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Chapter 16: Problem 86

Distances in space are often quoted in units of light years, the distance light travels in 1 year. (a) How many meters is a light-year? (b) How many meters is it to Andromeda, the nearest large galaxy, given that it is \(2.54 \times 10^{6}\) ly away? (c) The most distant galaxy yet discovered is \(13.4 \times 10^{9}\) ly away. How far is this in meters?

Short Answer

Expert verified
(a) A light-year is approximately \(9.46 \times 10^{15}\) meters. (b) The distance to Andromeda is about \(2.40 \times 10^{22}\) meters. (c) The distance to the most distant galaxy yet discovered is approximately \(1.27\times10^{26}\) meters.

Step by step solution

01

(Part a: Find the distance of 1 light-year in meters)

To calculate the distance of one light-year, first, we need to determine how many seconds are in a year. A year has: - 365.25 days (in a year, considering leap years) - 24 hours (in a day) - 60 minutes (in an hour) - 60 seconds (in a minute) Therefore, the number of seconds in a year is: \(Seconds_{year} = 365.25 \times 24 \times 60 \times 60 \) Next, we know that light travels at a speed of approximately \(3 \times 10^8 \mathrm{m/s}\). To find the distance light travels in one year, we can multiply its speed by the number of seconds in a year: \(Distance_{lightyear} = Speed_{light} \times Seconds_{year}\)
02

(Part b: Find the distance to Andromeda in meters)

Now that we have determined the distance of one light-year in meters, we can calculate the distance to Andromeda. Given that the distance to Andromeda is \(2.54\times10^6\) light-years away, we can multiply this value by the distance of one light-year: \(Distance_{Andromeda} = 2.54 \times 10^6 \times Distance_{lightyear}\)
03

(Part c: Find the distance to the most distant galaxy in meters)

Similarly, to find the distance to the most distant galaxy yet discovered, given that it is \(13.4 \times 10^9\) light-years away, we can multiply this value by the distance of one light-year: \(Distance_{distant\_galaxy} = 13.4 \times 10^9 \times Distance_{lightyear}\) Calculate the values in each step to find the respective distances in meters.

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Key Concepts

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

Light-Year Calculation
Understanding the concept of a light-year is crucial in astronomy as it's a common unit to measure vast distances. A light-year is the distance light travels in one year. To compute this, we first determine the total number of seconds in a year. An average year has approximately 365.25 days, considering leap years, multiplied by 24 hours per day, 60 minutes per hour, and 60 seconds per minute. The exact seconds per year would be:
  • 365.25 days/year
  • 24 hours/day
  • 60 minutes/hour
  • 60 seconds/minute
This results in roughly 31,557,600 seconds in a year.Knowing that the speed of light is about \(3 \times 10^8\) m/s, we multiply this speed by the number of seconds in a year to find the distance of a light-year:\[ \text{Distance}_{\text{lightyear}} = 3 \times 10^8 \times 31,557,600 \]This equals an astounding 9.46 trillion meters, precisely defining how far light travels in one year.
Andromeda Galaxy Distance
The Andromeda Galaxy is the nearest large galaxy to our Milky Way. It is located approximately 2.54 million light-years from Earth. To comprehend this in meters, we utilize our light-year calculation from before. Since one light-year equates to roughly 9.46 trillion meters, simply multiply this by 2.54 million to convert the distance to Andromeda:\[ \text{Distance}_{\text{Andromeda}} = 2.54 \times 10^6 \times 9.46 \times 10^{12} \]
  • Or roughly equating to about \(2.4 \times 10^{22}\) meters.
This extraordinary distance reiterates the enormity of cosmic scales and the interconnected luminescent journeys involved when viewing celestial bodies.
Distant Galaxy Distance
The universe hosts countless galaxies, some so distant that their light has taken billions of years to reach us. One such galaxy, observed as the most remote, is at a staggering 13.4 billion light-years away. To determine this distance in meters, multiply 13.4 billion by the distance light covers in one year, using our light-year value:\[ \text{Distance}_{\text{distant\_galaxy}} = 13.4 \times 10^9 \times 9.46 \times 10^{12} \]
  • This results in approximately \(1.27 \times 10^{26}\) meters.
These distances underscore not just the vastness of the cosmos, but also the universe's age, revealing light that has journeyed since ancient cosmic epochs to enrich our understanding of the universe's expanse.

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Most popular questions from this chapter

A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of \(7.50 \mu \mathrm{V} / \mathrm{m}\) (see below). (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of \(1.50 \times 10^{13} \mathrm{m}^{2}\) (a large fraction of North America), how much power does it radiate?

A Styrofoam spherical ball of radius 2 mm and mass \(20 \mu \mathrm{g}\) is to be suspended by the radiation pressure in a vacuum tube in a lab. How much intensity will be required if the light is completely absorbed the ball?

Radio station WWVB, operated by the National Institute of Standards and Technology (NIST) from Fort Collins, Colorado, at a low frequency of \(60 \mathrm{kHz}\), broadcasts a time synchronization signal whose range covers the entire continental US. The timing of the synchronization signal is controlled by a set of atomic clocks to an accuracy of \(1 \times 10^{-12} \mathrm{s}, \quad\) and repeats every 1 minute. The signal is used for devices, such as radio-controlled watches, that automatically synchronize with it at preset local times. WWVB's long wavelength signal tends to propagate close to the ground. (a) Calculate the wavelength of the radio waves from WWVB. (b) Estimate the error that the travel time of the signal causes in synchronizing a radio controlled watch in Norfolk, Virginia, which is 1570 mi ( \(2527 \mathrm{km}\) ) from Fort Collins, Colorado.

A parallel-plate capacitor has a plate area of \(A=0.250 \mathrm{m}^{2}\) and a separation of \(0.0100 \mathrm{m} .\) What must be must be the angular frequency \(\omega\) for a voltage \(V(t)=V_{0} \sin \omega t \quad\) with \(\quad V_{0}=100 \mathrm{V} \quad\) to produce \(\quad \mathrm{a}\) maximum displacement induced current of 1.00 A between the plates?

Accelerating electric charge emits electromagnetic radiation. How does this apply in each case: (a) radio waves, (b) infrared radiation.
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