/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 43 . Stars \(C\) and \(D\) both hav... [FREE SOLUTION] | 91Ó°ÊÓ

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. Stars \(C\) and \(D\) both have the same luminosity, but \(C\) is \(32 \mathrm{pc}\) from Earth while \(D\) is 128 pc from Earth. Which star appears brighter as seen from Earth? How many times brighter is it?

Short Answer

Expert verified
Star \(C\) appears brighter from Earth and is 16 times brighter than star \(D\).

Step by step solution

01

Understand the Problem

Two stars having the same luminosity are different distances from Earth. Star \(C\) is \(32 \mathrm{pc}\) away while star \(D\) is \(128 \mathrm{pc}\) away. The objective is to find out which star appears brighter and how much brighter it is.
02

Apply the Inverse Square Law

According to the inverse square law, the brightness of a star reduces with the square of its distance. So, the brightness of Star \(C\) and Star \(D\) relative to each other will depend on the square of their distances. Considering \(C\) as the reference point, we have: \( \frac{Brightness_{C}}{Brightness_{D}} = \left(\frac{Distance_{D}}{Distance_{C}}\right)^{2} \).
03

Substitute Values

Substitute the given values \(Distance_{D}=128 \mathrm{pc}\) and \(Distance_{C}=32 \mathrm{pc}\) into the equation above: \( \frac{Brightness_{C}}{Brightness_{D}} = \left(\frac{128 \mathrm{pc}}{32 \mathrm{pc}}\right)^{2} \).
04

Simplify

Simplify the right side of the equation obtained above: \( \frac{Brightness_{C}}{Brightness_{D}} = (4)^{2} =16 \). This means that Star \(C\) appears 16 times brighter than Star \(D\).

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Key Concepts

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

Stellar Brightness
When we gaze into the night sky, the stars we observe shine with varying degrees of brightness. However, this 'apparent brightness' is different from a star's true luminosity—its inherent brightness. Apparent brightness refers to how bright a star appears from Earth and depends not only on the star's intrinsic luminosity, but also on its distance from us.

Understanding the inverse square law is crucial for studying stellar brightness. The law states that the observed brightness of a star decreases proportionally to the square of the distance from the star. Mathematically, if a star's distance from the observer is doubled, its apparent brightness would fall by a factor of four (two squared).

In the exercise, Stars C and D have the same luminosity but different distances from Earth. Even without additional details, the inverse square law tells us that Star C, being closer, will naturally appear brighter to us. Simplifying the equation in Step 4 of the solution reveals that Star C is, in fact, 16 times brighter in appearance, from our terrestrial vantage point, than Star D.
Astronomical Distance Measurement
Measuring distances in the vast expanse of space is a complex task, but it is fundamental to the field of astronomy. Astronomers often use the unit 'parsec' symbolized as 'pc,' which stands for 'parallax second'.

A parsec is based on the method of parallax, a technique that involves measuring the apparent shift (relative to distant background stars) of a star's position as observed from opposite sides of Earth's orbit. One parsec is the distance at which one astronomical unit (AU) subtends an angle of one arcsecond. One AU is the average Earth-Sun distance, approximately 149.6 million kilometers.

As you may deduce from the exercise, Star D being 128 pc away compared to Star C at 32 pc, signifies that Star D is four times farther away from Earth. This is where the inverse square law plays a vital role in understanding how such distance measurements directly affect perceived stellar brightness.
Luminosity of Stars
Luminosity is an absolute measure of radiant power—the total amount of energy radiated by a star per unit of time. It is a fundamental property of a star that is intrinsic and does not change based on the observer's position. Measured in watts, luminosity accounts for all wavelengths of light emitted and therefore, can only be calculated with precise knowledge of a star’s distance and apparent brightness.

A star's temperature and size are the main factors that determine its luminosity. Stars with higher temperatures and larger radii emit more energy, making them more luminous. For instance, our Sun has a luminosity of about 3.828 x 10^26 watts. In the exercise provided, both Stars C and D have the same luminosity, which is why the concept of the inverse square law is exclusively used to describe their difference in apparent brightness from Earth.

It is important to understand that luminosity is not the same as brightness; it is a measure of energy output and does not vary with distance, while brightness is what we observe and can greatly vary depending on how far the star is from the observer.

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Most popular questions from this chapter

The solar constant, equal to \(1370 \mathrm{~W} / \mathrm{m}^{2}\), is the amount of light energy from the Sun that falls on 1 square meter of the Earth's surface in 1 second (see Section 17-2). What would the distance between the Earth and the Sun have to be in order for the solar constant to be 1 watt per square meter \(\left(1 \mathrm{~W} / \mathrm{m}^{2}\right)\) ?

Suppose that you want to determine the temperature, diameter, and luminosity of an isolated star (not a member of a binary system). Which of these physical quantities require you to know the distance to the star? Explain.

Some giant and supergiant stars are of the same spectral type (G2) as the Sun. What aspects of the spectrum of a G2 star would you concentrate on to determine the star's luminosity class? Explain what you would look for.

Observe the eclipsing binary Algol ( \(\beta\) Persei), using nearby stars to judge its brightness during the course of an eclipse. Algol has an orbital period of \(2.87\) days, and, with the onset of primary eclipse, its apparent magnitude drops from \(2.1\) to 3.4. It remains this faint for about 2 hours. The entire eclipse, from start to finish, takes about 10 hours. Consult the "Celestial Calendar" section of the current issue of Sky or Telescope for the predicted dates and times of the minima of Algol. Note that the schedule is given in Universal Time (the same as Greenwich Mean Time), so you will have to convert the time to that of your own time zone. Algol is normally the second brightest star in the constellation of Perseus. Because of its position on the celestial sphere (R.A. \(=3^{\mathrm{h}} 08.2^{\mathrm{m}}\), Decl. \(\left.=40^{\circ} 57^{\prime}\right)\), Algol is readily visible from northern latitudes during the fall and winter months.

The space velocity of a certain star is \(120 \mathrm{~km} / \mathrm{s}\) and its radial velocity is \(72 \mathrm{~km} / \mathrm{s}\). Find the star's tangential velocity.

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