How long will it take a satellite that is 260km from the moon to orbit the moon once?
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We need to find how long it takes for a satellite 260 kilometers above the Moon's surface to complete one orbit. This is a classic orbital mechanics problem that we can solve using Kepler's Third Law.
The first step is to calculate the orbital radius. We add the Moon's mean radius of 1737.4 kilometers to the satellite's altitude of 260 kilometers above the surface. This gives us a total orbital radius of 1997.4 kilometers.
Now we apply Kepler's Third Law, which relates orbital period to orbital radius. The formula is T equals 2 pi times the square root of a cubed divided by mu, where mu is the Moon's standard gravitational parameter, approximately 4.9 times 10 to the 12th cubic meters per second squared.
Now let's substitute our values into the formula. We have a equals 1,997,400 meters cubed, divided by mu equals 4.9 times 10 to the 12th. This gives us approximately 7997.6 seconds. Converting to more familiar units, this equals about 133 minutes, or approximately 2 hours and 13 minutes.
In conclusion, a satellite orbiting 260 kilometers above the Moon's surface will complete one full orbit in approximately 2 hours and 13 minutes. This demonstrates how Kepler's Third Law allows us to calculate orbital periods using just the orbital radius and the central body's gravitational parameter.