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(a) Neil A. Armstrong was the first person to walk on the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{m}\). Find the time it took for his voice to reach the earth via radio waves. (b) Someday a person will walk on Mars, which is \(5.6 \times 10^{10} \mathrm{m}\) from the earth at the point of closest approach. Determine the minimum time that will be required for a message from Mars to reach the earth via radio waves.

Short Answer

Expert verified
(a) 1.28 seconds, (b) 187 seconds.

Step by step solution

01

Understand the Speed of Radio Waves

Radio waves travel at the speed of light, which is approximately \( c = 3.00 \times 10^{8} \ \mathrm{m/s} \). We can use this speed to calculate the time it takes for a radio wave to travel a known distance.
02

Calculate Time for Radio Wave from Moon

Given that the distance from the Earth to the Moon is \( 3.85 \times 10^{8} \ \mathrm{m}\), use the formula for time, \( t = \frac{d}{c} \), where \( d \) is the distance and \( c \) is the speed of light:\[t_\text{moon} = \frac{3.85 \times 10^{8} \ \mathrm{m}}{3.00 \times 10^{8} \ \mathrm{m/s}} = 1.2833 \ \mathrm{s}\approx 1.28 \ \mathrm{s}.\]
03

Calculate Time for Radio Wave from Mars

Given that the distance from the Earth to Mars at the closest approach is \( 5.6 \times 10^{10} \ \mathrm{m}\), use the same formula for time:\[t_\text{Mars} = \frac{5.6 \times 10^{10} \ \mathrm{m}}{3.00 \times 10^{8} \ \mathrm{m/s}} = 186.67 \ \mathrm{s} \approx 187 \ \mathrm{s}. \]

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Key Concepts

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

Speed of Light
Understanding the speed of light is essential for calculating how long it takes for radio waves to travel through space. Radio waves are part of the electromagnetic spectrum, just like visible light, microwaves, and X-rays. They travel at the speed of light, which is approximately \( c = 3.00 \times 10^8 \ \mathrm{m/s} \). This enormous speed allows radio waves to cover vast distances in space relatively quickly. For example, the finite speed of light is what makes observing celestial bodies like stars a look into the past, as their light takes time to reach us. In calculations, the speed of light is used as a constant to determine how quickly signals can travel between celestial bodies, such as from the Earth to the Moon or Mars. This helps in estimating real-time communication delays between Earth and spacecraft or astronauts exploring distant locations.
Space Communication
Space communication relies heavily on radio waves due to their properties. They can travel through the vacuum of space, unlike sound waves, which require a medium to move through. Our ability to send and receive signals from space missions depends on this form of communication. For successful space communication:
  • Radio waves should be directed towards their target accurately.
  • There needs to be a receiver system on both ends, like a ground station on Earth and a communication device on a spacecraft.
  • Broadcasts must account for delays due to distances traveled.
Radio signals' travel times can lead to noticeable delays when communicating with spacecraft or astronauts on the Moon or Mars, influencing mission planning and operations. That's why understanding these times is crucial for space mission success.
Earth-Moon Distance
The average Earth-Moon distance is roughly \(3.85 \times 10^8 \ \mathrm{m} \). This is an important factor when calculating the time it takes for radio communication signals to travel between Earth and the Moon. Given this distance, and knowing that radio waves travel at the speed of light, the time it takes for a message to travel from the Moon to Earth is calculated using the formula \( t = \frac{d}{c} \), resulting in about 1.28 seconds.
  • This is quick enough for relatively smooth communication.
  • No significant delays are noticeable to the human perception.
  • However, during operations like remote-controlled vehicle commands, these delays must be considered.
Understanding the Earth-Moon distance helps us appreciate how fast radio waves can enable us to hear the first words from the Moon and ensures smooth operations for potential missions in the future.

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Most popular questions from this chapter

A politician holds a press conference that is televised live. The sound picked up by the microphone of a TV news network is broadcast via electromagnetic waves and heard by a television viewer. This viewer is seated \(2.3 \mathrm{m}\) from his television set. A reporter at the press conference is located \(4.1 \mathrm{m}\) from the politician, and the sound of the words travels directly from the celebrity's mouth, through the air, and into the reporter's ears. The reporter hears the words exactly at the same instant that the television viewer hears them. Using a value of \(343 \mathrm{m} / \mathrm{s}\) for the speed of sound, determine the maximum distance between the television set and the politician. Ignore the small distance between the politician and the microphone. In addition, assume that the only delay between what the microphone picks up and the sound being emitted by the television set is that due to the travel time of the electromagnetic waves used by the network.

Magnetic resonance imaging, or MRI (see Section 21.7 ), and positron emission tomography, or PET scanning (see Section 32.6 ), are two medical diagnostic techniques. Both employ electromagnetic waves. For these waves, find the ratio of the MRI wavelength (frequency \(=6.38 \times 10^{7} \mathrm{Hz}\) ) to the PET scanning wavelength (frequency \(=1.23 \times 10^{20} \mathrm{Hz}\) ).

A speeder is pulling directly away and increasing his distance from a police car that is moving at \(25 \mathrm{m} / \mathrm{s}\) with respect to the ground. The radar gun in the police car emits an electromagnetic wave with a frequency of \(7.0 \times 10^{9} \mathrm{Hz}\). The wave reflects from the speeder's car and returns to the police car, where its frequency is measured to be \(320 \mathrm{Hz}\) less than the emitted frequency. Find the speeder's speed with respect to the ground.

A laser emits a narrow beam of light. The radius of the beam is \(1.0 \times 10^{-3} \mathrm{m},\) and the power is \(1.2 \times 10^{-3} \mathrm{W} .\) What is the intensity of the laser beam?

An electromagnetic wave strikes a \(1.30-\mathrm{cm}^{2}\) section of wall perpendicularly. The rms value of the wave's magnetic field is determined to be \(6.80 \times 10^{-4} \mathrm{T}\). How long does it take for the wave to deliver \(1850 \mathrm{J}\) of energy to the wall?

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