Chapter 15--Our Sun - Geological Sciences

Chapter 15--Our Sun - Geological Sciences Chapter 15--Our Sun - Geological Sciences

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large decrease in rate of fusion slight decrease in core temperature slight rise in core temperature large rise in rate of fusion Solar Thermostat: Gravitational Equilibrium Because the energy supply is diminished, gravity starts to overcome thermal pressure. Increased energy output enables thermal pressure to overcome gravity. Gravity compresses the core, heats it up, and restores fusion rate to normal value. Increased thermal pressure causes the core to expand and then cool, which restores fusion rate to normal value. Figure 15.8 The solar thermostat. Gravitational equilibrium regulates the Sun’s core temperature. Everything is in balance if the amount of energy leaving the core equals the amount of energy produced by fusion. A rise in core temperature triggers a chain of events that causes the core to expand, lowering its temperature to its normal value. A decrease in core temperature triggers the opposite chain of events, also restoring the normal core temperature. Sun, observations of “sun quakes,” and observations of solar neutrinos. Mathematical Models The primary way we learn about the interior of the Sun and other stars is by creating mathematical models that use the laws of physics to predict the internal conditions. A basic model uses the Sun’s observed composition and mass as inputs to equations that describe gravitational equilibrium, the solar thermostat, and the rate at which solar energy moves from the core to the photosphere. With the aid of a computer, we can use the model to calculate the Sun’s temperature, pressure, and density at any depth. We can then predict the rate of nuclear fusion in the solar core by combining these calculations with knowledge about nuclear fusion gathered in laboratories here on Earth. Remarkably, such models correctly “predict” the radius, surface temperature, luminosity, age, and many other properties of the Sun. However, current models do not predict everything about the Sun correctly. Scientists are constantly working to discover what is missing from them. Successful prediction of so many observed characteristics of the Sun gives us confidence that the models are on the right track and that we really do understand what is going on inside the Sun. Sun Quakes A second way to learn about the inside of the Sun is to observe “sun quakes”—vibrations of the Sun that are similar to the vibrations of the Earth caused by earthquakes, although they are generated very differently. Earthquakes occur when Earth’s crust suddenly shifts, generating seismic waves that propagate through Earth’s interior [Section 10.2].We can learn about Earth’s interior by recording seismic waves on Earth’s surface with seismographs. Sun quakes result from waves of pressure (sound waves) that propagate deep within the Sun at all times. These waves cause the solar surface to vibrate when they reach it. Although we cannot set up seismographs on the Sun, we can detect the vibrations of the surface by measuring Doppler shifts [Section 6.5]. Light from portions of the surface that are rising toward us is slightly blueshifted, while light from portions that are falling away from us is slightly redshifted. The vibrations are relatively small but measurable (Figure 15.9). 504 part V • Stellar Alchemy

Figure 15.9 Vibrations on the surface of the Sun can be detected by Doppler shifts. In this schematic representation, red indicates falling gas, and blue indicates rising gas. The speckled region indicates the convection zone. The vibration pattern illustrated here is just one of many possible patterns. The overall vibration pattern of the Sun is a complex combination of patterns similar to this one. In principle, we can deduce a great deal about the solar interior by carefully analyzing these vibrations. (By analogy to seismology on Earth, this type of study of the Sun is called helioseismology—helios means “sun.”) Results to date confirm that our mathematical models of the solar interior are on the right track (Figure 15.10). At the same time, they provide data that can be used to improve the models further. Mathematical Insight 15.1 Mass-Energy Conversion in the Sun We can calculate how much mass the Sun loses through nuclear fusion by comparing the input and output masses of the proton– proton chain. A single proton has a mass of 1.6726 10 27 kg, so four protons have a mass of 6.690 10 27 kg. A helium-4 nucleus has a mass of only 6.643 10 27 kg, slightly less than the mass of the four protons. The difference is: 6.690 10 27 kg 6.643 10 27 kg 4.7 10 29 kg which is 0.7%, or 0.007, of the original mass. Thus, for example, when 1 kilogram of hydrogen fuses, the resulting helium weighs only 993 grams, while 7 grams of mass turns into energy. To calculate the total amount of mass converted to energy in the Sun each second, we use Einstein’s equation E mc 2 .The total energy produced by the Sun each second is 3.8 10 26 joules, so we can solve for the total mass converted to energy each second: E mc 2 E ⇒ m c 2 3.8 10 26 joules 3.0 10 8 m s 2 4.2 10 9 kg The Sun loses about 4 billion kilograms of mass every second, which is roughly equivalent to the combined mass of nearly 100 million people. Example: How much hydrogen is converted to helium each second in the Sun? Solution: We have already calculated that the Sun loses 4.2 10 9 kg of mass each second and that this is only 0.7% of the mass of hydrogen that is fused: 4.2 10 9 kg 0.007 mass of hydrogen fused We now solve for the mass of hydrogen fused: mass of hydrogen fused 4.2 10 9 kg 0.007 6.0 10 11 kg 1m etric ton 103 kg 6.0 10 8 metric tons The Sun fuses 600 million metric tons of hydrogen each second, of which about 4 million tons becomes energy. The remaining 596 million tons becomes helium. chapter 15Our Star 505

