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!

the answer is false. this is because for ice to melt, it requires latent heat of melting. latent heat means heat energy required to change phase without any increase in temperatures. therefore, the ice will take energy from the environment to melt hence atmospheric temperatures will reduce.