/*! 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 42 Suppose that an object with the ... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that an object with the mass of Earth \(\left(M_{\text {Earth }}=5.97 \times 10^{24} \mathrm{kg}\right)\) fell into a supermassive black hole with a 10 percent energy conversion. a. How much energy (in joules) would be radiated by the black hole? b. Compare your answer with the energy radiated by the Sun each second: \(3.85 \times 10^{26} \mathrm{J}\)

Short Answer

Expert verified
a. The energy radiated is \(5.373 \times 10^{40} \text{ J}\). b. This is approximately \(1.396 \times 10^{14}\) times the energy radiated by the Sun each second.

Step by step solution

01

Determine the mass of Earth

The mass of Earth is given as \[ M_{\text {Earth }}=5.97 \times 10^{24} \text{ kg} \].
02

Calculate the energy conversion for the mass

Using the 10 percent energy conversion rate, the energy radiated can be calculated using Einstein's mass-energy equivalence formula \[ E = \text{efficiency} \times M_{\text {Earth }} \times c^2 \]. Efficiency = 0.10 (or 10 percent)\[ c = 3 \times 10^8 \text{ m/s} \].
03

Plug in the values and compute

Substitute the values into the equation:\[ E = 0.10 \times 5.97 \times 10^{24} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \]. Simplify the equation:\[ E = 0.10 \times 5.97 \times 10^{24} \times 9 \times 10^{16} \]. \[ E = 0.10 \times 5.97 \times 9 \times 10^{40} \]. \[ E = 5.373 \times 10^{40} \text{ J} \].
04

Compare with the energy radiated by the Sun each second

The energy radiated by the Sun each second is given as \[ 3.85 \times 10^{26} \text{ J} \]. To compare, calculate the ratio: \[ \text{Ratio} = \frac{5.373 \times 10^{40} \text{ J}}{3.85 \times 10^{26} \text{ J}} \]. \[ \text{Ratio} = 1.396 \times 10^{14} \].

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Key Concepts

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

energy conversion
Energy conversion is the process of changing one form of energy into another. In the context of our exercise, the mass of Earth is being converted into energy as it falls into a supermassive black hole. This is explained by Einstein's famous mass-energy equivalence formula, which reveals that mass can be transformed into vast amounts of energy.
Imagine a scenario where something as massive as Earth falls into a black hole with a 10 percent energy conversion efficiency. Ten percent of that mass is converted into energy. That's a huge number! It underscores how powerful the process of energy conversion can be in astronomical events.
supermassive black hole
A supermassive black hole is an extremely massive black hole, typically found at the center of galaxies. They have masses ranging from millions to billions of times the mass of our Sun.
When objects fall into these cosmic giants, intense gravitational forces lead to extraordinary energy emissions.
In our exercise, Earth’s mass falling into a supermassive black hole results in significant energy radiation, demonstrating its vast energy conversion capabilities. These events are critical in shaping the dynamics of galaxies and influencing their evolution.
mass-energy equivalence formula
The mass-energy equivalence formula is expressed as \[ E = mc^2 \] This equation, proposed by Albert Einstein, indicates that energy (E) is equal to mass (m) times the speed of light (c) squared.
This formula shows a direct relationship between mass and energy, revealing that a small amount of mass can be converted into a tremendous amount of energy. To put this into practice, if we take Earth’s mass \(5.97 \times 10^{24} \text{ kg}\), and convert 10 percent of it to energy, we use the equation \[ E = 0.10 \times 5.97 \times 10^{24} \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \] It highlights how incredibly large numbers come into play when dealing with astronomical masses.
joules
Joules (J) are the unit of measurement for energy in the International System of Units (SI). One joule is the energy transferred when one newton of force is applied over a distance of one meter.
When we calculated the energy radiated by converting Earth’s mass into energy, we found the result to be \[5.373 \times 10^{40} \text{ J} \]. To give you a sense of scale, the energy radiated by the Sun each second is \[ 3.85 \times 10^{26} \text{ J} \]. Our result is significantly larger, demonstrating the immense energy associated with supermassive black holes and the process of mass-energy conversion.

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Most popular questions from this chapter

Some galaxies have regions that are relatively blue; other regions appear redder. What does this variation indicate about the differences between these regions?

Go to the Astronomy Picture of the Day app or website (http://apod.nasa.gov/apod) and look at some recent pictures of galaxies. In each case, consider the following questions: Was the picture taken from a large or small telescope; from the ground or from space? Are galaxies in the image face on, edge on, or at an angle? What wavelengths were used for making the image? Are any of the colors "false colors"? If the picture is a combination of images from several telescopes, what do the different colors indicate?

a. Go to the website for the NASA Swift Gamma-Ray observatory (swift.gsfc.nasa.gov), which studies gamma-ray bursts. Click on "Latest Swift News" and look for a story about supermassive black holes. What has been discovered? b. Go to the website for NuSTAR (Nuclear Spectroscopic Telescope Array- http://www.nustar.caltech.edu), a space telescope launched by NASA in 2012 . This mission is studying active galaxies hosting supermassive black holes. What type of telescope is this (wavelengths observed, general design)? What has been discovered?

Material ejected from the supermassive black hole at the center of galaxy \(\mathrm{M} 87\) extends outward from the galaxy to a distance of approximately 30,000 pc. \(\mathrm{M} 87\) is approximately 17 Mpc away. a. If this material were visible to the naked eye, how large would it appear in the nighttime sky? Give your answer in degrees \(\left(1 \text { radian }=57.3^{\circ}\right)\). b. Compare this size with the angular size of the Moon.

Estimate the Schwarzschild radius for a supermassive black hole with a mass of 26 billion \(M_{\text {Sun. }}\).

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