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(a) How much time does it take light to travel from the moon to the earth, a distance of \(384,000 \mathrm{~km}\) ? (b) Light from the star Sirius takes \(8.61\) years to reach the earth. What is the distance from earth to Sirius in kilometers?

Short Answer

Expert verified
(a) ~1.28 seconds; (b) ~81.4 billion km.

Step by step solution

01

Understand the Problem

We need to calculate the time it takes for light to travel a distance of 384,000 km from the Moon to the Earth and then find the distance light travels in 8.61 years from Sirius to the Earth.
02

Recall Light Speed

The speed of light in a vacuum is approximately 299,792 kilometers per second (km/s). We will use this value to calculate the time for part (a) and the distance for part (b).
03

Calculate Time for Moon to Earth

To find the time it takes for light to travel from the Moon to the Earth, use the formula \( \text{time} = \frac{\text{distance}}{\text{speed}} \). Substitute the values to get \( \text{time} = \frac{384,000 \text{ km}}{299,792 \text{ km/s}} \).
04

Solve for Time

Calculate the time using the formula from Step 3: \( \text{time} \approx 1.28 \text{ seconds} \).
05

Calculate Distance for Sirius to Earth

Convert the travel time from years to seconds for accurate calculation as we know the speed in km/s. There are 31,536,000 seconds in a year, so multiply by 8.61 years: \( 8.61 \times 31,536,000 = 271,034,160 \text{ seconds} \). Then, use \( \text{distance} = \text{speed} \times \text{time} \) to find the distance.
06

Solve for Distance

Substitute and calculate the distance: \( \text{distance} = 299,792 \text{ km/s} \times 271,034,160 \text{ s} \approx 81,395,688,832 \text{ km} \).

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Key Concepts

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

Light Travel Time
Light travels extremely fast at approximately 299,792 kilometers per second (km/s). Calculating the travel time of light involves understanding this speed and using it to determine how long light takes to move across different distances.

For example, when calculating how much time it takes for light to travel from the Moon to the Earth, which is a distance of 384,000 kilometers, we apply the formula for time, which is:
  • \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
Substitute the known values into the formula:
  • \( \text{time} = \frac{384,000 \text{ km}}{299,792 \text{ km/s}} \)
This results in a time of approximately 1.28 seconds. So, light from the Moon reaches Earth in just over a second. This quick travel time highlights how light traverses vast distances in space quickly, making it a critical constant for astronomical calculations.
Distance Calculation
Using the speed of light, we can also calculate astronomical distances when the travel time is known. For instance, if light from a star takes a certain number of years to reach Earth, we can find out how far away that star is in kilometers.

Suppose light from Sirius takes 8.61 years to reach Earth. First, we need to convert the time from years into seconds, since the speed of light is given in kilometers per second. There are 31,536,000 seconds in one year, so we multiply by 8.61 years:
  • \( 8.61 \times 31,536,000 = 271,034,160 \text{ seconds} \)
Next, we use the formula for distance, which is:
  • \( \text{distance} = \text{speed} \times \text{time} \)
Substituting the speed of light and the time in seconds, we get:
  • \( \text{distance} = 299,792 \text{ km/s} \times 271,034,160 \text{ seconds} \)
Through this calculation, we find that Sirius is approximately 81,395,688,832 kilometers away from Earth. These sorts of calculations are important for understanding how far different celestial bodies are based on the time it takes light to reach us.
Astronomical Distances
When discussing astronomical distances, we refer to the vast space between celestial bodies such as stars, planets, and galaxies. These distances are immense and often difficult to comprehend, hence why using the speed of light as a measure is so prevalent.

For example, astronomers frequently use light-years as a unit to express these distances. A light-year is the distance that light travels in one year, which equates to about 9.46 trillion kilometers. This unit makes it easier to grasp the scale of the universe.

Calculating these distances, like that to Sirius or from the Moon, involves accounting for both the powerful speed of light and the considerable time light takes to traverse these vast expanses. Understanding how to calculate these astronomical distances is crucial for cosmology and understanding the universe's structure and scale.

Moreover, knowing these distances can help in determining other properties of celestial bodies, such as their sizes and positions relative to Earth, essential for both navigation and exploration beyond our solar system.

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Most popular questions from this chapter

Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{~km}\), (ii) \(5.0 \mu \mathrm{m}\), (iii) \(5.0 \mathrm{~nm}\). (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{~Hz}\) and (ii) an AM station radio wave of frequency \(590 \mathrm{kHz}\) ?

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from WCCO is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

There are two categories of ultraviolet light. Ultraviolet A (UVA) has a wavelength ranging from \(320 \mathrm{~nm}\) to \(400 \mathrm{~nm}\). It is not harmful to the skin and is necessary for the production of vitamin D. UVB, with a wavelength between \(280 \mathrm{~nm}\) and \(320 \mathrm{~nm}\), is much more dangerous because it causes skin cancer. (a) Find the frequency ranges of UVA and UVB. (b) What are the ranges of the wave numbers for UVA and UVB?

Television Broadcasting. Public television station KQED in San Francisco broadcasts a sinusoidal radio signal at a power of \(316 \mathrm{~kW}\). Assume that the wave spreads out uniformly into a hemisphere above the ground. At a home \(5.00 \mathrm{~km}\) away from the antenna, (a) what average pressure does this wave exert on a totally reflecting surface, (b) what are the amplitudes of the electric and magnetic fields of the wave, and (c) what is the average density of the energy this wave carries? (d) For the energy density in part (c), what percentage is due to the electric field and what percentage is due to the magnetic field?

A satellite \(575 \mathrm{~km}\) above the earth's surface transmits sinusoidal electromagnetic waves of frequency \(92.4 \mathrm{MHz}\) uniformly in all directions, with a power of \(25.0 \mathrm{~kW}\). (a) What is the intensity of these waves as they reach a receiver at the surface of the earth directly below the satellite? (b) What are the amplitudes of the electric and magnetic fields at the receiver? (c) If the receiver has a totally absorbing panel measuring \(15.0 \mathrm{~cm}\) by \(40.0 \mathrm{~cm}\) oriented with its plane perpendicular to the direction the waves travel, what average force do these waves exert on the panel? Is this force large enough to cause significant effects?

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