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.)

1.02 · 10^3 m/s

Explanation:

Hi there!

The main formula we'll use for this problem is:

v=C/t

where v is the velocity of the object in orbit, C is the circumference of the orbital path, and t is the time it takes to complete one revolution around the orbital path.

We are given:

r (radius of the orbit) = 3.84 · 10^8 m

t (time to complete one revolution) = 27.3 days

Using this information, we can calculate C and t.

We are solving for v.

First, we calculate the circumference of the orbital path.

We start with this formula:

C=2πr

Now we substitute in our given value of r:

C=2π(3.84 · 10^8 m)

C=2.41 · 10^9 m

Now, let's convert the time from days to seconds, since the unit of t must be seconds if we want the unit of our answer to be in meters per second.

t = 27.3 days · 24hrs/day · 60 min/hr · 60 s/min

t = 2.36 · 10^6 seconds

Now we can plug in our values of C and t into the original formula and solve for v.

v=(2.41 · 10^9 m)/(2.36 · 10^6 s)

=1.02 · 10^3 m/s

Hope this helps!