Figure <strong>15</strong>.9 Vibrations on the surface of<br />

the <strong>Sun</strong> can be detected by Doppler shifts.<br />

In this schematic representation, red indicates<br />

falling gas, and blue indicates rising gas.<br />

The speckled region indicates the convection<br />

zone. The vibration pattern illustrated<br />

here is just one of many possible patterns.<br />

The overall vibration pattern of the <strong>Sun</strong> is a<br />

complex combination of patterns similar to<br />

this one.<br />

In principle, we can deduce a great deal about the<br />

solar interior by carefully analyzing these vibrations. (By<br />

analogy to seismology on Earth, this type of study of the<br />

<strong>Sun</strong> is called helioseismology—helios means “sun.”) Results<br />

to date confirm that our mathematical models of the<br />

solar interior are on the right track (Figure <strong>15</strong>.10). At the<br />

same time, they provide data that can be used to improve<br />

the models further.<br />

Mathematical Insight <strong>15</strong>.1 Mass-Energy Conversion in the <strong>Sun</strong><br />

We can calculate how much mass the <strong>Sun</strong> loses through nuclear<br />

fusion by comparing the input and output masses of the proton–<br />

proton chain. A single proton has a mass of 1.6726 10 27 kg,<br />

so four protons have a mass of 6.690 10 27 kg.<br />

A helium-4 nucleus has a mass of only 6.643 10 27 kg,<br />

slightly less than the mass of the four protons. The difference is:<br />

6.690 10 27 kg 6.643 10 27 kg 4.7 10 29 kg<br />

which is 0.7%, or 0.007, of the original mass. Thus, for example,<br />

when 1 kilogram of hydrogen fuses, the resulting helium weighs<br />

only 993 grams, while 7 grams of mass turns into energy.<br />

To calculate the total amount of mass converted to energy<br />

in the <strong>Sun</strong> each second, we use Einstein’s equation E mc 2 .The<br />

total energy produced by the <strong>Sun</strong> each second is 3.8 10 26 joules,<br />

so we can solve for the total mass converted to energy each second:<br />

E mc 2 E<br />

⇒ m c 2<br />

<br />

3.8 10 26 joules<br />

<br />

<br />

3.0 10 8 m s 2<br />

4.2 10 9 kg<br />

The <strong>Sun</strong> loses about 4 billion kilograms of mass every second,<br />

which is roughly equivalent to the combined mass of nearly<br />

100 million people.<br />

Example: How much hydrogen is converted to helium each<br />

second in the <strong>Sun</strong>?<br />

Solution: We have already calculated that the <strong>Sun</strong> loses 4.2 10 9 kg<br />

of mass each second and that this is only 0.7% of the mass of hydrogen<br />

that is fused:<br />

4.2 10 9 kg 0.007 mass of hydrogen fused<br />

We now solve for the mass of hydrogen fused:<br />

mass of<br />

hydrogen fused 4.2 10<br />

9 kg<br />

<br />

0.007<br />

6.0 10 11 kg 1m etric<br />

ton<br />

103<br />

kg<br />

6.0 10 8 metric tons<br />

The <strong>Sun</strong> fuses 600 million metric tons of hydrogen each second,<br />

of which about 4 million tons becomes energy. The remaining<br />

596 million tons becomes helium.<br />

chapter <strong>15</strong> • <strong>Our</strong> Star 505

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