Problem 86
Question
The Sun radiates energy into space at the rate of \(3.9 \times 10^{26} \mathrm{~J} / \mathrm{s}\). (a) Calculate the rate of mass loss from the Sun in \(\mathrm{kg} / \mathrm{s}\). (b) How does this mass loss arise? (c) It is estimated that the Sun contains \(9 \times 10^{56}\) free protons. How many protons per second are consumed in nuclear reactions in the Sun?
Step-by-Step Solution
Verified Answer
(a) The rate of mass loss from the Sun is approximately \(4.33 \times 10^9 kg/s\). (b) This mass loss arises due to nuclear fusion occurring within the Sun's core, where hydrogen is converted into helium, releasing energy in the form of photons and resulting in the corresponding mass loss. (c) Approximately \(2.59 \times 10^{36}\) protons are consumed per second in nuclear reactions in the Sun.
1Step 1: Calculate the rate of mass loss from the Sun
First, let's write down the given energy emission rate of the Sun: \(E = 3.9 \times 10^{26} J/s\). We can calculate the mass loss rate of the Sun using Einstein's mass-energy equivalence equation: \(E = mc^2\), where \(m\) is the mass, and \(c\) is the speed of light (\(3.0 \times 10^8 m/s\)). Let's now solve for the mass loss rate, \(m\):
\[m = \frac{E}{c^2}\]
\[m = \frac{3.9 \times 10^{26} J/s}{(3.0 \times 10^8 m/s)^2}\]
\[m = 4.33 \times 10^9 kg/s\]
The rate of mass loss from the Sun is approximately \(4.33 \times 10^9 kg/s\).
2Step 2: Explain how mass loss arises in the Sun
This mass loss arises due to nuclear fusion occurring within the Sun's core. In these nuclear reactions, hydrogen is converted into helium through fusion, releasing energy in the form of photons. During this process, some matter is transformed into energy according to Einstein's mass-energy equivalence equation, E=mc^2. As photons escape the Sun's surface, they carry away this energy, resulting in the corresponding mass loss.
3Step 3: Calculate the number of protons consumed per second
To calculate the number of protons consumed per second, we first need to find the mass of one proton. The mass of a proton is approximately \(1.67 \times 10^{-27} kg\). We now have the total mass loss rate and mass of a proton; hence we can estimate the number of protons consumed per second:
Number of protons consumed per second = \(\frac{\text{Mass loss rate}}{\text{Mass of one proton}}\)
Number of protons consumed per second = \(\frac{4.33 \times 10^9 kg/s}{1.67 \times 10^{-27} kg}\)
Number of protons consumed per second ≈ \(2.59 \times 10^{36} protons/s\)
Approximately \(2.59 \times 10^{36}\) protons are consumed per second in nuclear reactions in the Sun.
Key Concepts
Nuclear Fusion in the SunEinstein's Mass-Energy EquivalenceProton Consumption Rate
Nuclear Fusion in the Sun
The Sun, a massive nuclear furnace, is the source of vast amounts of energy that sustains life on Earth. It operates through a process known as nuclear fusion, which occurs in its core. During fusion, hydrogen atoms are compressed by immense gravitational forces, leading to collisions that fuse protons— the nuclei of hydrogen atoms—to form helium. Here's a fundamental breakdown of how it happens:
Through the fusion process, the Sun loses an infinitesimally small fraction of its mass, yet over astronomical time scales, this leads to significant changes. This understanding is crucial for comprehending the longevity and lifecycle of stars like our Sun.
The Proton-Proton Chain Reaction
This is the primary fusion process within the Sun. It involves several steps where protons combine, release particles, produce helium, and liberate energy. Crucially, this chain reaction converts some of the mass of the protons into energy as per Einstein's famous formula. That energy is eventually emitted as sunlight, after countless interactions within the Sun's layers.Energy Release and the Solar Luminosity
One might wonder: If the Sun continually fuses protons, why doesn't it change dramatically or even explode? This is because the rate of fusion is incredibly steady. The amount of energy produced through fusion is directly related to the Sun's luminosity— the total amount of energy emitted into space. The energy released by the Sun is nearly constant, providing a stable amount of heat and light to our planet.Through the fusion process, the Sun loses an infinitesimally small fraction of its mass, yet over astronomical time scales, this leads to significant changes. This understanding is crucial for comprehending the longevity and lifecycle of stars like our Sun.
Einstein's Mass-Energy Equivalence
One of the most revolutionary concepts in physics is encapsulated in the simple equation, \(E=mc^2\). This is Einstein's mass-energy equivalence principle, indicating that mass and energy are interchangeable. It operates at the core of the processes that power the stars. In the context of the Sun, the equation explains how the Sun can lose mass by radiating energy into space.
Understanding the Key Components
The \(E\) in the equation represents energy, while \(m\) stands for mass, and \(c\), the speed of light, is the constant that relates the two. Given that the speed of light is incredibly large, a small amount of mass can be converted into a tremendous amount of energy. This is the fundamental principle behind both the power of the Sun and nuclear energy on Earth.Practical Implications in Astrophysics
In the Sun, the mass lost through nuclear fusion is converted into energy radiated away. On the scale of the Sun's total mass, this loss is minuscule, but over billions of years, it amounts to a significant portion of mass. This equation isn't just a theoretical concept; it's observable in the changing dynamics of celestial bodies and is a cornerstone of how we understand the universe.Proton Consumption Rate
To comprehend the grandeur of the Sun's power, one can look at the rate at which it consumes protons. At a staggering rate of approximately \(2.59 \times 10^{36}\) protons per second, the Sun relentlessly converts mass into energy. Each proton contributes to the Sun's luminosity and eventual mass loss.
Scale of Proton Consumption
The number is nearly incomprehensible, but it's a testament to the efficiency of nuclear fusion as an energy source. Considering that every second, multiple earth masses worth of protons are fused, this highlights the massive scale of the Sun's energy production.Lifespan of the Sun
By calculating the proton consumption rate, scientists can estimate just how long the Sun can continue to burn. Given that the Sun has a finite supply of hydrogen, its fusion fuel, this rate ultimately dictates the Sun's lifespan. Once the hydrogen in the Sun's core is nearly exhausted, the Sun will enter the next phase of its lifecycle—as most stars do—leading to significant changes in its structure and the solar system.Other exercises in this chapter
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