/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 (a) In May 2008 , the United Sta... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) In May 2008 , the United States successfully landed a spacecraft named Phoenix near the northern polar regions of Mars. Immediately upon landing, the craft sent a message indicating that all had gone well. Using the astronomical data in the appendix of this book, determine the shortest amount of time it took this signal to reach the Earth. (b) If the Phoenix transmitter sent out spherical electromagnetic waves with a power of \(100 \mathrm{~W}\), how many watts per square meter would arrive at the Earth, assuming that Mars was in its closest location to Earth? (c) A radio signal was sent to a deep space probe traveling in the plane of the solar system. Earth received a response 3.5 days later. Assuming the probe computers took 4.5 hours to process the signal instructions and to send out the return message, was the probe within the solar system? (Assume a solar system radius of about 40 times the Earth-Sun distance.)

Short Answer

Expert verified
(a) 182 seconds. (b) 2.68 x 10^{-17} W/m^2. (c) No, the probe was outside the solar system.

Step by step solution

01

Understanding the Speed of Light

The speed of light, which is the speed at which electromagnetic waves travel, is approximately \(299,792,458\) meters per second (m/s). We will use this information to calculate the time it takes for a signal to travel from Mars to Earth.
02

Calculate Distance between Mars and Earth at Closest Approach

The closest approach of Mars to Earth is approximately 54.6 million kilometers, or \(54.6 \times 10^9\) meters. This distance is referred to as opposition, when Mars and Earth are aligned on the same side of the Sun.
03

Calculate Time for Signal to Reach Earth

Using the speed of light and the closest distance between Earth and Mars, the time \(t\) is calculated using the formula \(t = \frac{d}{c}\), where \(d\) is the distance and \(c\) is the speed of light. Substituting the values, \(t = \frac{54.6 \times 10^9}{299,792,458} \approx 182\) seconds.
04

Calculating Intensity of Electromagnetic Wave

Electromagnetic waves spread out as spheres centered on the source. The intensity \(I\) at a distance \(d\) from the source is given by \(I = \frac{P}{4\pi d^2}\), where \(P\) is the power. Substituting \(P = 100 \mathrm{~W}\) and \(d = 54.6 \times 10^9 \mathrm{~m}\), we find \(I = \frac{100}{4\pi (54.6 \times 10^9)^2} \approx 2.68 \times 10^{-17} \mathrm{~W/m^2}\).
05

Calculate Total Time for Signal Travel to Deep Space Probe

The total time for the round trip signal is given: 3.5 days, which equals \(3.5 \times 24 \times 3600\) seconds. The signal travel time has two legs: to the probe and back.
06

Determine Signal Travel Time to Probe

Subtract the processing time (4.5 hours or 16,200 seconds) from the total time to find the actual travel time: \((3.5 \times 24 \times 3600) - 16200 = 288600\) seconds.
07

Calculate Distance to Deep Space Probe

Signal travel time one-way is half of 288600 seconds, \(144300\) seconds. Use \(d = ct\) to find one-way distance: \(d = 299,792,458 \times 144300 \approx 4.32 \times 10^{13}\) meters.
08

Compare Probe Distance with Solar System Radius

The distance from the Sun to the boundary of the solar system (40 AU) is about \(6 \times 10^{12}\) meters. The probe is at \(4.32 \times 10^{13}\) meters, which is beyond the solar system boundary defined as 40 AU.

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Key Concepts

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

Electromagnetic Waves
Electromagnetic waves are an essential part of our understanding of physics, allowing for the transmission of information over vast distances. These waves are oscillations of electric and magnetic fields that travel through space at the speed of light. They include a broad range of frequencies, such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

The property of electromagnetic waves that enables communication, like sending signals from Mars to Earth, is their ability to travel through the vacuum of space. This is crucial for interplanetary communication, where there is no air or medium to carry signals. The waves propagate spherically from a point source, diminishing in intensity as they spread further into space. The calculation of signal intensity at large distances, as in the exercise, relies on understanding the nature of these waves.
Speed of Light
The speed of light is a fundamental constant in physics, often denoted by the symbol \(c\). Its approximate value is \299,792,458\ meters per second (m/s) in a vacuum. This speed is not only crucial for physics calculations but also serves as a universal limit for the speed at which information and matter can travel.

