This problem is based on problems from Carroll & Ostlie and uses data from Chabrier et al.(2000, ApJ, 542, 464) and Schaller et al. (1992, AASS, 96, 269). The lower end of the main= 3.23 and log1o(L/Lo) = -4.3. On the othersequence occurs near 0.072M0, with log10 Teffhand, an 85MO star near the upper end of the main sequence has an effective temperatureand luminosity of log10 Teff = 4.705 and log10(L/Lo) = 6.006, respectively. Assume that the0.072M. star is entirely convective so that, through convective mixing, all of its hydrogen,rather than just the inner 10%, becomes available for burning.(a)ends of the main sequence. If the less massive star is Star A and the more massive star isStar B, what is the ratio tA/tB of their lifetimes? How do these lifetimes compare to theEstimate the hydrogen-burning lifetimes of stars near the lower and upperhydrogen-burning lifetime of the Sun?(b)RB/RA of their radii?Calculate the radii of both stars in units of solar radii. What is the ratio

The expression for the life-time be defined as, tMS=EhydrogenL=0.007fXMc2L Where, f be defined as the fraction of hydrogen fused into helium. X be defined as the mass fraction of the hydrogen. L be defined as luminosity. M be defined as the mass of the sun c be defined as the speed of light.The luminosity for the upper and lower end calculated as, Upper End:log10LL⊙=6.006LL⊙=106.006L=1.01×106L⊙L=1.01×1063.846×1026 W=3.88×1032 WLower End:log10LL⊙=-4.3LL⊙=10-4.3L=5.01×10-53.846×1026 WL=19.27×1021 WTherefore, for lower-end, the time period be calculated as, tMSupper=0.007fXMc2L=0.00710.790M⊙3.0×108 m.s-12Lupper=0.00710.7901.989×1030 kg3.0×108 m.s-123.88×1032 W=7.89×10463.88×1032=2.033×1014 s1 Yr3.154×107 s=6.45×106 YrtMSlower=0.007fXMc2L=0.0070.10.70.072M⊙3.0×108 m.s-12Lupper=0.0070.10.70.0721.989×1030 kg3.0×108 m.s-1219.27×1021 W=0.00063154×104619.27×1021=3.277×1015 s1 Yr3.154×107 s=1.04×108 YrThe ratio of the time period of the less massive star to the more massive star be calculated as, tAtB=1.04×108 Yr6.45×106 Yr=0.1612×102 Yr=16.12 Yr The life-span of the sun began roughly 4.6 billion years ago and will continue for another 4.5-5.5 billion years. Therefore, it will be about, 10.1 billion years.(b) The luminosity of the star in terms of radius and temperature be expressed as, L=4πR2σTe4 Where, L be defined as luminosity. R be defined as the radius of a star. Te is temperature. σ be defined as the Stefan-Boltzmann constant.Therefore, the radii of lower end star are calculated as, L=4πR2σTe4LL⊙=4πR2σTe44πR2σT⊙e4R2R⊙2=LL⊙T⊙4T⊙e4=5.01×10-51.11×10158.31×1012=66.920R=8.18R⊙ The radii of the upper end are calculated as, L=4πR2σTe4LL⊙=4πR2σTe44πR2σT⊙e4R2R⊙2=LL⊙T⊙4T⊙e4=1.01×1061.11×10156.60×1018=169.8R=13.03R⊙The ratio of their radii be calculated as, RBRA=RupperRlowerRBRA=13.03R⊙8.18R⊙RBRA=1.59

Answered: This problem is based on problems from…