ASTR 1040                                                                      Spring, 2005

### ANSWERS - HOMEWORK NO. 3

1.a.  Estimate the hydrogen-burning lifetime of a star of 15 solar masses and a luminosity 10,000 times greater than the solar luminosity.

This is easiest to do by comparing with the Sun, using the relationship t = to(M/L), where t_o is the Sun's lifetime of 10^10 years, and M and L are the mass and luminosity of the star, in solar units.  Then we get t = 10^10(15/10,000) = 1.5 X 10^7 yrs.

b.  For a star having a mass of 0.2 M_o and a luminosity of 0.008 L_o,

Using the same method yields

t = 10^10(0.2/0.008) = 2.5 X 10^11 yrs.

c.  If 10 percent of the star's  original mass is now in the form of helium in the core, calculate the helium-burning lifetime of the 15-solar mass star.

The mass converted to energy is the difference between the sum of the masses of three helium nuclei (i.e. 3 X 4.002603 u = 12.007809 u) and the mass of the resulting carbon nucleus (12.000000 u), which is 0.007809 u. This represents a fractional loss of mass of 0.007809/12.007809 = 0.00065 of the original mass.  Thus 0.00065 of one-tenth of the star's mass of 1.99 X 1030 kg will be converted into energy during triple-a burning, or E = mc2 = 0.1(15)1.99 X 10^30)(0.00065)(3 X 10^8)2 = 1.8 X 10^44 J.

The helium-burning lifetime is t = E/L = 1.8 X 10^44/100(3.83 X 10^26) = 4.6 X 10^15 sec = 1.4 X 10^7 yrs.

2. Suppose that when the Sun becomes a red giant it will increase its radius by a factor of 100 while its surface temperature drops to half of its current value.

a.  What will be the Sun's luminosity at that point, compared to its current luminosity (i.e., expressed as a ratio) and in basic units (i.e., watts)?

This is easiest to do if we take the ratio of the Stefan-Boltzmann equations representing the Sun as a main sequence star (subscript m, below) and as a red giant (subscript g).  Then we get Lg/Lm = (Rg/Rm)^2(Tg/Tm)^4 = (10)^2(1/2)^4 = 625; i.e., as a red giant the Sun will have 625 times its main sequence luminosity.  This is equivalent to L = 2.39 X 10^29 W.

b.  How far away could this red giant be detected with a telescope whose limiting magnitude is mv = +20, if the star's bolometric correction is B.C. = -2.0?

This requires several steps, because we need to know the star's visual absolute magnitude before we can use the distance modulus relation.  First we find the bolometric absolute magnitude and then apply the bolometric correction to find Mv.  The easiest way to get Mbol is to use the magnitude equation to compare bolometric absolute magnitudes and luminosities of the Sun and the red giant: Mbol(m) - Mbol(g) = 2.5log(Lg/Lm), where again g represents the red giant and m the main sequence star (which has the Sun's bolometric absolute magnitude of +4.75).  Solving for Mbol(g), we find Mbol(g) = Mbol(m) - 2.5log((Lg/Lm) = 4.75 - 2.5log(625) = -2.24.

The visual absolute magnitude is Mv = Mbol - B.C. = -2.24 -(-2.0) = -0.24.

Now we can use the distance modulus relation to find the distance to this star if its visual apparent magnitude is mv = 20:

mv - Mv = 5log(d/10) yields d = 10[(m - M)/5] + 1 = 1.1 X 10^5 parsecs.

c.  What will the surface gravity be (expressed as a ratio with the Sun's current surface gravity)?

Surface gravity is proportional to M/R^2; in this case M is constant while R increases by a factor of 100, so the new surface gravity is g = (1/100)^2 = 10^4 times less than it was before expansion to the red giant phase (this low surface gravity helps red giants to lose mass through winds; the outer layers are only very weakly bonded to the star, and escape easily).

d.  What will the escape speed be from the surface of the Sun as a red giant?  Compare with the escape speed now.

Escape speed is given by v_e = sqrt[ 2GM/R]; for M = the solar mass and R = 100 times the solar radius, we get v_e = sqrt[2(6.67 X 10-11)(1.99 X 1030/(100)(6.96 X 108)] = 61,800 m/sec = 61.8 km/sec.

