Math Problem Statement

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?

Solution

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:

θmin=1.22λD\theta_{\text{min}} = \frac{1.22 \lambda}{D}

where:

  • θmin\theta_{\text{min}} is the minimum angular separation (in radians),
  • λ\lambda is the wavelength of observation (in meters),
  • DD 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:

θ=sd\theta = \frac{s}{d}

where:

  • s=5.5AUs = 5.5 \, \text{AU} is the separation between the stars,
  • d=35pcd = 35 \, \text{pc} is the distance to the system.

We need to express both ss and dd in the same units. Recall:

  • 1AU=1.496×1011m1 \, \text{AU} = 1.496 \times 10^{11} \, \text{m},
  • 1pc=3.086×1016m1 \, \text{pc} = 3.086 \times 10^{16} \, \text{m}.

So: s=5.5AU=5.5×1.496×1011m=8.228×1011m,s = 5.5 \, \text{AU} = 5.5 \times 1.496 \times 10^{11} \, \text{m} = 8.228 \times 10^{11} \, \text{m}, d=35pc=35×3.086×1016m=1.08×1018m.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 θ\theta: θ=8.228×10111.08×10187.62×107radians.\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 λ=580nm=580×109m\lambda = 580 \, \text{nm} = 580 \times 10^{-9} \, \text{m}.

From the Rayleigh criterion: θmin=1.22λD\theta_{\text{min}} = \frac{1.22 \lambda}{D} To resolve the stars, θminθ\theta_{\text{min}} \leq \theta, so: D=1.22λθD = \frac{1.22 \lambda}{\theta}

Substitute the values of λ\lambda and θ\theta: D=1.22×580×1097.62×1070.929m.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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Angular Separation
Rayleigh Criterion
Small Angle Approximation
Astronomical Units
Parsecs

Formulas

Rayleigh criterion: θ_min = 1.22 * λ / D
Angular separation: θ = s / d

Theorems

Rayleigh Criterion

Suitable Grade Level

Undergraduate Astronomy/Physics