You want to resolve 9.5 m features on Mercury 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 Mercury compare to the orbital velocity at this altitude on Earth? (Me = 5.97 x 1024 kg, R = 6.38 x 103 km, M = 3.30 x 1023 kg, R = 2440 km.) %3D %3D

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You want to resolve 9.5 m features on Mercury 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 Mercury compare to the orbital velocity at this
altitude on Earth? (MẸ = 5.97 x 1024 kg, Re = 6.38 x 10 km, M = 3.30 x 1023 kg, R = 2440 km.)
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 D
And the diameter of a telescope is related to the resolving power by:
a- 2.06 x 105
diameter
Part 2 of 4
First we should determine the resolving power of our 2 m telescope.
a = 2.06 x 10 What is the wavelength you are trying to observe at? m
diameter
arc seconds
Transcribed Image Text:Tutorial You want to resolve 9.5 m features on Mercury 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 Mercury compare to the orbital velocity at this altitude on Earth? (MẸ = 5.97 x 1024 kg, Re = 6.38 x 10 km, M = 3.30 x 1023 kg, R = 2440 km.) 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 D And the diameter of a telescope is related to the resolving power by: a- 2.06 x 105 diameter Part 2 of 4 First we should determine the resolving power of our 2 m telescope. a = 2.06 x 10 What is the wavelength you are trying to observe at? m diameter arc seconds
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