Problems

#01

Must be referring to section 16.2 and table 16.2. Let's first figure out what mass we are talking about. The sun has a total mass of about $2\times 10^30$ kg. Ten percent of this will under go fusion to make helium, or $2\times 10^29$ kg. Of this 0.7% (see pg 411 of the text) actually goes into making energy, or $m=1.4\times 10^27$ kg. The total energy generated is $E=mc^2 = 1.3\times 10^{44}$ Joules. Is this reasonable? The Sun will live for about 10 billion years, or about $10^{17}$ sec. Its luminosity is about $10^{26}$ Joules/sec, so the total energy output over this period is the product, or about $10^{43}$ Joules. Close enough, given that we dropped the constants!

#05

At 25 million K, the difference is a factor of 100. At abundances of 1/10th solar, the factor should be 1/10th as large, or a factor of just 10 instead of 100.

#13

The Sun will live for 10 billion years on the main sequence, and the luminosity $L$ of a star on the main sequence depends on the 4th power of a star's mass, or $M^4$. Now the lifetime also depends on mass, with $t \propto M/L$. Since $L\propto M^4$, this means the main sequence lifetime goes as $t\propto M/M^4 = 1/M^3$. So as the mass goes up, the lifetime decreases as the cube of mass. So the Sun is $1M_\odot$ and will live for $10^{10}$ years, hence the lifetime $t$ for any other star will be $t = 10^{10} \times (M/M_\odot)^{-3}$ years. This makes sense, since more massive stars are expected to live shorter lives.

The problem is giving us cluster ages and asking us to find the mass of the star just leaving the main sequence. So we have to turn the equation around to find mass in terms of age. This means taking the quarter power of both sides of the equation, giving

$$\frac{M}{M_\odot} = \sqrt[3]{\frac{10^{10}}{t}}$$

(a) For 400 million years, the mass will be $M/M_\odot = \sqrt[3]{10^{10}/4\times10^8} \approx 3$, or 3 solar masses.

(b) Now $M/M_\odot = \sqrt[3]{10^{10}/4\times10^9} = 1.7$, or 1.7 solar masses.