When the Sun becomes a red giant, its radius will increase by a factor of 100; since the escape speed is inversely proportional to the square root of the radius, the escape speed in the red giant phase is sqrt[1/(100)] = 0.1 times as fast as the main sequence escape speed (hence the main sequence escape speed is 10 times the red giant escape speed, or 618 km/sec).

3.  Now suppose that after red giant mass-loss phase the Sun consists of a core (a white dwarf) having a mass equal to half the original main sequence mass, a  radius equal to 0.01 of the original radius, and a surface temperature equal to twice the original temperature of 5780 K.

a.  What are the luminosity and bolometric absolute magnitude of the white dwarf remnant?

Compare with the Sun, using the Stefan-Boltzmann equation:

Lwd/Lm = (Rwd/Rm)^2(Twd/Tm)^4 = (1/100)^2(2)^4 = 0.0016.

In basic units, this means that the white dwarf's luminosity is 0.0016(3.83 X 10^26) = 6.13 X 10^23W.  To find the bolometric absolute magnitude, use the same technique as in Problem 2.b., above:

Mbol(wd) = Mbol(m) - 2.5log(L_wd/Lm) = 4.75 - 2.5log(0.0016) = 11.74.

b.  If the star's bolometric correction is B.C. = -0.4, how far away can this remnant be detected with a telescope whose limiting visual magnitude is +20?  Compare your answer with the distance you found in 2.b., above.

The visual absolute magnitude is Mv = Mbol - B.C. = 11.74 - (-0.4) = 12.11.  Using the distance modulus relation (again, as we did in Problem 2.b.), d = 10[(m - M)/5] + 1 = 10[(20.00 - 12.11)/5] + 1  = 378 parsecs.  This is a much smaller distance than we found in 2.b., showing that white dwarfs cannot be detected as far away as red giants, not surprisingly.

c.  Calculate the average density (converted to units of grams/cm3) and the escape speed for the white dwarf.

Density is given by r = M/V (mass divided by volume).  For a sphere, the volume is V = 4pR^3/3, so the density is r = 3M/4pR^3.  For 0.5 solar mass and 0.01 solar radius, this becomes r = 3(0.5)(1.99 X 10^30)/4p[(0.01)(6.96 X 10^8)]3 = 7.05 X 10^8 kg/m^3 = 7.05 X 10^5 gm/cm^3.

The escape speed is

ve = sqrt[2(6.67 X 10^-11)(0.5)(1.99 X 10^30/(0.01)(6.96 X 10^8)] = 4.37 X 10^6 m/sec = 4370 km/sec.

d.  Calculate the wavelength of maximum emission for the white dwarf.  What kind of telescope would be best for observing it?

To find (lambda)max you need to use Wien's law:  (lambda)max = 0.0029/T.  In this case T = 2 X 5780 = 11,560 K, so

(lambda)max = 0.0029/11560 = 2.51 X 10^-7 m = 2510Å.

This requires an ultraviolet telescope.

4.  A star of initial mass 30 solar masses loses mass at a rate of 5 X 10^-6 solar masses per year during its main sequence lifetime of 3 X 10^6 years.  Then it blows up in a supernova explosion, leaving behind a remnant whose mass is 1.6 solar masses.  The absolute visual magnitude of the supernova at its peak is Mv = -19.0.

a.  What is the star's mass just before it blows up?

The mass is the initial mass of 30 solar masses minus the mass that is lost at a rate of 5 X 10^-6 solar masses per year over a time of 3 X 10^6 years.  Thus the final mass is M = 30 - (5 X 10^-6)(3 X 10^6) = 30 - 15 = 15 solar masses.

b.  If the mass blown off in the explosion travels outward with an average speed  of 2000 km/sec, what is the kinetic energy of the explosion?

