Answer
See answers.
Work Step by Step
a. The sun’s power is distributed uniformly over the surface of a sphere. Consider a radius equal to the orbital radius of the Earth. Multiply the sun’s intensity by this surface area to find the sun’s power.
$$(1300W/m^2)4 \pi (1.496\times10^{11}m)^2\approx 3.7\times10^{26}W$$
b. Equation 31–7 tells us that the reaction releases 26.2 MeV for every 4 protons consumed. Calculate the number of protons consumed per second.
$$(3.656\times10^{26}J/s)\frac{4\;protons}{26.2MeV}\frac{1MeV}{1.6\times10^{-13}J}\approx 3.5\times10^{38}protons/s$$
c. Convert the Sun’s mass to a number of protons. Finally, use the number of protons consumed per second to estimate the Sun’s lifetime.
$$(2.0\times10^{30}kg)\frac{proton}{1.673\times10^{-27}kg}\frac{1s}{3.489\times10^{38}protons}\frac{yr}{3.156\times10^7 s}$$
$$\approx 1.1\times10^{11}yr$$
