Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S2 (Astrophysics: from planets to the cosmos)

For two bodies of masses $M_1,M_2$, the relative Kepler orbit has semimajor axis $a$ and orbital period $P$ satisfying Kepler's third lawHere $a$ is the semimajor axis of the separation vector, not either star's individual barycentric orbit. In conventional stellar units the same relation is approximately $(P/\mathrm{yr}{)}^2=(a/\mathrm{AU}{)}^3/[(M_1+M_2)/M_{\odot }]$, with year, astronomical unit and solar mass as the units.

Equal radial velocity amplitudes imply equal stellar masses. Under the circular, edge-on eclipsing binary interpretation, each star moves at $K=100\mathrm{km}{\mathrm{s}}^{-1}$, and the relative speed is $2K$. ThusUsing $P=5.45$ days givesMore generally, spectroscopic binary amplitudes determine $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi )$ and $(M_1+M_2){\sin }^{3}i=P(K_1+K_2{)}^3(1-e^2{)}^{3/2}/(2\pi G)$. Thus these numerical masses and separation use $e=0$, $\sin i=1$; eclipses alone imply high inclination rather than its exact value.

TODO: the source light curve, its numerical contact phases and the meanings of the four plotted time labels are absent, so unique numerical stellar radii cannot be recovered. The central eclipse contact-radius relations explain what the missing measurements would determine.

If $t_1,\dots ,t_4$ denote the usual first through fourth contacts of a central eclipse, define $T_{14}=t_4-t_1$ and $T_{23}=t_3-t_2$, and let $R_L\ge R_S$ be the two radii. For a circular edge-on orbit,Taking half their sum and difference gives the radii. For short eclipses the familiar approximation is $R_L\simeq v_{\mathrm{rel}}(T_{14}+T_{23})/4$ and $R_S\simeq v_{\mathrm{rel}}(T_{14}-T_{23})/4$, with $v_{\mathrm{rel}}=2K$. At first and fourth contact the discs touch externally; at second and third the smaller disc is just internally tangent to the larger. Noncentral eclipses require the projected impact parameter as well.

Build the star by adding spherical shells. With $m(r)$ the enclosed mass, its gravitational binding energy isFor a nondegenerate extended star, $r(m)<R_{\odot }$ on the interior, soThis is a gravitational binding-energy lower bound for a spherical star. The Kelvin-Helmholtz cooling time is of orderThe displayed energy bound gives $|\Omega |/L_{\odot }>1.6\times 10^7$ years; it supplies a lower bound on that reservoir-to-luminosity ratio. A stellar structure model sets its coefficient. For an ideal monatomic, gas-supported star the virial theorem puts about half the released gravitational energy into heat, leaving half available for radiation; a uniform-density model would give about $9\times 10^6$ years. These contraction estimates are all of order $10^7$ years, much shorter than the nuclear lifetime.

Use the supplied atomic masses consistently. Their mass defect per four hydrogen atoms is $4(1.0078)-4.0026=0.0286$ atomic mass units. The conversion efficiency isThe atomic-mass bookkeeping includes the associated positron-annihilation energy; an exact photon-heating budget also subtracts energy carried away by neutrinos. With the stated ten-percent fuel fraction and constant present solar luminosity, the main sequence lifetime estimate isThe much longer timescale explains why stellar nuclear fusion can power the Sun throughout its long main-sequence history.

A Hertzsprung-Russell diagram plots stellar luminosity upward and effective temperature increasing toward the left. The main sequence runs from hot luminous stars at upper left to cool faint stars at lower right, with the Sun near $T_{\mathrm{eff}}\simeq 5800\mathrm{K}$, $L=L_{\odot }$.

In the evolution of a solar-mass star, core hydrogen fusion initially supplies the luminosity. As core hydrogen is exhausted, hydrogen-shell fusion around a contracting helium core carries the star through the subgiant and red giant phases toward higher luminosity and lower surface temperature. Helium ignition is followed by core helium fusion with a hydrogen-burning shell, placing a solar-mass star in the red clump. The later Asymptotic giant branch has hydrogen- and helium-burning shells around a carbon-oxygen core. Envelope ejection exposes a hot central star of a planetary nebula; residual shell burning ends and the resulting white dwarf radiates stored thermal energy as it cools.

Thermal interstellar dust produces a bright galactic foreground concentrated toward the Milky Way plane, with additional cloud structure. Its emission spectrum differs from the approximately $2.73\mathrm{K}$ blackbody spectrum of the cosmic microwave background, so comparisons across frequencies help separate it. In the sketch below this is the bright equatorial band.

Observer motion produces the cosmic microwave background dipole: to first order in $\beta =v/c$,The direction of motion is hotter and the opposite direction cooler. Its full-sky pattern has one broad positive hemisphere and one broad negative hemisphere; it is not concentrated in the galactic plane.

After subtracting the mean, the motion-induced dipole and galactic foregrounds, intrinsic Cosmic microwave background anisotropy appears as angular hot and cold patches, typically at the $10^{-5}$ fractional scale. These spatial temperature fluctuations encode early density and velocity structure. The displayed patch pattern is illustrative, with no claim to reproduce an observed sky map.

Measure the absolute spectral radiance over a range of frequencies and fit the Planck spectrumFor example, the COBE FIRAS blackbody spectrum experiment used a differential Michelson interferometer to compare the sky with an internal blackbody; an external calibrated blackbody could cover the sky entrance. Cold bolometers measured the difference spectrum, and foreground separation distinguished Milky Way emission from the nearly isotropic background. The fitted background temperature is approximately $\boxed{T_0=2.73\mathrm{K}}$.^[ref] The result relies on an absolute spectral comparison, rather than merely measuring a map's hot-to-cold contrast.

The stated one-sided temperature contrast is $\beta =0.00105$, so the Doppler dipole givesThe motion is toward the hotter region. The contrast already gives one side relative to the mean; the hot-to-cold difference would be twice as large.

Adiabatic expansion preserves the blackbody form while each photon's energy redshifts as $a^{-1}$. Thus the cosmic microwave background temperature obeys $T(z)=T_0(1+z)$, giving

Convert both inputs to energy density. The matter contribution, including rest energy, is $u_{M,0}={\rho }_{M,0}{c}^2=2.25\times 10^{-10}\mathrm{J}{\mathrm{m}}^{-3}$. The given photon value is $u_{\gamma ,0}=5\times 10^5(1.602\times 10^{-19})=8.01\times 10^{-14}\mathrm{J}{\mathrm{m}}^{-3}$. HenceMatter energy density scales as $(1+z{)}^3$, whereas the cosmic microwave background scales as $(1+z{)}^4$. The photon-only matter-radiation equality condition is thereforeThis uses the supplied CMB photon density. Additional relativistic species change equality for total radiation.