In practical terms, understanding the speed of light allows us to calculate how long it takes for signals to travel across the vast distances of space. For communication between Earth and Mars, knowing that the speed of light remains constant helps us predict transmission times accurately. Such calculations form the basis for determining when a spacecraft's message will reach Earth, as seen in the exercise with the Phoenix spacecraft.
Astronomical Distance Calculation
Astronomical distances are challenging to comprehend due to the enormous scales involved. The closest approach of Mars to Earth, for instance, is about 54.6 million kilometers or \(54.6 \times 10^9\) meters. Distances between astronomical bodies like planets are often in millions or billions of kilometers, requiring specific techniques to measure and understand them.

One such technique involves calculating times based on the speed of light, as shown in the exercise. By understanding the distance and the speed at which electromagnetic waves travel, we can determine the time for signals to travel across these distances. This information is critical for space missions and for determining the feasibility and timing of communication with probes, satellites, and spaceships.
Signal Transmission Time
Signal transmission time refers to the duration it takes for a signal to travel between two points. In the context of space exploration, this involves calculating how long a message sent from a spacecraft takes to reach Earth or vice versa. The exercise highlights the method to determine this time using the distance between celestial bodies and the speed of light.

For instance, the signal travel time from Mars to Earth can be computed by dividing the distance by the speed of light, resulting in a transmission time of approximately 182 seconds under certain conditions. Additionally, for missions beyond the solar system, these calculations become more complex due to larger distances, potentially affecting the time-sensitive nature of the communications.

Understanding transmission times helps us plan and execute space missions more effectively, taking into account both round-trip delays and the processing times required for interpreting and responding to the received signals.

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Most popular questions from this chapter

A circular loop with an area of \(0.015 \mathrm{~m}^{2}\) is in a uniform magnetic field of \(0.30 \mathrm{~T}\). What is the flux through the loop's plane if it is (a) parallel to the field, (b) at an angle of \(37^{\circ}\) to the field, and (c) perpendicular to the field?

The transformer in the power supply for a computer's 500-GB external hard drive changes a \(120-\mathrm{V}\) input voltage (from a regular house line) to the 5.0 - \(\mathrm{V}\) output voltage that is required to power the drive. (a) Find the ratio of the number of turns in the primary coil to the number of turns in the secondary coil. (b) If the drive is rated at \(10 \mathrm{~W}\) when running and the transformer is ideal, what is the current in the primary and secondary when the drive is in operation?

An ideal transformer steps \(8.0 \mathrm{~V}\) up to \(2000 \mathrm{~V}\), and the 4000-turn secondary coil carries 2.0 A. (a) Find the number of turns in the primary coil. (b) Find the current in the primary coil.

A small plant produces electric energy and, through a transformer, sends it out over the transmission lines at \(50 \mathrm{~A}\) and \(20 \mathrm{kV}\). The line reaches a small town over \(25-\mathrm{km}\) -long transmission lines whose resistance is \(1.2 \Omega / \mathrm{km} .\) (a) What is the power loss in the lines if the energy is transmitted at \(20 \mathrm{kV} ?\) (b) What should be the output voltage of the transformer to decrease the power loss by a factor of 15 ? Assume the transformer is ideal. (c) What would be the current in the lines in part (b)?

A boy is traveling due north at a constant speed while carrying a metal rod. The rod's length is oriented in the east-west direction and is parallel to the ground. (a) There will be no induced emf when the rod is (1) at the equator, (2) near the Earth's magnetic poles, (3) somewhere between the equator and the poles. Why? (b) Assume that the Earth's magnetic field is \(1.0 \times 10^{-4} \mathrm{~T}\) near the North Pole and \(1.0 \times 10^{-5} \mathrm{~T}\) near the equator. If the boy runs with a speed of \(1.3 \mathrm{~m} / \mathrm{s}\) northward near each location, and the rod is \(1.5 \mathrm{~m}\) long, calculate the induced emf in the rod in each location.

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