Before anyone knew about nuclear energy, physicists asked what keeps the Sun shining. Chemical burning would last only thousands of years. In the 1850s and 1860s Hermann von Helmholtz and Lord Kelvin proposed the best idea then available: slow contraction, releasing gravitational energy, as the virial theorem describes.
Why it works
A star of mass $M$ and radius $R$ has gravitational energy of order $GM^2/R$. By the virial theorem, about half of what contraction releases can be radiated. Divide the available energy by the rate at which it is spent, the luminosity $L$, and you get a time: $$t_{KH} \sim \frac{GM^2}{RL}.$$ For a uniform sphere the radiated energy available from collapsing from far away is $\tfrac{3}{10}GM^2/R$, which shortens the time by about a factor of 3; the order of magnitude is what matters.
Key idea
The Kelvin–Helmholtz time $t_{KH} \approx GM^2/(RL)$ is about 30 million years for the Sun, far shorter than Earth's age of about 4.5 billion years. Gravity cannot be the Sun's main power source, though it does power protostars before fusion begins.
Worked examples
Worked example: The Sun
Find $GM^2/(RL)$ for the Sun.
- $$\frac{GM^2}{R} = \frac{(6.67 \times 10^{-11})(1.99 \times 10^{30})^2}{6.96 \times 10^{8}} = 3.80 \times 10^{41}\text{ J}.$$
- Divide by $L = 3.83 \times 10^{26}$ W: $9.9 \times 10^{14}$ s.
- Divide by $3.16 \times 10^{7}$ s per year.
Answer: About 31 million years.
Worked example: Fusion's supply
The Sun will fuse about 10% of its mass, and fusion turns 0.7% of that mass into energy. How long can that last at today's luminosity?
- $$E = (0.007)(0.1)(1.99 \times 10^{30})(3.00 \times 10^{8})^2$$ which is $1.25 \times 10^{44}$ J.
- $E/L = 1.25 \times 10^{44}/3.83 \times 10^{26} = 3.3 \times 10^{17}$ s.
Answer: About 10 billion years, roughly 300 times the Kelvin–Helmholtz time.
Worked example: How fast would the Sun have to shrink?
If contraction alone powered the Sun (uniform sphere, radiated energy $\tfrac{3}{10}GM^2/R$), how fast would its radius shrink?
- $L = \dfrac{3GM^2}{10R^2}\left|\dfrac{dR}{dt}\right|$, so $\left|\dfrac{dR}{dt}\right| = \dfrac{10LR^2}{3GM^2}$.
- $= 2.3 \times 10^{-6}$ m/s.
Answer: About 74 m per year: too slow to notice in a human lifetime, but fatal over geological time.
Common mistake
Thinking the Sun could shine for billions of years just by shrinking. The gravitational energy is large but not large enough: at the Sun's luminosity it lasts only a few tens of millions of years. That clashed with geology and with Darwin's estimates of the time evolution needed, and radioactive dating later gave Earth an age of about 4.5 billion years. Check: compare $GM^2/(RL)$ with the age of Earth; they differ by a factor of over 100.