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Chapter 32: Problem 1

(a) How much time does it take light to travel from the moon to the earth, a distance of 384,000 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) Approximately 1.28 seconds. (b) Approximately \(8.161 \times 10^{13}\) kilometers.

Step by step solution

01

Understand the Speed of Light

The speed of light in a vacuum is approximately 299,792,458 meters per second (m/s). For convenience, this can be approximated to 300,000 kilometers per second (km/s) to simplify calculations.
02

Convert Distance from Kilometers to Kilometers per Second (Part a)

You are given a distance of 384,000 km from the Moon to the Earth. To find the time, use the formula:\[\text{Time} = \frac{\text{Distance}}{\text{Speed of Light}}\]Substitute the values into the formula:\[\text{Time} = \frac{384,000 \text{ km}}{300,000 \text{ km/sec}}\]
03

Compute the Time for Light to Travel from the Moon to the Earth (Part a)

Calculate the time required for light to travel from the Moon to the Earth:\[\text{Time} \approx 1.28 \text{ seconds}\]
04

Convert Time from Years to Seconds (Part b)

Light from Sirius takes 8.61 years to reach Earth. First, convert this time from years to seconds, knowing there are about 31,536,000 seconds in a year.\[\text{Time in seconds} = 8.61 \text{ years} \times 31,536,000 \text{ sec/year}\]
05

Calculate Distance Using Time and Speed of Light (Part b)

Using the time in seconds and the speed of light to find the distance:\[\text{Distance} = \text{Time in seconds} \times \text{Speed of Light} = (8.61 \times 31,536,000 \text{ sec}) \times 300,000 \text{ km/sec}\]
06

Compute the Distance from Earth to Sirius (Part b)

Perform the multiplication to get the distance:\[\text{Distance} \approx 8.161 \times 10^{13} \text{ km}\]

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Key Concepts

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

Distance Calculation
Understanding how to calculate distances using the speed of light is an essential skill in physics and astronomy. When light travels from one point to another, such as from the Moon to Earth, the distance can be calculated using a simple formula:
  • First, we know that the speed of light is approximately 300,000 kilometers per second.
  • The formula we'll use is: \( \text{Time} = \frac{\text{Distance}}{\text{Speed of Light}} \).
  • By applying the distance from the Moon to Earth, which is 384,000 km, and dividing by the speed of light, we get the time light takes to travel that distance.
This method is a straightforward approach that helps us visualize how incredibly fast light is, allowing us to calculate time based on vast yet reachable astronomical distances.
Light Travel Time
Light travel time is a fascinating concept that demonstrates just how fast light travels over enormous distances. When we talk about light traveling, we're discussing how long it takes for light to go from one celestial body to another.
  • This concept can be seen in practice when calculating how long it takes light to travel from the Moon to Earth, which takes only about 1.28 seconds.
  • For stars much farther away, like Sirius, light can take years to reach us, showing the vastness of space.
  • For example, from Sirius, light takes 8.61 years to reach the planet Earth.
This shows us the immense distance light covers, stretching over trillions of kilometers, a testament to the vastness and wonder of the universe. When making such calculations, convert all units appropriately to avoid errors.
Astronomical Distances
Astronomical distances often involve incomprehensibly large numbers. These distances are frequently measured in light years, which is the distance light travels in one year. By knowing the time light takes to travel from Sirius to Earth, we can calculate this massive distance.
  • First, convert the years into seconds, using the known conversion of 31,536,000 seconds per year, to comprehend just how large a single light year is.
  • Next, multiply this time in seconds by the speed of light to find the total distance covered by light in kilometers.
  • In the exercise example, light from Sirius covers roughly \( 8.161 \times 10^{13} \) kilometers.
These distances reveal just how immense the universe is. Tools like the speed of light and conversion of time enhance our understanding and capability to calculate these vast stretches of space. Such insights help deepen our understanding of the cosmos and all it encompasses.

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Most popular questions from this chapter

Medical x rays are taken with electromagnetic waves having a wavelength of around 0.10 nm in air. What are the frequency, period, and wave number of such waves?

The energy flow to the earth from sunlight is about 1.4 kW/m\(^2\). (a) Find the maximum values of the electric and magnetic fields for a sinusoidal wave of this intensity. (b) The distance from the earth to the sun is about 1.5 \(\times\) 10\(^{11}\) m. Find the total power radiated by the sun.

A monochromatic light source with power output 60.0 W radiates light of wavelength 700 nm uniformly in all directions. Calculate \(E_{max}\) and \(B_{max}\) for the 700-nm light at a distance of 5.00 m from the source.

A small helium-neon laser emits red visible light with a power of 5.80 mW in a beam of diameter 2.50 mm. (a) What are the amplitudes of the electric and magnetic fields of this light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a 1.00-m length of the beam?

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electricfield vector \(\vec{E}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (\(Hint\): Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.
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