Chapter 7.1, Problem 6PP

a.

To determine

Orbital period of satellite.

Answer to Problem 6PP

  $0.06147\text{days}$

Explanation of Solution

Given:

Moon’s orbital period, $T_M$  is 27.3 days.

Moon’s center to center distance from Earth, $r_M$   $3.9\times {10}^5\text{km}$

Satellite distance from Earth’s center $r_S$  is $\text{6}\text{.7}\times {10}^3\text{km}$

Formula Used:

From Kepler’s third law of planetary motion:

  $T^2\propto r^3$

Where, T is the orbital period and r is the orbital radius.

Calculations:

Let the satellite’s orbital period be $T_S$ .

  $T_S{}^2=T_M{\left(\frac{r_S}{r_M}\right)}^3$

Now, put the given values:

  $\begin{array}{c}{T}_S{}^2=T_M{}^2{\left(\frac{r_S}{r_M}\right)}^3\\ T_S{}^2={\left(27.3\right)}^2{\left(\frac{6.7\times {10}^3}{3.9\times {10}^5}\right)}^3\\ T_S=0.06147\text{days}\end{array}$

Conclusion:

Orbital period of satellite is $0.06147\text{days}$ .

b.

To determine

Distance of satellite from Earth’s surface.

Answer to Problem 6PP

  $\text{332}\text{km}$

Explanation of Solution

Given:

Satellite distance from earth’s center $r_S$  is $\text{6}\text{.70}\times {10}^3\text{km}$

Formula Used:

Altitude = satellite distance from Earth’s center − Earth’s radius

Calculations:

As radius of Earth is 6378 km

So, altitude of satellite will be

  $\begin{array}{c}h=6700-6378\text{km}\\ \text{=322}\text{km}\end{array}$

Conclusion:

Distance of satellite from Earth’s surface is $\text{332}\text{km}$ .