Chapter 7.2, Problem 16PP

a.

To determine

Speed of satellite.

Answer to Problem 16PP

  $v=7.81\times {10}^3\text{m/s}$

Explanation of Solution

Given:

Satellite distance from earth’s surface, $h=150\text{km}$ .

Formula Used:

Orbital speed:

  $v=\sqrt{G\frac{M_E}{R+h}}$

Where, G is the gravitational constant, M is the mass of Earth, R is the radius and his the height above Earth’s surface.

Calculation:

Gravitational acceleration:

  $\begin{array}{l}g=G\frac{M_{\text{E}}}{R^2}\\ \Rightarrow G{M}_E=g{R}^2\end{array}$

So, $v=R\sqrt{\frac{g}{R+h}}$ …… (1)

  $\begin{array}{l}g=9.8{\text{m/s}}^{\text{2}}\\ R=6.37\times {10}^6\text{m}\end{array}$

Substitute the value in (1)

  $\begin{array}{l}v=6.37\times {10}^6\times \sqrt{\frac{9.8}{6.37\times {10}^6+0.15\times {10}^6}}\\ =6.37\times {10}^6\times 1.226\times {10}^{-3}\\ =7.81\times {10}^3\text{m/s}\end{array}$

Conclusion:

Speed of satellite is $v=7.81\times {10}^3\text{m/s}$ .

b.

To determine

Time in minutes and seconds to return to cannon.

Answer to Problem 16PP

Time to return $T=5.2427\times {10}^3\text{s}$ or $T=87.37\min$

Explanation of Solution

Given:

Satellite distance from earth’s surface $h=150\text{km}$

Formula Used:

Time to return to cannon is:

  $T=\frac{2\pi \left(R+h\right)}{v}$

Calculations:

Substitute:

  $\begin{array}{l}R=6.37\times {10}^6\text{m}\\ h=0.15\times {10}^6\text{m}\\ v=7.81\times {10}^3\text{m/s}\\ \Rightarrow T=\frac{2\times 3.14\times \left(6.37\times {10}^6+0.15\times {10}^6\right)}{7.81\times {10}^3}\\ =5.2427\times {10}^3\text{s}\end{array}$

And in minutes

  $\begin{array}{l}T=\frac{5.2427\times {10}^3}{60}\min \\ =87.37\min \end{array}$

Conclusion:

Time to return $T=5.2427\times {10}^3\text{s}$ or $T=87.37\min$