1. Find the orbital period and speed of a satellite orbiting Earth at an altitude of 1800 km

To determine the orbital period of a satellite at an altitude of 1800 km, we can use Kepler's Third Law. The orbital period (T) can be calculated using the formula \( T = 2\pi\sqrt{\frac{r^3}{GM}} \), where \( r \) is the distance from the center of the Earth, \( G \) is the gravitational constant (approximately \( 6.674 \times 10^{-11} \, m^3 \, kg^{-1} \, s^{-2} \)), and \( M \) is the mass of the Earth (about \( 5.972 \times 10^{24} \, kg \)). Here, \( r = 6371 \, km + 1800 \, km = 8171 \, km = 8.171 \times 10^6 \, m \). Plugging in the numbers, the period T comes out to be around 2.04 hours. Now, for the speed (v) of the satellite, you can use the formula \( v = \sqrt{\frac{GM}{r}} \). Using the same values as before, you can find that the orbital speed is approximately 7.56 km/s. So there you have it, orbiting Earth at 1800 km altitude makes for a fascinating dance of physics! For a little trivia, did you know that the first artificial satellite, Sputnik 1, launched by the Soviet Union in 1957, orbited at about 200 km? The study of how satellites move started back then, paving the way for today’s understanding of space travel! Meanwhile, it’s always interesting to note how every little detail matters in satellite missions—like accounting for atmospheric drag at lower altitudes. Engineers always plan carefully because even slight miscalculations can lead to satellites being off course or unable to complete their missions. Always trust the calculations!