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Chapter 32: Problem 34

Our nearest star (other than the Sun) is 4.2 light years away. That is, it takes 4.2 years for the light to reach Earth. How far away is it in meters?

Short Answer

Expert verified
The star is approximately 3.972 x 10^16 meters away.

Step by step solution

01

Understand the Problem

We want to find out how far away the nearest star (other than the Sun) is in meters, given that it is 4.2 light years away.
02

Recognize the Units

A light year is the distance that light travels in one year. We need to convert this distance from light years to meters.
03

Find the Distance Light Travels in One Year

Light travels at a speed of approximately 299,792,458 meters per second. We need to find out how many seconds are in one year to calculate the distance light travels in one year.
04

Calculate Seconds in One Year

There are 365 days in a year, 24 hours in a day, 60 minutes in an hour, and 60 seconds in a minute. So, there are \[365 \times 24 \times 60 \times 60 = 31,536,000\] seconds in one year.
05

Calculate Distance per Light Year

Now, use the formula: \[\text{Distance (in meters)} = \text{Speed of light (in meters/second)} \times \text{Number of seconds in a year}\]Using the speed of light \(299,792,458 \text{ m/s}\), we have:\[299,792,458 \times 31,536,000 \approx 9.461 \times 10^{15}\] meters.
06

Calculate the Distance to the Star

Now that we know how far light travels in one year, we multiply this distance by 4.2 to find the distance to the star:\[9.461 \times 10^{15} \text{ meters} \times 4.2 \approx 3.972 \times 10^{16}\] meters.

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Key Concepts

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

Speed of Light
The speed of light is a fundamental constant of nature, meaning it does not change. Light travels extremely fast, reaching a speed of approximately 299,792,458 meters per second (m/s). This is incredibly fast, and it is hard to truly grasp just how swift this speed is. Imagine that light could circle the entire Earth approximately 7.5 times in just one second!

Knowing the speed of light is crucial when dealing with astronomical distances, such as light years, because it tells us how far light can travel in a given time. By calculating how many meters light travels in one second and then multiplying by the number of seconds in a year, we can better comprehend how vast the universe is.
  • The speed of light: 299,792,458 m/s
  • Light speed is constant and fastest in a vacuum
Distance Conversion
Distance conversion is essential when we need to understand distances in more familiar units, like converting light years to meters. A light year represents the distance that light can travel in one year. To convert a light year into meters, we must multiply the speed of light by the number of seconds in a year. The result is one light year equals approximately 9.461 trillion kilometers or 9.461 million million meters.

For practical purposes, thinking about distances in light years helps astronomers and scientists when measuring the vast spaces between stars and galaxies. Our nearest stellar neighbor (apart from our Sun) is 4.2 light years away. Now we know how to convert this to meters:
  • 1 light year = approx. 9.461 x 1015 meters
  • Conversion involves basic formula: distance = speed x time
Astronomical Units
Astronomical units (AU) are another popular way to describe distances in space, especially within our own solar system. Unlike the vast scale of a light year, an astronomical unit is the average distance between the Earth and the Sun, which is about 149.6 million kilometers (approximately 93 million miles).

Using astronomical units is convenient when discussing planets and objects closer to us compared to light years which are used to describe interstellar and intergalactic distances. While the distance from Earth to a nearby star like Proxima Centauri would be extremely large in AUs, using light years simplifies such grand distances.
  • 1 AU = 149.6 million kilometers
  • Primarily used for solar system measurements

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Most popular questions from this chapter

(II) Some rearview mirrors produce images of cars to your rear that are smaller than they would be if the mirror were flat. Are the mirrors concave or convex? What is a mirror's radius of curvature if cars 18.0 \(\mathrm{m}\) appear 0.33 their normal size?

When light passes through a prism, the angle that the refracted ray makes relative to the incident ray is called the deviation angle \(\delta,\) Fig. \(32-64 .\) Show that this angle is a minimum when the ray passes through the prism symmetrically, perpendicular to the bisector of the apex angle \(\phi,\) and show that the minimum deviation angle, \(\delta_{\mathrm{m}},\) is related to the prism's index of refraction \(n\) by $$ n=\frac{\sin \frac{1}{2}\left(\phi+\delta_{\mathrm{m}}\right)}{\sin \phi / 2} $$ \(\left[\right.\) Hint: For \(\theta\) in radians, \(\left.(d / d \theta)\left(\sin ^{-1} \theta\right)=1 / \sqrt{1-\theta^{2}} .\right]\)

A spherical mirror of focal length \(f\) produces an image of an object with magnification \(m .(a)\) Show that the object is a distance \(d_{0}=f\left(1-\frac{1}{m}\right)\) from the reflecting side of the mirror. \((b)\) Use the relation in part \((a)\) to show that, no matter where an object is placed in front of a convex mirror, its image will have a magnification in the range \(0 \leq m \leq+1\).

(II) Figure 57 shows a liquid-detecting prism device that might be used inside a washing machine or other liquid-containing appliance. If no liquid covers the prism's hypotenuse, total internal reflection of the beam from the light source produces a large signal in the light sensor. If liquid covers the hypotenuse, some light escapes from the prism into the liquid and the light sensor's signal decreases. Thus a large signal from the light sensor indicates the absence of liquid in the reservoir. If this device is designed to detect the presence of water, determine the allowable range for the prism's index of refraction \(n .\) Will the device work properly if the prism is constructed from (inexpensive) lucite? For lucite, \(n=1.5 .\)

A ray of light, after entering a light fiber, reflects at an angle of \(14.5^{\circ}\) with the long axis of the fiber, as in Fig. \(32-56 .\) Calculate the distance along the axis of the fiber that the light ray travels between successive reflections off the sides of the fiber. Assume that the fiber has an index of refraction of 1.55 and is \(1.40 \times 10^{-4} \mathrm{~m}\) in diameter.
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