Question

Calculate the speed with which the moon orbits the earth given the distance from earth to moon as R = 3.84 · 108 m. (Astronomers note that the true orbital period of the moon, is 27.3 Earth days. Interestingly, this would mean that there are approximately 13 months in a year. Use the 27.3 days/month for T - the time required for one revolution in your calculation.)

Answer 1

This sounds pretty easy, in fact. The orbital motion can be assumed to be circular and with constant speed. Then, the period is the time to do one revolution. The distance is the length of a revolution. That is 2*pi*R, where R is the distance between the Moon and the Earth (the respective centers to be precise). In summary, it's like a simple motion with constant speed:

v = 2*pi*R/T,

you have R in m and T is days, which multiplied by 86,400 s/day gives T in seconds.

Then v = 2*pi*3.84*10^8/(27.3*86,400) = 1,022.9 m/s ~ 1 km/s (about 3 times the speed of sound :)

For the Earth around the Sun, it would be v = 2*pi*149.5*10^9/(365*86,400)~ 29.8 km/s!

I know it's not in the problem, but it's interesting to know how fast the Earth moves around the Sun! And yet we do not feel it (that's one of the reasons some ancient people thought crazy the Earth not being at the center, there would be such strong winds!)