The

ANS:

a = 2"

R = 384400km

a/(360*3600)*(2pi*R)

2/(360*3600)*(2pi*384400) #I multiply 360 with 3600 for coversion to arcsec

= 3.7272km

there is another way:

(a/3600)*(pi/180)*R

(2/3600)*(pi/180)*384400

=3.7272km

*resolution*of a telescope, as we will learn, is the minimal angular size of features it can resolve. If we have a telescope with a resolution of 2'' (2 arc-seconds) what is the size,**in kilometers (km)**, of the smallest crater we can make out on the moon, at a distance of 384,400 km? (The Moon's diameter is 3480 km, for comparison.) Round your answer to two significant figures.ANS:

a = 2"

R = 384400km

a/(360*3600)*(2pi*R)

2/(360*3600)*(2pi*384400) #I multiply 360 with 3600 for coversion to arcsec

= 3.7272km

there is another way:

(a/3600)*(pi/180)*R

(2/3600)*(pi/180)*384400

=3.7272km