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Chapter 5: Problem 7

The life expectancy \(E\) of a main-sequence star \({ }^{12}\) depends on its mass \(M\). The relation is given by $$ E=M^{-2.5}, $$ where \(M\) is solar masses and \(E\) is solar lifetimes. The sun is thought to be at the middle of its life, with a total life expectancy of about 10 billion years. Thus the value \(E=1\) corresponds to a life expectancy of 10 billion years. a. Does a more massive star have a longer or a shorter life expectancy than a less massive star? b. Spica is a main-sequence star that is about \(7.3\) solar masses. What is the life expectancy of Spica? c. Express using functional notation the life expectancy of a main-sequence star with mass equal to \(0.5\) solar mass, and then calculate that value. d. Vega is a main-sequence star that is expected to live about \(6.36\) billion years. What is the mass of Vega? e. If one main-sequence star is twice as massive as another, how do their life expectancies compare?

Short Answer

Expert verified
a. More massive stars have shorter life expectancies. b. Spica has a life expectancy of 0.123 billion years. c. 0.5 solar mass star: 56.6 billion years. d. Vega's mass is 1.73 solar masses. e. Twice as massive stars have about 17.7% the life expectancy.

Step by step solution

01

Understanding the Relationship

The life expectancy, \(E\), of a star is inversely related to its mass, \(M\), raised to the power of \(-2.5\) as given by the equation:\[ E = M^{-2.5} \]This means increasing the mass will decrease the life expectancy and vice versa.
02

Evaluate Life Expectancy for Spica

Given Spica has a mass of \(7.3\) solar masses, substitute \(M = 7.3\) into the equation:\[ E = 7.3^{-2.5} \]Calculate to find \(E \approx 0.0123\). Thus, Spica's life expectancy is approximately \(0.0123\) solar lifetimes, which is \(0.0123 \times 10\) billion = \(0.123\) billion years.
03

Determine Life Expectancy for 0.5 Solar Mass

Using the functional notation, express the life expectancy when \(M = 0.5\) as \(E(0.5) = 0.5^{-2.5}\).Calculate to find \(E(0.5) \approx 5.66\). Thus, a star with \(0.5\) solar mass has a life expectancy of \(5.66\) solar lifetimes, or \(5.66 \times 10\) billion = \(56.6\) billion years.
04

Find the Mass of Vega

For Vega, we know \(E = 6.36\) billion years, which is \(0.636\) solar lifetimes since \(E = 1\) corresponds to \(10\) billion years.Set up the equation: \(0.636 = M^{-2.5}\) and solve for \(M\) by taking both sides of the equation to the power of \(-\frac{1}{2.5}\) (i.e., -0.4):\[ M = 0.636^{-0.4} \]Calculating gives \(M \approx 1.73\). Thus, Vega has a mass of approximately \(1.73\) solar masses.
05

Compare Life Expectancies for Stars with Different Masses

If one star is twice as massive as another star, say with masses \(M_1 = M\) and \(M_2 = 2M\), compare their life expectancies:\[ E_1 = M^{-2.5} \]\[ E_2 = (2M)^{-2.5} = \frac{1}{2^{2.5}} M^{-2.5} = \frac{1}{5.657} E_1 \]Thus, if one star is twice as massive as another, its life expectancy is approximately \(\frac{1}{5.657}\) or \(17.7\%\) of the less massive star's life expectancy.

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Key Concepts

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

Exponential Functions
Exponential functions are mathematical expressions in which a variable appears in the exponent. This means the function grows rapidly as the input value increases or decreases. In our case, the function describing a star's life expectancy is an exponential function because the mass of the star is raised to the power of -2.5. This particular exponent means the function decreases sharply with increasing mass, illustrating how small changes in mass can lead to significant variations in expected life span.
  • An exponent of -2.5 means that the function shows inverse exponential decay.
  • Exponential functions are essential in various fields like finance, biology, and physics, where they model growth, decay, or other rapid changes.
Understanding the nature of exponential behavior helps predict how quickly a star will deplete its nuclear fuel based on its mass.
Inverse Proportionality
Inverse proportionality describes a relationship between two variables in which their product is constant. If one variable increases, the other decreases so that their product remains unchanged. In this exercise, the life expectancy \(E\) of a star is inversely proportional to its mass \(M\) to the -2.5 power (\(E = M^{-2.5}\)).
  • As the mass increases, the life expectancy decreases.
  • This property is evident in massive stars that burn out quickly, having shorter life spans compared to smaller, longer-living stars.
Inverse proportionality is a fundamental concept in physics, where it helps to understand phenomena like the decrease in gravitational force with increased distance.
Astrophysics Models
Astrophysics models help scientists understand complex relationships between various cosmic entities. These models use mathematical equations to simulate the behavior and properties of objects in space based on theoretical and observational data.
Specific to stars, models like the one in this exercise relate the life expectancy of a star to its mass. They reveal insights into the mechanisms governing star life cycles.
  • Mass determines a star's energy production and longevity.
  • Such models help astrophysicists predict star behaviors, lifecycles, and eventual fate.
  • Understanding these dynamics assists in broader cosmological theories and improves predictions about stellar populations in galaxies.
By incorporating scientific principles and data, astrophysics models become crucial tools in unraveling the mysteries of the universe.
Scientific Notation
Scientific notation is a method for expressing very large or very small numbers in a compact form. It simplifies calculations and readability by representing numbers as a product of a decimal and a power of ten. For instance, instead of writing 10 billion, we write it as \(1 \times 10^{10}\). This is helpful in scientific fields like astrophysics where numbers can be extremely large or small.
In this exercise, the life expectancy of stars is often expressed in terms of solar lifetimes (e.g., 10 billion years as one solar lifetime).
  • Scientific notation uses powers of ten to represent magnitude efficiently.
  • It is a universal language in science that enables easier communication and computation.
Learning to use scientific notation helps in managing vast or infinitesimal quantities common in fields such as astronomy and chemistry.