Kinetic energy is given by E_kin = mv^2/2.  In this case the mass in motion is 13.4 solar masses (i.e. the final mass of the star minus the 1.6 solar masses that remains behind in the remnant), and the speed is 2000 km/sec (= 2 X 10^6 m/sec), so the kinetic energy is

Ekin = (13.4)(1.99 X 10^30)(2 X 10^6)2/2 = 5.3 X 10^43 J.

c.  If the bolometric correction at peak brightness is B.C. = -1.0, and the duration of the peak is 3 days, how much energy is radiated away during this time?  Compare this with the kinetic energy you found in part a.

To find the total luminous energy radiated during the 3 days (= 2.6 X 10^5 sec), we need to know the luminosity.  We have the visual absolute magnitude and the bolometric correction, so we can get the luminosity by first finding the bolometric absolute magnitude and then comparing with the Sun to get the luminosity in solar units.  The bolometric absolute magnitude is

Mbol = Mv + B.C. = -19.0 + (-1.0) = -20.0.

Using the Sun as a comparison, we get

Mbol(sun) - Mbol(SN) = 2.5log(LSN/LSun), or

(LSN/LSun) = 10[4.75 - (-20)]/2.5 = 7.9 X 10^9,

which means that the luminosity of the supernova is (7.9 X 10^9)(3.83 X 10^26) = 3.0 X 10^36 W.  Now the total energy emitted during the three days is

Elum = Lt = (3.0 X 10^36)(2.6 X 10^5) = 7.8 X 10^41 J.

This is only about 0.015 times the kinetic energy of the explosion.

d.  Assume that the energy released in the form of neutrinos is 100 times greater than the sum of the kinetic plus luminous energy of the explosion.  How does  the total energy (including all three forms) released in the supernova explosion             compare with the total energy the Sun can produce over its entire hydrogen-burning lifetime?

The total energy is

Etot = 100(E_kin + E_lum) = 100(5.3 X 10^43 + 7.8 X 10^41) = 5.4 X 10^45 J.

In its lifetime of t = 10^10 yrs = 3.2 X 10^17 sec, the Sun will produce a total energy of

E = tL = (3.2 X 10^17)(3.83 X 10^26) = 1.2 X 10^44 J.

So the supernova's total energy is about a factor of 50 greater, and is released instantaneously.  (Note:  You could have calculated the Sun's total lifetime energy by using E = mc^2, assuming that 0.007 of the innermost 10 percent of the solar mass is converted to energy; this is a bit more tedious, but produces the same result.)

e.  If the remnant has a radius of 10 km, calculate its average density (expressed in  units of grams/cm3).  Compare your value with those for the white dwarf and  for the Sun before it became a red giant.

The mass is 1.6 solar masses and the radius is 10 km = 1 X 10^4 m.  Hence the density is

r = 3M/4pR^3 = 3(1.6)(1.99 X 10^30)/4p(1 X 10^4)3 = 5.07 X 10^17 kg/m3 =  5.07 X 10^14 gm/cm3.

This is comparable to the density of an atomic nucleus!

f.  If the remnant's surface temperature is 106 K, what is its luminosity?  If its bolometric correction is B.C. = -3.0, how far away can this remnant be detected  by our telescope whose limiting visual apparent magnitude is +20?

We can get the luminosity from the Stefan-Boltzmann equation:

L = 4(pi)R^2(sigma)T^4 = 4(pi)(1 X 10^4)^2(5.67 X 10^-8)(1 X 10^6)^4 = 7.1 X 10^25 W.

In solar units, this is Lns/Lsun = (7.1 X 10^25)/(3.83 X 10^26) = 0.185.  Now we can use the magnitude equation to find the bolometric absolute magnitude of the neutron star:

Mbol(ns) = Mbol(m) - 2.5log(Lns/Lsun) = 4.75 - 2.5log(0.185) = 6.58.

Given the bolometric correction B.C. = -3.0, the visual absolute magnitude is

Mv = Mbol - B.C. = 6.58 - (-3.0) = 9.58.

Finally, using the distance modulus relation, we find the distance

d = 10[m - M]/5 + 1 = 10[20 - 9.58]/5 + 1 = 1210 parsecs.

This is comparable to the distance to which we found white dwarfs to be detectable (in Problem 3.b.), but we don't detect as many neutron stars directly (exactly 1 so far!), because they are far more rare, and they don't stay this hot for long.