Problems

#05

This is a nice challenging problem. The reaction sequence of Fig. 16.5 generates $4.3\times 10^{-12}$ J, including 2 neutrinos. The question is, how many solar neutrinos pass through the Earth every second.

Let's break this down. We know how much luminosity is produced at the Sun. This energy per second. If we divide that luminosity by the amount energy created in each reaction, then we have the number of reactions occurring per second. Ah, but for every reaction, we get 2 neutrinos, so we can find the number of neutrinos created every second (or neutrino generation rate).

But we need to know the number of neutrinos passing through the Earth. If we assume that all the neutrinos created in the core travel out isotropically, or in equal amounts in each direction, then the fraction that intercept the Earth will be the ratio of the circular cross-section of the Earth divided by the total area of the sphere of radius the Earth's orbit.

Let's do the first part. The Sun's luminosity is about $3.86\times 10^{26}$ J/s. The reaction rate is thus $3.86\times 10^{26} / 4.3\times 10^{-12} \approx 10^{38}$ per sec. (Remember the problem asks us to estimate.) We get 2 neutrinos for every reaction, so to order of magnitude, this number is also the neutrino generation rate.

Now what fraction pass through the Earth. The cross-sectional area of the Earth (i.e., the area of the Earth as seen in projection) is $\pi R_E^2$, for $R_E=6400$ km. The spherical area of the Earth's orbit is $4\pi d^2$, where $d$ is 1 AU, or $d=1.5\times 10^8$ km. The ratio of the two areas is about $10^{-10}$. That means that only 1 in 10 billion of all the neutrinos produced in the Sun will pass through the Earth.

Multiplying those two numbers - the neutrino generation rate by the fraction intercepted - give approximately $10^{28}$ neutrinos/second.

#09

This problem amounts to taking the ratios of the two temperatures and then taking the fourth power of that number. The answer is $(4500/ 5500)^4=0.45$, meaning a sunspot is just 45% as bright (per unit area) as the regular photosphere.

#11

The solar escape speed is listed as 618 km/s. The More Precisely 8-1 gives an equation for the speed of particles for a given temperature. The molecular mass of a proton is just one hydrogen mass, so the denominator of that equation is 1. In the corona that is 2 million K, protons move at a speed of $0.157 \sqrt{T} = 220$ km/s. This is considerably less than the escape speed. But the 618 km/s value is the escape speed at the photosphere. Gravity weakens farther from the Sun and so does the escape speed. Recall that $v_{esc} =\sqrt{2GM/r}$. To get the escape speed down by about 3 times (from 618 to 220), the radius must increase by about 9 times, or 9 solar radii out. At this radius the protons can leave the Sun not to return.