Problem 98 - A Graphical Approach to Precalculus with Limits

Question

Solve each problem. Hubble Telescope The brightness or intensity of starlight varies inversely with the square of its distance from Earth. The Hubble Telescope can see stars whose intensities are \(\frac{1}{50}\) 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.

Step-by-Step Solution

Answer

The Hubble Telescope can see about 7.07 times farther into space than ground-based telescopes.

Step 1: Understanding the Relationship

The problem tells us that the intensity of starlight varies inversely with the square of its distance from Earth. This means if we denote the intensity by \( I \) and the distance by \( d \), then the relationship can be written as \( I \propto \frac{1}{d^2} \) or \( I = \frac{k}{d^2} \), where \( k \) is a constant.

Step 2: Establish the Equation for Hubble Telescope

For the Hubble Telescope, the intensity is given to be \( \frac{1}{50} \) of the faintest intensity seen by ground-based telescopes, \( I_g \). If \( d_h \) is the distance the Hubble can see, then \( \frac{k}{d_h^2} = \frac{I_g}{50} \).

Step 3: Establish the Equation for Ground-Based Telescopes

For ground-based telescopes, we denote the distance as \( d_g \), and we have \( I_g = \frac{k}{d_g^2} \).

Step 4: Relate the Two Equations

Setting the two equations equal based on \( I_g \), we have \( \frac{k}{d_h^2} = \frac{1}{50} \times \frac{k}{d_g^2} \).

Step 5: Simplify and Solve for the Distance Ratio

Cancel out \( k \) from both sides to get \( \frac{1}{d_h^2} = \frac{1}{50 \cdot d_g^2} \). Rearrange to get \( d_h^2 = 50 \cdot d_g^2 \). Take the square root of both sides to find the distance factor: \( d_h = \sqrt{50} \cdot d_g \).

Step 6: Final Calculation

Since \( \sqrt{50} = \sqrt{25 \times 2} = 5 \times \sqrt{2} \approx 7.07 \), the Hubble Telescope can see approximately 7.07 times farther into space than ground-based telescopes.

Key Concepts

Hubble Telescope

The Hubble Telescope is an iconic device launched into low Earth orbit in 1990. It revolutionized our understanding of space with its ability to capture images of distant celestial objects. Unlike ground-based telescopes, the Hubble is not affected by atmospheric interferences, which allows it to have a clearer and uninterrupted view of the universe.
Deploying the Hubble in space allows astronomers to observe astronomical phenomena with unparalleled clarity and detail. Due to its location outside Earth's atmosphere, the Hubble can detect faint stars and galaxies that are difficult or impossible to see from the ground.
The advancements in digital sensors and optics in the Hubble contributed greatly to its capability. This enhanced ability gives it a significant advantage in space exploration compared to traditional observatories on Earth.

Starlight Intensity

Starlight intensity, or the brightness we perceive from stars, depends on multiple factors. One crucial factor is the star's distance from the observer here on Earth.
The concept of inverse variation describes how starlight intensity decreases as the distance from the light source increases. This means the farther a star is, the dimmer it appears. Mathematically, this relationship is expressed as the intensity, \( I \), being inversely proportional to the square of the distance, \( d \), i.e., \( I \propto \frac{1}{d^2} \).
By understanding this relationship, astronomers can determine how far stars are from us by measuring their intensity. Instruments like the Hubble Telescope use this principle to discern stars at great distances that would otherwise be too faint to detect with ground-based telescopes.

Distance Formula

In astronomical measurements, understanding the distance formula is essential for calculating distances between celestial bodies. Distance in space is often measured using the concept of light years or astronomical units (AU).
The starlight intensity problem involves the use of a formula that relates the distance to inverse variation. Specifically, if the intensity is known, the distance can be calculated using \( d = \sqrt{\frac{k}{I}} \), where \( k \) is a constant unique to the system being measured.
For the Hubble Telescope exercise, knowing the intensity ratio allows for calculation of how much farther it can see than ground-based telescopes. The simplified relationship \( d_h = \sqrt{50} \cdot d_g \) illustrates this concept. Here, \( d_h \) is the distance the Hubble can see and \( d_g \) is the ground-based telescope's distance.

Square Root

The square root is a fundamental mathematical operation often used in solving equations involving squared distance, such as in the case of starlight intensity and inverse variation.
To find how much farther the Hubble Telescope can see compared to a ground-based telescope, the calculation involves finding the square root of a ratio, specifically \( \sqrt{50} \).
This precise operation results in an approximate value of 7.07. Thus, expressing the increased observing capacity of the Hubble, it indicates the telescope's ability to see about 7.07 times farther than its ground-based counterparts.
Understanding how to manipulate and calculate square roots is crucial in both scientific calculations and everyday mathematical problems. Each step in these calculations helps break down complex relationships, making them accessible and understandable.