Blog #13 - Kelvin-Helmholtz Timescale

Kelvin-Helmholtz Timescale

We know that the Sun started from the gravitational collapse of a giant cloud of gas. Let's hypothesize that the Sun is powered solely by this gravitational contraction, as was once posited by astronomers long ago. As it shrinks, its internal thermal energy increases, increasing its temperature due to the virial theorem and thereby causing it to radiate. How long would the Sun last if it was thermally radiating its current power output, \(L_{\odot} = 4*10^{33} erg \space s^{-1}\)? This is known as the Kelvin-Helmholtz timescale. How does this timescale compare to the age of the oldest Moon rocks (about 4.5 billion years old, also known as Gyr)?

We begin with the formula for internal energy derived earlier in the worksheet \[U = -\frac{3GM^2}{5R}\] For this problem we know that the M and R are the mass and radius of the Sun, respectively. We solve the problem with \[M_{Sun} = 2*10^{30} kg \space \space R_{Sun} = 7*10^8 m \]Which gives us \[U_{Sun} = 2.27*10^{41} Joules\] Using the conversion of joules to ergs as \[1 Joule = 10^7 ergs\]We get \[U_{Sun} = 2.27*10^{48} erg\]We know that luminosity is in a scale of \([\frac{erg}{s}] \) meaning we need to divide our \(U_{Sun}\) by its luminosity in order to get the Kelvin-Helmholtz Timescale. \[KHT = \frac{U}{L} = \frac{2.27*10^{48} erg}{4*10^{33} \frac{erg}{s}} = 5*10^{14} seconds\]Converting from seconds to years we get \[KHT \approx 16 \space million \space years\]We're told that the oldest moon rocks are 4.5 billion years old, much older than the expected age of the Sun, thus there is a flaw in the assumptions and astronomers were forced to reassess how the Sun's energy is created.