SCCC 116
Homework 4 - Due 28 February 2006
Prof. Christina Lacey
R&D: 21-18, 22-7, 22-17, 22-18
P: 22-4, 22-8

R&D 21-18

Why is supernova 1987A so important?

The host galaxy, the Large Magellanic Cloud (LMC), is the nearest galaxy to us; its distance is very well determined. Knowing, the distance to the supernova allowed astronomers to immediately determine its brightness and total energy output. The LMC is also near enough that the supernova was bright and easy to see. Between us and the LMC is very little dust, so the supernova was not obscured. Lastly, because the supernova was the product of a supergiant star exploding, for the first time astronomers had observed a star and knew its basic properties before it exploded.

SN 1987A will continue to be important as astronomers follow the evolution from its progenitor stage through its development as a Type II supernova.

R&D 22-7

Why do you think astronomers were surprised to find planets around a pulsar?

A supernova explosion should completely destroy any planetary system. However, if the star that goes supernova is in a binary system, the effects of the explosion may destroy most, if not all, of the companion. What is left behind could form a disc of material that eventually accretes to form planetary-sized objects in orbit around the pulsar. This theory is still unproved.

R&D 22-17

What makes Cygnus X-1 a good black hole candidate?

Cygnus X-1 is a good black hole candidate because it is a binary system; the black hole candidate's mass has been determined to be in the range of 5 to 10 solar masses. Mass transfer is occurring and produces X-rays from the black hole candidate. The X-rays vary at a rate suggesting the candidate is small, less than 300 km.

R&D 22-18

What evidence is there for black holes much more massive than the Sun?

X-ray images from the Chandra satellite show a number of sources in M82 that are likely candidates for black holes with masses in the range of 100-1000 solar masses. How these are formed is still not understood.

P 22-4

Find luminosity for a neutron star with a radius of 10 km with a temperature of $10^5$ K, $10^7$ K, and $10^9$ K. At what wavelength does each star radiate and where is it located on the H-R diagram for each temperature, respectively?

$L (L_\odot)=R^2 (R_\odot)T^4  (T_\odot)$

$R_{neutron star}=10 km \frac{1 _\odot}{7\times 10^5 km}=1.4\times 10^{-5} R_\odot$

All temperatures would radiate in the X-ray part of the EM spectrum.

T=$10^5$ K: $L=(1.4\times 10^{-5} R_\odot)^2 (1\times 10^5 K)^4(\frac{1 T_\odot}{5800 K})^4=1.6\times 10^{-5} L_\odot$

This luminosity corresponding to T=$10^9$ is too low to be placed on the H-R diagram.

T=$10^7$ K: $L=(1.4\times 10^{-5} R_\odot)^2 (1\times 10^7 K)\frac{1 T_\odot}{5800 K}=1600 L_\odot$

The luminosity corresponding to T=$10^7$ would fit on the H-R diagram, but the temperature is too high.

T=$10^9$ K: $L=(1.4\times 10^{-5} R_\odot)^2 (1\times 10^9 K)\frac{1 T_\odot}{5800 K}=1.6\times 10^{11} L_\odot$

The luminosity corresponding to T=$10^9$ is too hot and too luminous to appear on the H-R diagram.

P 22-8

What is the Schwarzschild radius of black holes of 1 million and 1 billion solar masses?

$R_{Schwarzschild} (km)=M_\odot\times 3 km$

For 1 million $M_\odot$:
$R_{Schwarzschild} (km)=1\times 10^6 M_\odot \times 3 km =3\times 10^6 km$

Compared to the radius of the Sun:
$R_{Schwarzschild}=3\times 10^6 km\frac{1 R_\odot}{7\times 10^5 km}=4.3 R_\odot$
The black hole is roughly 4 times bigger than the Sun.

For 1 billion $M_\odot$:
$R_{Schwarzschild} (km)=1\times 10^9 M_\odot \times 3 km =3\times 10^9 km$

Compared to the radius of the Solar system (100 A.U.):
$R_{Schwarzschild}=3\times 10^9~km\frac{1~A.U.}{1.5\times 10^8~km}=20~A.U.$
The black hole is smaller than the solar system.