 Solution

Hydrostatic balance is $dP/dr=-GM(r)\rho /r^2$. A free-fall estimate gives $t_{\mathrm{dyn}}\sim \sqrt{R_{\odot }^3/(G{M}_{\odot })}\simeq 1.6\times 10^3$ s; a test particle falling in a fixed point-mass field adds factor $\pi /(2\sqrt{2})$, giving about 30 minutes.

For an ideal nonrelativistic gas with negligible boundary pressure, the virial theorem is $2U+\Omega =0$, so total energy is $\Omega /2$. Since $\Omega =-{\int }_0^{M}Gmdm/r(m)$ and $r\le R$, $|\Omega |\ge G{M}^2/(2R)$. This is a minimum magnitude, not a lower bound on the negative energy: unrestricted central concentration can make the latter arbitrarily negative. A uniform-density estimate gives $\Omega =-3G{M}^2/(5R)$ and available contraction energy $3G{M}^2/(10R)$. Its release at present solar luminosity lasts $\boxed{t\sim 10^7\mathrm{yr}}$ (order $10^7$–$10^8$ years for alternative structural estimates), much less than a nuclear-burning lifetime.