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It has been proposed to use an array of infrared telescopes spread over thousands of kilometers of space to observe planets orbiting other stars. Consider such an array that has an effective diameter of 6000 km and observes infrared radiation at a wavelength of 10 \(\mu\)m. If it is used to observe a planet orbiting the star 70 Virginis, which is 59 light-years from our solar system, what is the size of the smallest details that the array might resolve on the planet? How does this compare to the diameter of the planet, which is assumed to be similar to that of Jupiter (1.40 \(\times\) 10\(^{5}\) km)? (Although the planet of 70 Virginis is thought to be at least 6.6 times more massive than Jupiter, its radius is probably not too different from that of Jupiter. Such large planets are thought to be composed primarily of gases, not rocky material, and hence can be greatly compressed by the mutual gravitational attraction of different parts of the planet.)

Step by step solution

01

Understand the Problem

We need to determine the smallest details that an infrared telescope array can resolve on a distant planet orbiting 70 Virginis. The telescope array's effective diameter is 6000 km, and it observes at a wavelength of 10 \(\mu\)m. The distance to the star is 59 light-years. We will also compare this resolution to the diameter of a planet similar to Jupiter.

02

Calculate Angular Resolution

The angular resolution \( \theta \) of a telescope can be calculated using the formula \( \theta = 1.22 \times \frac{\lambda}{D} \), where \( \lambda \) is the wavelength and \( D \) is the diameter of the telescope's aperture. Here, \( \lambda = 10 \times 10^{-6} \) m and \( D = 6000 \times 10^{3} \) m. Calculate \( \theta \) using these values.

03

Convert Angular Resolution to Linear Resolution

To find the smallest detail size \( s \) that can be resolved, use \( s = d \times \theta \), where \( d \) is the distance to the star in meters. The distance \( d = 59 \) light-years = \( 59 \times 9.461 \times 10^{15} \) m. Multiply \( \theta \) from Step 2 by \( d \) to get \( s \).

04

Compare with Jupiter's Diameter

Jupiter's diameter is approximately \( 1.40 \times 10^{5} \) km. Convert the resolved detail size \( s \) from meters to kilometers and compare it with Jupiter's diameter.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Telescope Array

Infrared telescope arrays are powerful tools for astronomers. They consist of multiple individual telescopes spread over vast distances, working together to function as a single massive telescope. This setup significantly enhances their observing power and resolution.

Using a telescope array can capture more information from distant objects in the universe. The effective diameter of such an array, like the proposed 6000 km one, determines its capability to gather light. The larger the effective diameter, the finer the details it can observe.

Here are some key features of telescope arrays:

• Increased sensitivity to faint objects due to the combined collection of light.
• Enhanced ability to improve resolution and image clarity.
• Flexibility to observe multiple parts of the sky simultaneously.

Understanding how telescope arrays function is fundamental to exploring distant celestial bodies with great precision.

Angular Resolution

Angular resolution is a critical aspect of astronomical observations. It refers to the telescope's ability to distinguish between two closely spaced objects. The finer the angular resolution, the more details we can discern in the images obtained.

The formula for angular resolution is given by \[ \theta = 1.22 \times \frac{\lambda}{D} \]where

• \(\lambda\) is the wavelength of the observed light (in meters),
• \(D\) is the diameter of the telescope or the effective diameter of a telescope array (in meters),

In our scenario, the angular resolution helps determine the smallest detail on a distant exoplanet that can be observed. A smaller value of \(\theta\) indicates a better ability to resolve fine details. Conversion of this angular measurement into a linear resolution allows astronomers to compare it directly with known dimensions, such as the size of Jupiter.

Exoplanet Observation

Observing exoplanets far from our solar system has become increasingly feasible thanks to advances in infrared astronomy. Using telescope arrays capable of capturing infrared wavelengths opens a new realm of information about these planets.

Infrared telescopes are particularly effective in exoplanet studies because:

• They can detect the warmth of planets not visible in optical light.
• They penetrate dust clouds that might obscure optical observations.
• They provide data on the composition and atmospheric conditions of distant planets.

When observing an exoplanet, such as one orbiting the star 70 Virginis, astronomers utilize the power of infrared arrays to detect minute details. This allows for a more in-depth understanding of the planet's characteristics, even if it is merely a tiny point of light from vast distances.

Wavelength

The choice of wavelength in astronomical observations has profound implications on the details we can observe. Wavelengths are categorized based on their size, from radio waves to gamma rays, with visible light falling somewhere in between.

For infrared astronomy, as in the case of studying exoplanets, the wavelength used here is 10 \(\mu\)m. This specific range helps astronomers uncover different properties of celestial objects.

Some important aspects of observing with different wavelengths include:

• Infrared wavelengths are better for observing cool, faint objects like exoplanets.
• Different wavelengths reveal various types of celestial phenomena.
• An understanding of wavelength helps astronomers calibrate instruments for precise measurements.

By selecting an appropriate wavelength, scientists can maximize the information they gather from distant objects, allowing for discoveries that wouldn't be possible with visible light alone.