Suppose you wanted to build a telescope capable of resolving a planet the size of Earth that is 18 light-years away.
 
a. Calculate the angular diameter of Earth if it were that far. To see any detail on the surface, we would want to see an angular size about ten times smaller than this.
 
b. Calculate the diameter of a telescope that would be needed to resolve this angular size if ​observing at 560 nm. Is this feasible?
 
a. The angular diameter is  × 10-9 degree.
 
b. The diameter of the telescope would have to be *D* =    km.
 
It   (Click to select)   is   is not  feasible to build a telescope this size.

To solve this problem, we need to follow these steps:

### Part (a): Calculating the Angular Diameter of Earth

The angular diameter \(\theta\) of an object is given by the formula:

\[
\theta = \frac{d}{D}
\]

where:
- \(d\) is the actual diameter of the object.
- \(D\) is the distance to the object.

The diameter of Earth (\(d\)) is approximately 12,742 km.

The distance to the object (\(D\)) is 18 light-years. To convert this into kilometers:
\[
D = 18 \times 9.461 \times 10^{12} \text{ km}
\]

Thus,
\[
D = 1.70298 \times 10^{14} \text{ km}
\]

Plugging in the values:
\[
\theta = \frac{12,742 \text{ km}}{1.70298 \times 10^{14} \text{ km}}
\]

\[
\theta \approx 7.485 \times 10^{-11} \text{ radians}
\]

To convert radians to degrees, we use the conversion factor \(1 \text{ radian} = \frac{180}{\pi} \text{ degrees}\):

\[
\theta \approx 7.485 \times 10^{-11} \times \frac{180}{\pi}
\]

\[
\theta \approx 4.29 \times 10^{-9} \text{ degrees}
\]

So, the angular diameter of Earth if it were 18 light-years away is \(4.29 \times 10^{-9} \text{ degrees}\).

### Part (b): Calculating the Diameter of the Telescope

To resolve an angular size about ten times smaller, the desired angular resolution would be:

\[
\theta_{desired} = \frac{4.29 \times 10^{-9}}{10} \approx 4.29 \times 10^{-10} \text{ degrees}
\]

The angular resolution \(\theta_{res}\) of a telescope is given by:

\[
\theta_{res} = 1.22 \frac{\lambda}{D}
\]

where:
- \(\lambda\) is the wavelength of light.
- \(D\) is the diameter of the telescope.

Rearranging to solve for \(D\):

\[
D = 1.22 \frac{\lambda}{\theta_{res}}
\]

Given:
\(\lambda = 560 \text{ nm} = 560 \times 10^{-9} \text{ m}\)

\(\theta_{res} = 4.29 \times 10^{-10} \text{ degrees} \times \frac{\pi}{180} \text{ radians/degree} \approx 7.48 \times 10^{-12} \text{ radians}\)

Plugging in the values:

\[
D = 1.22 \frac{560 \times 10^{-9}}{7.48 \times 10^{-12}} \approx 91.32 \times 10^3 \text{ m}
\]

\[
D \approx 91.32 \text{ km}
\]

So, the diameter of the telescope would have to be approximately 91.32 km.

### Feasibility

A telescope with a diameter of 91.32 km is not feasible with current technology. It is far beyond the scale of any existing or planned telescopes.

Would you like more details or have any questions?

Here are some questions you might consider next:

1. What are the limits of current telescope technology in terms of diameter?
2. How do different wavelengths of light affect telescope resolution?
3. What advancements in technology could make larger telescopes feasible?
4. How does the atmosphere impact ground-based telescope observations?
5. What are interferometric techniques, and how can they be used to achieve higher resolutions?

**Tip:** Understanding the principles of angular resolution and the limitations of current technology is essential when considering the feasibility of astronomical observations.

Learn how to calculate the requirements for a telescope capable of resolving Earth-sized planets 18 light-years away. Includes calculations of angular diameter and necessary telescope diameter at 560 nm wavelength. Explore the feasibility of building such a large telescope.

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