An astronomer is observing a binary star that is 35 pc away. The two stars have an orbital separation of 5.5 AU. If the astronomer is observing the binary star at a wavelength of 580 nm, what is the minimum aperture of the telescope that is needed in order to resolve the two stars?

To resolve the two stars in the binary system, we can use the **Rayleigh criterion**, which gives the minimum angular separation that can be resolved by a telescope. The Rayleigh criterion is given by:

\[
\theta_{\text{min}} = \frac{1.22 \lambda}{D}
\]

where:
- \(\theta_{\text{min}}\) is the minimum angular separation (in radians),
- \(\lambda\) is the wavelength of observation (in meters),
- \(D\) is the aperture of the telescope (in meters).

### Step 1: Angular Separation of the Binary Stars
First, we need to calculate the angular separation between the two stars in radians. The orbital separation is given as 5.5 AU, and the distance to the binary system is 35 parsecs (pc). Using the small-angle approximation:

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

where:
- \(s = 5.5 \, \text{AU}\) is the separation between the stars,
- \(d = 35 \, \text{pc}\) is the distance to the system.

We need to express both \(s\) and \(d\) in the same units. Recall:
- \(1 \, \text{AU} = 1.496 \times 10^{11} \, \text{m}\),
- \(1 \, \text{pc} = 3.086 \times 10^{16} \, \text{m}\).

So:
\[
s = 5.5 \, \text{AU} = 5.5 \times 1.496 \times 10^{11} \, \text{m} = 8.228 \times 10^{11} \, \text{m},
\]
\[
d = 35 \, \text{pc} = 35 \times 3.086 \times 10^{16} \, \text{m} = 1.08 \times 10^{18} \, \text{m}.
\]

Now, calculate the angular separation \(\theta\):
\[
\theta = \frac{8.228 \times 10^{11}}{1.08 \times 10^{18}} \approx 7.62 \times 10^{-7} \, \text{radians}.
\]

### Step 2: Aperture of the Telescope
The astronomer is observing at a wavelength of \(\lambda = 580 \, \text{nm} = 580 \times 10^{-9} \, \text{m}\).

From the Rayleigh criterion:
\[
\theta_{\text{min}} = \frac{1.22 \lambda}{D}
\]
To resolve the stars, \(\theta_{\text{min}} \leq \theta\), so:
\[
D = \frac{1.22 \lambda}{\theta}
\]

Substitute the values of \(\lambda\) and \(\theta\):
\[
D = \frac{1.22 \times 580 \times 10^{-9}}{7.62 \times 10^{-7}} \approx 0.929 \, \text{m}.
\]

### Final Answer:
The minimum aperture of the telescope needed to resolve the two stars is approximately **0.93 meters**.

Would you like more details on this? Here are five related questions you might be interested in:

1. How does the Rayleigh criterion change for different wavelengths of light?
2. What is the effect of atmospheric turbulence on resolving binary stars?
3. How does the aperture of the telescope affect its resolution power?
4. How do astronomers measure the angular separation between stars in practice?
5. What other factors, besides aperture, influence a telescope's ability to resolve objects?

**Tip:** Increasing the aperture of a telescope improves its resolving power and allows it to capture finer details in distant objects.

Discover how to calculate the minimum aperture of a telescope required to resolve two stars in a binary system located 35 parsecs away, with an orbital separation of 5.5 AU, using the Rayleigh criterion and small-angle approximation. This solution is perfect for undergraduate astronomy and physics students.

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