitute $1990\times {10}^3$ for $r$ , $0.012$ for $k$ in the planet velocity equation $v$ .

  $\begin{array}{c}v=\sqrt{\frac{3.99\times {10}^{14}\times 0.012}{1990\times {10}^3}}\\ =\sqrt{\frac{0.04788}{19900}}\times \frac{{10}^7}{10}\\ =\sqrt{0.00000240603}\times {10}^6\\ \approx 0.001551\times {10}^6\\ \approx 1551\end{array}$

Hence, the velocity of the lunar module is $1551$ m/sec.

b)

To find:the length of time required for the lunar module to circle the moon once.

The time taken is $2$ hours and $14$ minutes.

Given information:

The velocity of the lunar module is $1551$ m/sec.(Calculated in the previous section)

Formula used:

Time taken can be calculated using the formula: $\frac{2\pi r}{v}$

Calculation:

Substitute $1990\cdot {10}^3$ for $r$ , $3.14$ for $\pi$ and $1551$ for $v$ in the formula to find the time taken.

  $\frac{2\cdot 3.14\cdot 1990\cdot {10}^3}{1551}\approx 8057$

The time is in seconds. So, convert it into hours by dividing it by $3600$ .

  $\frac{8057}{3600}\approx 2.24$

To write the time taken in hours and minutes, multiply $.24$ by $60$ .

  $0.24\times 60=14.4$

Hence, the time taken is $2$ hours and $14$ minutes.