Chapter 5, Problem 58P

To determine

The period of an artificial satellite orbiting very near the surface of the Earth’s surface.

Answer to Problem 58P

Solution:

84.5minutes

Explanation of Solution

Given:

Period of Moon, $T_m=27.4\text{days}$

Calculation:

  $\begin{array}{c}{T}_m=27.4\text{days}\\ =27.4\times 24\text{hours}\\ =27.4\times 24\times 3600\text{s}\end{array}$

  $T_m=2367.36\times {10}^{\text{3}}\text{s}$

The radius of the near Earth’s orbit, $r_e=6.38\times {10}^6\text{m}$ and the orbit radius of the Moon, $r_m=3.84\times {10}^8\text{m}$

According to Kepler’s third law of planetary motion:

  ${\left(\frac{T_e}{T_m}\right)}^2={\left(\frac{r_e}{r_m}\right)}^3$

  $T_e=T_m{\left(\frac{r_e}{r_m}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Substitute the values and solve:

  $T_e=2367.36\times {10}^3{\left(\frac{6.38\times {10}^6}{3.84\times {10}^8}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

  $\begin{array}{l}{T}_e=5069.89\text{s}\\ T_e=84.5\min .\end{array}$

Conclusion:

Therefore, the period of an artificial satellite orbiting very near the Earth’s surface is 84.5minutes.