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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.) 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 Part 2 of 4 First we should determine the resolving power of our 2 m telescope. a = 2.06 x 105 What is the wavelength you are trying to observe at? m diameter m a = Use the resolving power formula to calculate the angular resolution. arc seconds

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.) 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 Part 2 of 4 First we should determine the resolving power of our 2 m telescope. a = 2.06 x 105 What is the wavelength you are trying to observe at? m diameter m a = Use the resolving power formula to calculate the angular resolution. arc seconds

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### 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**? \[ (M_E = 5.97 \times 10^{24} \text{ kg}, R_E = 6.38 \times 10^3 \text{ km}, M_M = 7.35 \times 10^{22} \text{ kg}, R_M = 1740 \text{ km}) \] #### Part 1 of 4 The small angle formula tells us how distance and linear size are related to the angular size of an object. \[ \frac{\theta}{2.06 \times 10^5} = \frac{d}{D} \] And the diameter of a telescope is related to the resolving power by: \[ \alpha = 2.06 \times 10^5 \frac{\lambda}{\text{diameter}} \] #### Part 2 of 4 First, we should determine the resolving power of our 2 m telescope. \[ \alpha = 2.06 \times 10^5 \frac{\text{What is the wavelength you are trying to observe at? m}}{\text{diameter m}} \] \[ \alpha = \text{Use the resolving power formula to calculate the angular resolution. arc seconds} \]
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**? \[ (M_E = 5.97 \times 10^{24} \text{ kg}, R_E = 6.38 \times 10^3 \text{ km}, M_M = 7.35 \times 10^{22} \text{ kg}, R_M = 1740 \text{ km}) \] #### Part 1 of 4 The small angle formula tells us how distance and linear size are related to the angular size of an object. \[ \frac{\theta}{2.06 \times 10^5} = \frac{d}{D} \] And the diameter of a telescope is related to the resolving power by: \[ \alpha = 2.06 \times 10^5 \frac{\lambda}{\text{diameter}} \] #### Part 2 of 4 First, we should determine the resolving power of our 2 m telescope. \[ \alpha = 2.06 \times 10^5 \frac{\text{What is the wavelength you are trying to observe at? m}}{\text{diameter m}} \] \[ \alpha = \text{Use the resolving power formula to calculate the angular resolution. arc seconds} \]
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