You want to resolve 5.5 m features on the Moon with a 2 m telescope using 550 nm light. How close (in km) do you need to be? How does the orbital velocity (in km/s) at this altitude on the Moon compare to the orbital velocity at this altitude on Earth? (M = 5.97 x 1024 kg, R. = 6.38 x 103 km, M, = 7.35 x 1022 kg, R = 1740 km.) M.

icon
Related questions
Question
Tutorial
You want to resolve 5.5 m features on the Moon with a 2 m telescope using 550 nm light. How close (in km)
do you need to be?
How does the orbital velocity (in km/s) at this altitude on the Moon compare to the orbital velocity at this
altitude on Earth? (M, = 5.97 x 1024 kg, R- = 6.38 x 103 km, M = 7.35 x 1022 kg, Ry = 1740 km.)
%3!
M.
Part 1 of 4
The small angle formula tells us how distance and linear size are related to the angular size of an object.
2.06 x 105
And the diameter of a telescope is related to the resolving power by:
a = 2.06 x 105
diameter
Transcribed Image Text:Tutorial You want to resolve 5.5 m features on the Moon with a 2 m telescope using 550 nm light. How close (in km) do you need to be? How does the orbital velocity (in km/s) at this altitude on the Moon compare to the orbital velocity at this altitude on Earth? (M, = 5.97 x 1024 kg, R- = 6.38 x 103 km, M = 7.35 x 1022 kg, Ry = 1740 km.) %3! M. Part 1 of 4 The small angle formula tells us how distance and linear size are related to the angular size of an object. 2.06 x 105 And the diameter of a telescope is related to the resolving power by: a = 2.06 x 105 diameter
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 3 steps with 2 images

Blurred answer
Similar questions