Question

The Sun radiates energy into space at the rate of $3.9\times {10}^{26}\mathrm{J}/\mathrm{s}.(\mathrm{a})$ 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?

Short answer

The rate of mass loss from the Sun is approximately $4.33\times {10}^9\mathrm{kg}/\mathrm{s}$, arising due to nuclear fusion reactions in its core. Approximately $2.6\times {10}^{36}$ protons are consumed per second in the Sun's nuclear fusion reactions.

Step-by-step solution

Calculate the rate of mass loss from the Sun

To calculate the rate of mass loss from the Sun, we can use the famous equation by Einstein: E=mc^2, where E is the energy, m is the mass, and c is the speed of light (approximately $3\times {10}^8\mathrm{m}/\mathrm{s}$). The energy radiated by the Sun every second is given as $3.9\times {10}^{26}\mathrm{J}/\mathrm{s}$. To find the corresponding mass loss per second, we can rearrange the equation as: $m=\frac{E}{c^2}$ Now, substitute the given values: $m=\frac{3.9\times {10}^{26}\mathrm{J}/\mathrm{s}}{(3\times {10}^8\mathrm{m}/\mathrm{s}{)}^2}=4.33\times {10}^9\mathrm{kg}/\mathrm{s}$ The rate of mass loss from the Sun is approximately $4.33\times {10}^9\mathrm{kg}/\mathrm{s}$.

Explain the reason for mass loss

The mass loss from the Sun arises due to nuclear fusion reactions happening in its core. During these reactions, hydrogen nuclei combine to form helium nuclei, releasing a tremendous amount of energy in the form of light and heat. This release of energy causes a small fraction of the mass to be converted into energy, following the principle of mass-energy equivalence (E=mc^2). This energy is radiated into space, and consequently, the mass of the Sun decreases.

Estimate the number of protons consumed per second

It is given that the Sun has approximately $9\times {10}^{56}$ free protons. The energy produced in the Sun's core comes from the nuclear fusion of these protons into helium nuclei. This process consists of several steps, but in the end, it consumes 4 protons to form a helium nucleus. Thus, to produce 1 helium nucleus, 4 protons are needed. Since we calculated the mass loss from the Sun to be $4.33\times {10}^9\mathrm{kg}/\mathrm{s}$, we can find the number of helium nuclei produced by dividing this mass by the mass of one helium nucleus: Number of helium nuclei produced per second = $\frac{4.33\times {10}^9\mathrm{kg}/\mathrm{s}}{6.646\times {10}^{-27}\mathrm{kg}}=6.51\times {10}^{35}\mathrm{Helium}/\mathrm{s}$ Now, to find the number of protons consumed per second, multiply the number of helium nuclei produced by 4 (since 4 protons are needed for each helium nucleus): Number of protons consumed per second = $\text{number of helium nuclei per second }\times 4=6.51\times {10}^{35}\cdot 4=2.6\times {10}^{36}\mathrm{protons}/\mathrm{s}$. Approximately $2.6\times {10}^{36}$ protons are consumed per second in the Sun's nuclear fusion reactions.

Key concepts

Nuclear Fusion

Nuclear fusion is a powerful process that occurs in the core of the Sun and other stars. It is the engine that fuels the Sun's immense energy output. In nuclear fusion, light atomic nuclei combine to form heavier nuclei. This process releases tremendous amounts of energy.

The most common nuclear fusion process in the Sun involves converting hydrogen nuclei into helium. Specifically, four hydrogen nuclei (protons) combine through a series of reactions to form one helium nucleus. This fusion occurs under extremely high pressure and temperature, allowing protons to overcome their positive charge repulsion.

• Hydrogen nuclei merge, forming a helium nucleus.
• This process releases a significant amount of energy.
• Energy release comes in the form of light and heat.

Nuclear fusion not only powers the Sun but also results in a small conversion of mass into energy, demonstrating the principle of mass-energy equivalence.

Proton Consumption

The concept of proton consumption is central to understanding how the Sun generates energy. During nuclear fusion, a substantial number of hydrogen protons are consumed every second to form helium nuclei. Each helium nucleus requires four protons to be formed.

Given that the Sun contains roughly $9\times {10}^{56}$ free protons, the dynamics of proton consumption directly affect the longevity and energy production of our star.

• Four protons are needed to form one helium nucleus.
• Approximately $2.6\times {10}^{36}$ protons are consumed every second.
• This process ensures the Sun continues to shine brightly.

Proton consumption highlights how the Sun's core is a bustling center of continuous nuclear reactions, ensuring Earth's life with energy and light.

Solar Mass Loss

Solar mass loss is a phenomenon resulting from the Sun's energy output through nuclear fusion and the mass-energy equivalence principle. As the Sun shines, it loses mass in the form of energy radiated into space. This can be understood using Einstein's equation, $E=m{c}^2$, where energy $E$ is equivalent to mass $m$ times the speed of light squared $c^2$.

The Sun radiates energy at about $3.9\times {10}^{26}\text{ J/s}$, translating into a mass loss of approximately $4.33\times {10}^9$ kg per second. Over time, this mass loss affects the Sun's structure but has a minimal immediate impact due to the Sun's vast size.

• The Sun loses around $4.33\times {10}^9$ kg per second due to energy radiation.
• This is because a small fraction of its mass changes to energy.
• Mass loss depicts the continual energy provided to the solar system.

Understanding solar mass loss helps explain how the Sun's internal reactions translate to the energy and life experienced on Earth.