The Sun's hydrogen consumption

Problem

The Sun radiates a power $L \approx 3{,}8 \cdot 10^{26}\ \text{W}$ produced by the proton-proton cycle, which converts hydrogen into helium according to the scheme $4\,^{1}\text{H} \to\ ^{4}\text{He}$, releasing energy $Q \approx 26\ \text{MeV}$ per cycle. Estimate (a) the kilograms of hydrogen consumed every second and (b) in how many years the Sun will exhaust the hydrogen present in its core, knowing that $M_{\text{core}} \approx 10\%$ of $M_{\odot} \approx 2 \cdot 10^{30}\ \text{kg}$. Use $1\ \text{MeV} = 1{,}6 \cdot 10^{-13}\ \text{J}$, $m_p \approx 1{,}67 \cdot 10^{-27}\ \text{kg}$, $1\ \text{year} \approx 3{,}156 \cdot 10^{7}\ \text{s}$.

Solution

Energy per cycle. Each cycle consumes 4 protons and releases $Q = 26\ \text{MeV}$:

$Q = 26 \cdot 1{,}6 \cdot 10^{-13}\ \text{J} = 4{,}16 \cdot 10^{-12}\ \text{J}$

(a) Hydrogen consumed per second. The number of cycles per second is obtained by dividing the power by the energy of one cycle:

$\frac{\text{cycles}}{\text{s}} = \frac{L}{Q} = \frac{3{,}8 \cdot 10^{26}}{4{,}16 \cdot 10^{-12}} \approx 9{,}1 \cdot 10^{37}\ \text{s}^{-1}$

Each cycle burns 4 protons, each of mass $m_p$, so the mass of hydrogen consumed per second is:

$\frac{\dd m}{\dd t} = 9{,}1 \cdot 10^{37} \cdot 4 \cdot 1{,}67 \cdot 10^{-27}\ \text{kg} \approx 6 \cdot 10^{11}\ \text{kg/s}$

$\ev{\frac{\dd m}{\dd t} \approx 6 \cdot 10^{11}\ \text{kg/s} \sim 10^{12}\ \text{kg/s}}$

(b) Time to exhaustion. The mass of hydrogen available in the core is:

$M_{\text{core}} = 0{,}1 \cdot 2 \cdot 10^{30} = 2 \cdot 10^{29}\ \text{kg}$

The time is obtained by dividing by the rate of consumption:

$t = \frac{M_{\text{core}}}{\dd m/\dd t} = \frac{2 \cdot 10^{29}}{6 \cdot 10^{11}} \approx 3{,}3 \cdot 10^{17}\ \text{s}$

Converting to years:

$t = \frac{3{,}3 \cdot 10^{17}}{3{,}156 \cdot 10^{7}} \approx 1 \cdot 10^{10}\ \text{years} \quad\Rightarrow\quad \ev{t \sim 10^{10}\ \text{years}}$

The result is of the same order of magnitude as the estimated remaining lifetime of the Sun ($\sim 5$ billion years): the Fermi estimate correctly captures the timescale.

Connections

Topics: Nuclear physics Concepts: Nuclear fusion · Mass-energy equivalence Skills: Fermi estimates