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Most popular questions from this chapter

A rock is thrown downward, and the distance \(D\), in feet, that it falls in \(t\) seconds is given by \(D=16 t^{2}+3 t\). Find how long it takes for the rock to fall 400 feet by using a. the quadratic formula. b. the crossing-graphs method.

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When a road is being built, it usually has straight sections, all with the same grade, that must be linked to each other by curves. (By this we mean curves up and down rather than side to side, which would be another matter.) It's important that as the road changes from one grade to another, the rate of change of grade between the two be constant. \({ }^{56}\) The curve linking one grade to another grade is called a vertical curve. Surveyors mark distances by means of stations that are 100 feet apart. To link a straight grade of \(g_{1}\) to a straight grade of \(g_{2}\), the elevations of the stations are given by $$ y=\frac{g_{2}-g_{1}}{2 L} x^{2}+g_{1} x+E-\frac{g_{1} L}{2} . $$ Here \(y\) is the elevation of the vertical curve in feet, \(g_{1}\) and \(g_{2}\) are percents, \(L\) is the length of the vertical curve in hundreds of feet, \(x\) is the number of the station, and \(E\) is the elevation in feet of the intersection where the two grades would meet. (See Figure 5.105.) The station \(x=0\) is the very beginning of the vertical curve, so the station \(x=0\) lies where the straight section with grade \(g_{1}\) meets the vertical curve. The last station of the vertical curve is \(x=L\), which lies where the vertical curve meets the straight section with grade \(g_{2}\). Assume that the vertical curve you want to design goes over a slight rise, joining a straight section of grade \(1.35 \%\) to a straight section of grade \(-1.75 \%\). Assume that the length of the curve is to be 500 feet (so \(L=5\) ) and that the elevation of the intersection is \(1040.63\) feet. a. What phrase in the first paragraph of this exercise assures you that a quadratic model is appropriate? b. What is the equation for the vertical curve described above? Don't round the coefficients. c. What are the elevations of the stations for the vertical curve? d. Where is the highest point of the road on the vertical curve? (Give the distance along the vertical curve and the elevation.)

Table \(5.3\) gives the length \(L\), in inches, of a flying animal and its maximum speed \(F\), in feet per second, when it flies. \({ }^{27}\) (For comparison, 10 feet per second is about \(6.8\) miles per hour.) $$ \begin{array}{|l|c|c|} \hline \text { Animal } & \text { Length } L & \begin{array}{c} \text { Flying speed } \\ F \end{array} \\ \hline \text { Fruit fly } & 0.08 & 6.2 \\ \hline \text { Horse fly } & 0.51 & 21.7 \\ \hline \begin{array}{l} \text { Ruby-throated } \\ \text { hummingbird } \end{array} & 3.2 & 36.7 \\ \hline \text { Willow warbler } & 4.3 & 39.4 \\ \hline \text { Flying fish } & 13 & 51.2 \\ \hline \text { Bewick's swan } & 47 & 61.7 \\ \hline \text { White pelican } & 62 & 74.8 \\ \hline \end{array} $$ a. Judging on the basis of this table, is it generally true that larger animals fly faster? b. Find a formula that models \(F\) as a power function of \(L\). c. Make the graph of the function in part b. d. Is the graph you found in part c concave up or concave down? Explain in practical terms what your answer means. e. If one bird is 10 times longer than another, how much faster would you expect it to fly? (Use the homogeneity property of power functions.)
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