Problem 43 - University Physics with Modern Physics

Question

The sun emits energy in the form of electromagnetic waves at a rate of \(3.9 \times 10^{26} \mathrm{W}\) . This energy is produced by nuclear reactions deep in the sun's interior. (a) Find the intensity of electromagnetic radiation and the radiation pressure on an absorbing object at the surface of the sun (radius \(r=R=6.96 \times 10^{5} \mathrm{km}\)) and at \(r=R / 2\) , in the sun's interior. Ignore any scattering of the waves as they move radially outward from the center of the sun. Compare to the values given in Section 32.4 for sunlight just before it enters the earth's atmosphere. (b) The gas pressure at the sun's surface is about \(1.0 \times 10^{4} \mathrm{Pa}\) ; at \(r=R / 2,\) the gas pressure is calculated from solar models to be about \(4.7 \times 10^{33} \mathrm{Pa}\) Comparing with your results in part (a), would you expect that radiation pressure is an important factor in determining the structure of the sun? Why or why not?

Step-by-Step Solution

Answer

Radiation pressure is negligible compared to gas pressure in the sun, hence it's not a significant factor in the structure.

Step 1: Understanding the given data

We begin by identifying the problem's requirements. The sun emits energy at a rate (power) of \(P = 3.9 \times 10^{26} \mathrm{W}\). We need to find the intensity of radiation at the surface (\(r = R = 6.96 \times 10^{8}\) m) and at \(r = R/2\). Intensity \(I\) is defined as \(I = \frac{P}{A}\), where \(A\) is the surface area of a sphere (\(A = 4 \pi r^2\)).

Step 2: Calculate intensity at sun's surface

At the sun's surface (\(r = R\)), we use the formula \(I = \frac{P}{4 \pi R^2}\). Substituting \(R = 6.96 \times 10^{8}\) m and \(P = 3.9 \times 10^{26} \mathrm{W}\), we calculate: \[ I = \frac{3.9 \times 10^{26}}{4 \pi (6.96 \times 10^{8})^2} \approx 6.33 \times 10^7 \, \text{W/m}^2 \]

Step 3: Calculate intensity at r = R/2

At \(r = R/2 = 3.48 \times 10^{8}\) m, the intensity is: \[ I = \frac{P}{4 \pi (R/2)^2} = \frac{3.9 \times 10^{26}}{4 \pi (3.48 \times 10^{8})^2} \approx 2.53 \times 10^8 \, \text{W/m}^2 \]

Step 4: Calculate radiation pressure on absorbing object

Radiation pressure \(p\) is given by \(p = \frac{I}{c}\) for an absorbing object, where \(c\) is the speed of light (\(3 \times 10^8 \, \text{m/s}\)). At the surface, \(p_{surface}\) is:\[ p_{surface} = \frac{6.33 \times 10^7}{3 \times 10^8} \approx 0.211 \, \text{Pa} \]At \(r = R/2\), \(p_{R/2}\) is:\[ p_{R/2} = \frac{2.53 \times 10^8}{3 \times 10^8} \approx 0.843 \, \text{Pa} \]

Step 5: Comparing radiation and gas pressures

At the sun's surface, gas pressure is \(1.0 \times 10^4\) Pa, significantly larger than radiation pressure (0.211 Pa). At \(r = R/2\), gas pressure is \(4.7 \times 10^{33}\) Pa, again much larger than radiation pressure (0.843 Pa).

Step 6: Conclusion on radiation pressure's importance

Radiation pressure is much smaller compared to gas pressure at both the surface and deeper in the sun. Thus, radiation pressure is not a significant factor in determining the sun's structure compared to gas pressure.

Key Concepts

Solar Radiation Pressure

Solar radiation pressure is a direct consequence of the sun emitting energy in the form of electromagnetic waves. When these waves strike an object, like a spacecraft or dust particle, they exert a force. This force is due to the momentum of the photons.

The intensity of solar radiation is pivotal in calculating this force. At the sun's surface, we calculate the radiation pressure using the formula \( p = \frac{I}{c} \), where \( I \) is the intensity and \( c \) is the speed of light. With intensity levels being highest near the sun, the pressure can have substantial effects on particles close to the sun.

However, as distance increases, like Earth, this pressure diminishes, affecting smaller scales like solar sails in spacecraft rather than larger celestial mechanics. For an absorbing object at the sun's surface, radiation pressure is approximately 0.211 Pa, which is rather minimal compared to the pressures found within the sun's interior.

Gas Pressure in the Sun

Gas pressure within the sun is enormous due to the intense gravitational forces compressing the gas and plasma. At the sun's surface, the pressure is about \(1.0 \times 10^4 \) Pa; however, deeper inside at \(r = R/2 \), it's about \(4.7 \times 10^{33} \) Pa. This immense pressure results from the dense plasma, primarily hydrogen and helium, undergoing nuclear fusion.

Gas pressure provides the outward force necessary to counteract the gravitational pull trying to collapse the sun. It's crucial in maintaining the sun's structure, ensuring hydrostatic equilibrium is maintained. Without this balance, the sun would not sustain its current state, ultimately affecting its lifespan and energy production.

Radiation Pressure Significance

While radiation pressure is a fascinating phenomenon, its significance in the sun's overall structural integrity is relatively minor compared to gas pressure.

The calculated radiation pressures of 0.211 Pa at the surface and 0.843 Pa at \(r = R/2\) are negligible if compared to the vast gas pressures. This comparison indicates that radiation pressure does not significantly influence the sun's structure.

However, it's important to note that radiation pressure is still significant in other cosmic scenarios. Outside any strong gravitational or high-pressure environments like the sun, radiation pressure can influence small particles or light-weighted objects, impacting their trajectories and dynamics.

Physics Problem Solving

Physics problem-solving involves breaking down given data and leveraging known formulas to arrive at desired outcomes. With issues like solar radiation and pressure calculations, understanding each quantity and their relationships is crucial.

Start by identifying the requirements and applying relevant equations. For instance, calculating intensity involves understanding power distribution over spherical surfaces. Next, implementing the formula for radiation pressure through known constants like the speed of light helps determine the forces exerted by electromagnetic radiation effectively.

Practicing step-by-step resolutions with calculations and comparisons as seen in this exercise is vital for mastering physics concepts and their applications. Furthermore, comparing calculated values with known data enhances understanding and validates the calculations, ensuring accuracy and deeper learning insight.