Problem 109 - College Algebra with Modeling and Visualization

Question

The brightness, or intensity, of starlight varies inversely as the square of its distance from Earth. The Hubble Telescope can see stars whose intensities are \(\frac{1}{50}\) that of the faintest star now seen by ground-based telescopes. Determine how much farther the Hubble Telescope can see into space than ground based telescopes. (Sounce: National Aeronautics and Space Administration.)

Step-by-Step Solution

Answer

The Hubble Telescope can see approximately 7.07 times farther than ground-based telescopes.

Step 1: Define the relationship

The problem states that brightness \( B \) varies inversely as the square of the distance \( d \). This can be expressed with the equation \( B = \frac{k}{d^2} \), where \( k \) is a constant.

Step 2: Understand the intensity ratio

It is given that the intensity of the stars seen with the Hubble Telescope is \( \frac{1}{50} \) of the intensity of the faintest star visible with ground-based telescopes. Let's denote the intensity of the faintest visible star with ground-based telescopes as \( B_g = \frac{k}{d_g^2} \), and the Hubble's as \( B_h = \frac{k}{d_h^2} \). It is given \( B_h = \frac{1}{50} B_g \).

Step 3: Set the equation with given intensities

Using the expression from the intensity ratio, write the equation: \( \frac{k}{d_h^2} = \frac{1}{50} \times \frac{k}{d_g^2} \).

Step 4: Simplify the equation

Divide both sides by \( \frac{k}{d_g^2} \): \( \frac{k}{d_h^2} \times \frac{d_g^2}{k} = \frac{1}{50} \). Since \( k \) cancels out, this simplifies to \( \frac{d_g^2}{d_h^2} = \frac{1}{50} \).

Step 5: Solve for distance ratio

Rearrange the equation from Step 4 to find \( \left( \frac{d_g}{d_h} \right)^2 = \frac{1}{50} \). Taking the square root of both sides gives \( \frac{d_g}{d_h} = \frac{1}{ ext{sqrt}(50)} \).

Step 6: Find how much farther Hubble can see

The reciprocal indicates how much farther Hubble can see: \( \frac{d_h}{d_g} = \text{sqrt}(50) \). Calculating this gives \( \text{sqrt}(50) \approx 7.07 \).

Key Concepts

Stellar Brightness

Stellar brightness, also known as intensity, describes how bright a star appears from Earth. The brightness of a star is affected by both its inherent luminosity and its distance from us. There is an interesting relationship called inverse variation at play here. This means that the brightness of a star seen from Earth will decrease as the distance from the star increases. The intensity of starlight is inversely proportional to the square of the distance from the Earth. In mathematical terms, this can be expressed as \( B = \frac{k}{d^2} \). Here, \( B \) is the brightness, \( d \) is the distance from the star, and \( k \) is a constant. This relationship helps astronomers understand and measure how far away stars are, based only on how bright they appear.

Hubble Telescope

The Hubble Telescope is a revolutionary space telescope that significantly advanced our capabilities to observe distant objects in space. Unlike ground-based telescopes, the Hubble is positioned above Earth's atmosphere, eliminating atmospheric interference and allowing it to capture extraordinarily faint objects in the far reaches of the universe. It can detect stars that are much dimmer than those seen with ground-based telescopes. For instance, it can observe stars up to \( \frac{1}{50} \) of the intensity of the faintest star visible from ground-based telescopes, meaning it can "see" much farther into space. This expanded viewing capability enables astronomers to explore farther galaxies, better understand cosmic phenomena, and gain insights into the origins of the universe.

Distance and Intensity Relationship

In astronomical observations, the relationship between the distance of a star and its observed intensity is crucial for understanding how we perceive distant celestial objects. This is explained through an inverse square law, indicating that the intensity of light (or other forms of radiation) from a source diminishes with the square of the distance from that source. In simpler terms, if a star is twice as far away, it will appear four times less bright. In the case of the Hubble Telescope and a ground-based telescope, the difference in observed intensity is \( \frac{1}{50} \), which mathematically relates to how much further the Hubble can see compared to ground-based telescopes. By resolving the equation \( \frac{d_g^2}{d_h^2} = \frac{1}{50} \), we learn that Hubble sees \( \sqrt{50} \) times further.

Algebraic Problem Solving

Algebra plays a pivotal role in solving problems related to inverse variation and stellar observations. In this exercise, students are guided to manipulate equations to understand how far the Hubble Telescope can see compared to ground-based telescopes. First, you set up the equations representing the inverse relationship between distance and brightness. With the given intensity ratio, \( B_h = \frac{1}{50} B_g \), and after simplifying the proportional relationship, \( \frac{d_g^2}{d_h^2} = \frac{1}{50} \), algebra is used to rearrange and solve for \( \frac{d_h}{d_g} = \sqrt{50} \). This is a typical example of how algebraic techniques help resolve practical scientific queries, turning abstract equations into clear, meaningful results that inform our understanding of the universe.