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Star A and star B appear equally bright in the sky. Star A is twice as far away from Earth as star B. How do the luminosities of stars \(A\) and \(B\) compare? a. \(\operatorname{Star} \mathrm{A}\) is 4 times as luminous as star \(\mathrm{B}\). b. Star A is 2 times as luminous as star \(B\). c. \(\operatorname{star} \mathrm{B}\) is 2 times as luminous as star \(\mathrm{A}\) d. \(\operatorname{Star} \mathrm{B}\) is 4 times as luminous as star \(\mathrm{A}\)

Short Answer

Expert verified
a. Star A is 4 times as luminous as star B.

Step by step solution

01

- Understand the Inverse Square Law

The brightness observed from a star, also known as its apparent brightness, is related to the star's luminosity and the distance from the observer using the inverse square law. The relationship is given by: \[ b = \frac{L}{4 \pi d^2} \] where \(b\) is the apparent brightness, \(L\) is the luminosity, and \(d\) is the distance to the star.
02

- Analyze the Given Information

It is given that star A and star B appear equally bright in the sky. Therefore, the apparent brightness of both stars, \(b_A\) and \(b_B\), are equal: \[ b_A = b_B \] Also, it is given that star A is twice as far away from Earth as star B, which means: \[ d_A = 2d_B \]
03

- Set Up the Equations

Using the inverse square law for both stars: \[ b_A = \frac{L_A}{4 \pi d_A^2} \] \[ b_B = \frac{L_B}{4 \pi d_B^2} \] And since \(b_A = b_B\), we can equate the two equations: \[ \frac{L_A}{4 \pi d_A^2} = \frac{L_B}{4 \pi d_B^2} \]
04

- Substitute the Distance Values

Substitute \( d_A = 2 d_B \) into the equation: \[ \frac{L_A}{4 \pi (2 d_B)^2} = \frac{L_B}{4 \pi d_B^2} \] Simplifying: \[ \frac{L_A}{16 \pi d_B^2} = \frac{L_B}{4 \pi d_B^2} \]
05

- Solve for \(L_A/L_B\)

Since the denominators are equal, we can simplify: \[ L_A \times 4 = L_B \times 16 \] \[ L_A = 4 \times L_B \] So, star A is 4 times as luminous as star B.

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Key Concepts

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

inverse square law
The inverse square law is a fundamental principle in physics that describes how a physical quantity decreases with the square of the distance from its source. When it comes to light, this law explains that the apparent brightness of a star becomes dimmer as it moves further away. The formula for this relationship is: \[ b = \frac{L}{4 \pi d^2} \] where:
  • \( b \) is the apparent brightness,
  • \( L \) is the luminosity (the actual brightness of the star),
  • \( d \) is the distance from the star to the observer.
This law tells us that if you double the distance to a star, its apparent brightness becomes four times smaller because the denominator \( 4 \pi d^2 \) increases by a factor of four. Understanding this relationship is crucial for interpreting astronomical observations and comparing the luminosities of different stars based on their apparent brightness and distances.
apparent brightness
Apparent brightness is how bright a star appears to an observer on Earth. It differs from luminosity, which is the star's intrinsic brightness. Apparent brightness depends on both the luminosity of the star and its distance from Earth as dictated by the inverse square law: \[ b = \frac{L}{4 \pi d^2} \] Here are key points:
  • A star with high luminosity will appear brighter.
  • A star that is closer to Earth will also appear brighter.
In our exercise, star A and star B have the same apparent brightness even though star A is farther away. This can only happen if star A has a higher luminosity. Specifically, if star A is twice as distant as star B but appears equally bright, then it must be significantly more luminous. We can use the formula for apparent brightness to show that star A must be four times more luminous than star B to appear equally bright despite being twice as far away.
distance-luminosity relation
The distance-luminosity relation is a useful concept in astronomy that connects a star's distance to its luminosity and its apparent brightness. It helps astronomers understand how the light we see from stars changes with distance. For our problem, we know:
  • Star A is twice as far from us as star B (\( d_A = 2d_B \)).
  • Both stars have the same apparent brightness (\( b_A = b_B \)).
Using the equation provided by the inverse square law, we can see how the luminosities compare: \[ \frac{L_A}{4 \pi (2 d_B)^2} = \frac{L_B}{4 \pi d_B^2} \] After simplifying, you will find: \[ \frac{L_A}{16 \pi d_B^2} = \frac{L_B}{4 \pi d_B^2} \ L_A = 4 L_B \] This shows that star A must be four times as luminous as star B to compensate for being twice the distance and still have the same apparent brightness when viewed from Earth. Understanding this relationship allows us to infer properties of stars we observe, even if their distance varies.

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