An artificial satellite is in a circular orbit $5.50 \times 10^{2} \mathrm{~km}$ from the surface of a planet of radius $5.90 \times 10^{3} \mathrm{~km}$. The period of revolution of the satellite around the planet is 4.00 hours.
What is the average density $\rho_{\text {avg }}$ of the planet?
\[
\rho_{\mathrm{avg}}=
\]

Step 1 ：Given values are the distance from the surface of the planet to the satellite, which is \(5.50 \times 10^{6} \) m, the radius of the planet, which is \(5.90 \times 10^{6} \) m, and the period of revolution of the satellite around the planet, which is 4.00 hours.

Step 2 ：First, convert the period of revolution from hours to seconds. This gives \(4.00 \times 60 \times 60 = 14400 \) seconds.

Step 3 ：Next, calculate the distance from the center of the planet to the satellite. This is the sum of the radius of the planet and the distance from the surface of the planet to the satellite. So, \(5.90 \times 10^{6} + 5.50 \times 10^{6} = 1.14 \times 10^{7} \) m.

Step 4 ：Using the gravitational constant \(G = 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2}\), we can calculate the average density of the planet using the formula \(\rho_{\text{avg}} = \frac{3r^3}{GT^2R^3}\).

Step 5 ：Substituting the given values into the formula, we get \(\rho_{\text{avg}} = \frac{3 \times (1.14 \times 10^{7})^3}{6.674 \times 10^{-11} \times (14400)^2 \times (5.90 \times 10^{6})^3} = 1563.75 \, \text{kg/m}^3\).

Step 6 ：Final Answer: The average density of the planet is \(\boxed{1563.75 \, \text{kg/m}^3}\).