Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2016 / S2 / 2

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.

Using the supplied atomic masses consistently includes the net electron bookkeeping after positron annihilation. The released fraction is $\eta =(4m_H-m_{He})/(4m_H)\simeq 0.0071$, corresponding to about 26.7 MeV per helium nucleus. If 10% of the solar mass is processed and luminosity stays constant, $t\simeq 0.1\eta M_{\odot }{c}^2/L_{\odot }\simeq 10^{10}$ years.

At separation one femtometre the proton Coulomb energy is 1.44 MeV. At $2\times 10^7$ K, mean thermal kinetic energy $3k_{B}T/2\simeq 2.59$ keV, making their ratio about $5.6\times 10^2$. Classical over-barrier collisions are therefore extremely improbable. Quantum tunnelling through the Coulomb barrier and the high-energy distribution tail allow fusion, with the slow weak-interaction step regulating the proton-proton chain. The reaction summary is a net energy/matter balance, not one simultaneous four-proton collision.