/*! 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 21 A space probe \(2.0 \times 10^{1... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A space probe \(2.0 \times 10^{10} \mathrm{~m}\) from a star measures the total intensity of electromagnetic radiation from the star to be \(5.0 \times 10^{3} \mathrm{~W} / \mathrm{m}^{2}\). If the star radiates uniformly in all directions, what is its total average power output?

Short Answer

Expert verified
The total average power output of the star is approximately \(5.03 \times 10^{26}\) watts.

Step by step solution

01

Understand the Problem Context

We need to determine the total power output of the star, given its intensity at a certain distance. The given intensity is measured by a space probe located at a distance of \(2.0 \times 10^{10} \text{ m}\) from the star.
02

Identify the Relevant Formula

We know that intensity \(I\) at a distance from a star can be given by the formula \(I = \frac{P}{A}\), where \(P\) is the power output of the star and \(A\) is the area over which the power is spread. Since the star radiates uniformly in all directions, \(A\) will be the surface area of a sphere.
03

Calculate the Area of the Sphere

The surface area \(A\) of a sphere with radius \(r\) is given by the formula \(A = 4\pi r^{2}\). Here, \(r = 2.0 \times 10^{10} \text{ m}\). Substitute the radius value into the formula:\[A = 4\pi (2.0 \times 10^{10})^2\].
04

Substitute into Intensity Formula

We have the formula \(I = \frac{P}{A}\). Rearrange it to solve for the power \(P\):\[P = I \times A\].
05

Insert Values and Calculate

Substitute \(I = 5.0 \times 10^{3} \text{ W/m}^2\) and \(A = 4\pi (2.0 \times 10^{10})^2\) into the formula:\[P = 5.0 \times 10^{3} \text{ W/m}^2 \times 4\pi (2.0 \times 10^{10})^2\]. Calculate this expression to find \(P\).
06

Evaluate the Total Power

The expression evaluates to:\[P = 5.0 \times 10^{3} \times 4\pi (2.0 \times 10^{10})^2 \text{ watts}\]. Calculating gives us \(P \approx 5.03 \times 10^{26} \text{ watts}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Intensity of Electromagnetic Radiation
The intensity of electromagnetic radiation is a measure of how much energy the radiation transfers per unit area. It's like understanding how light shines from a bulb or the sun onto a surface. The farther you are from the source (e.g., a star), the spread out the radiation becomes over a larger area, which means less intensity in one spot. For a space probe measuring the intensity at a certain distance from the star, it reads the energy received per square meter. This vital measurement helps determine how much total energy the star emits. In this scenario, intensity is noted as being very high at a staggering 5000 watts per square meter at a particular distance.
Surface Area of a Sphere
The surface area of a sphere plays a crucial role when calculating how energy, like light or radiation, spreads out in space. Picture a star sending out light uniformly in all directions, filling a spherical area around it. This area can be calculated precisely as it acts like a giant invisible ball surrounding the star.The formula to find this area is: \[ A = 4\pi r^2 \] where \( r \) is the distance from the center of the star to the point of interest, in this case, the space probe's location. The larger the sphere's radius, the bigger the surface area over which the power disperses.
Inverse Square Law
The inverse square law is a principle that describes how the intensity of light or other energy spreads out as you move away from the source. If you double the distance from a star, the light intensity becomes one-fourth (1/4) as strong per unit area. This principle arises because the surface area of the sphere increases with the square of the radius: - Distance doubles, area quadruples - Therefore, intensity decreases to a quarter of its original value This fundamental concept explains why distant stars appear dimmer than those nearby, despite possibly having the same total power output.
Sphere Geometry
In sphere geometry, understanding how objects and their properties change relative to a sphere's surface is essential. The sphere is the shape used when visualizing how light and radiation move through space. This is because when you spread something uniformly from a point in three dimensions, you naturally form a spherical wave front. Knowing how to calculate the surface area of a sphere helps you understand where the total power of a star goes. It's not just about the radius but how this radius forms an area through which the energy flows, giving context to the star's radiation measures.
Power Radiated by Stars
When we talk about the power radiated by stars, we're referring to the total energy output the star emits per unit of time, often measured in watts. This power can be thought of as the sum of all the energy spread across the entire spherical surface around the star. To find the total power output: 1. Measure the intensity of electromagnetic radiation at a distance.2. Calculate the spherical surface area over that distance using geometry.3. Use the formula \( P = I \times A \) to find the power, where \( I \) is the intensity and \( A \) is the surface area.This calculation tells us the staggering amount of energy stars emit, even when they're many light-years away, illuminating the universe efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a certain experiment, a radio transmitter emits sinusoidal electromagnetic waves of frequency \(110.0 \mathrm{MHz}\) in opposite directions inside a narrow cavity with reflectors at both ends, causing a standing-wave pattern to occur. (a) How far apart are the nodal planes of the magnetic field? (b) If the standing-wave pattern is determined to be in its eighth harmonic, how long is the cavity?

An electromagnetic wave with frequency \(65.0 \mathrm{~Hz}\) travels in an insulating magnetic material that has dielectric constant \(3.64\) and relative permeability \(5.18\) at this frequency. The electric field has amplitude \(7.20 \times 10^{-3} \mathrm{~V} / \mathrm{m}\). (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

Flashlight to the Rescue. You are the sole crew member of the interplanetary spaceship \(T: 1339\) Vorga, which makes regular cargo runs between the earth and the mining colonies in the asteroid belt. You are working outside the ship one day while at a distance of \(2.0 \mathrm{AU}\) from the sun. [1 AU (astronomical unit) is the average distance from the earth to the sun, \(149,600,000 \mathrm{~km} .]\) Unfortunately, you lose contact with the ship's hull and begin to drift away into space. You use your spacesuit's rockets to try to push yourself back toward the ship, but they run out of fuel and stop working before you can return to the ship. You find yourself in an awkward position, floating \(16.0 \mathrm{~m}\) from the spaceship with zero velocity relative to it. Fortunately, you are carrying a 200-W flashlight. You turn on the flashlight and use its beam as a "light rocket" to push yourself back toward the ship. (a) If you, your spacesuit, and the flashlight have a combined mass of \(150 \mathrm{~kg}\), how long will it take you to get back to the ship? (b) Is there another way you could use the flashlight to accomplish the same job of returning you to the ship?

A small helium-neon laser emits red visible light with a power of \(4.60 \mathrm{~mW}\) in a beam that has a diameter of \(2.50 \mathrm{~mm}\). (a) What are the amplitudes of the electric and magnetic fields of the 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 \(1.00-\mathrm{m}\) length of the beam?

A circular loop of wire has radius \(7.50 \mathrm{~cm}\). A sinusoidal electromagnetic plane wave traveling in air passes through the loop, with the direction of the magnetic field of the wave perpendicular to the plane of the loop. The intensity of the wave at the location of the loop is \(0.0195 \mathrm{~W} / \mathrm{m}^{2}\), and the wavelength of the wave is \(6.90 \mathrm{~m}\). What is the maximum emf induced in the loop?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.