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Observations of Supernova 1987a
R. D. Gehrz
University of MInnesota
Theories Confirmed by Observations of
Supernova 1987a
• Neutrino pulse: right size and duration
• Infrared line emission and visual light curve decay
rate: reveal formation of heavy elements
• Spectrum of precursor star: blue supergiant
• Thermal IR emission: formation of dust grains
• Light echoes: illuminated shells ejected by the
precursor star and also illuminated a “phantom”
nebula caused by reflections of the explosion off
interstellar clouds
• Central remnant: Is a pulsar or black hole left
over in the center of the system?
The Supernova Explosion
• The core collapses until the neutrons touch: The collapse
stops because of neutron degeneracy pressure
• A shock wave rebounds through the outer layers of the star:
some of the neutrons fly out through the ejecta and make
heavy elements by 56Fe + Nn  56+NX
• The core continues to collapse: it either halts as a stable
neutron star or becomes a black hole depending upon the
mass remaining in the core after the rebound
Core Collapse of a Massive Star
• Formation of the iron core: nuclei lighter than iron give off energy when they are formed
by fusion, nuclei heavier than iron absorb energy when they are formed by fusion.
Iron is on the dividing line between these processes.
• Photo-disintegration of iron and helium:  + 56Fe  134He + 4n
and  + 4He  4n
• Neutronization: p+ + e-  n + 
• Formation of a neutron star or a black hole:
A neutron star forms if Mc < 1.44 Solar Masses (the Chandrasekhar mass
limit)
A black hole forms if M exceeds Mc RBlack Hole= 2GM/c2
The Escape Velocity Derivation
When an object is thrown up into the air there is a battle between the object's
kinetic energy and the gravitational pull on the object (its gravitational potential
energy). If our object is to escape the gravitational pull of the Earth, its kinetic
energy must balance the gravitational force acting upon it.
The kinetic energy = ½ mv2, where m is the mass of the object and v is its
velocity.
The gravitational potential energy of an object = GmM/r, where G is the
gravitational constant, m is the mass of the object, M is the mass of the Earth and
r is the Earth's radius.
For a balance between kinetic energy and gravitational potential energy we
must have: ½ mv2 = GmM/r
_______
Doing simple algebra, we obtain our escape velocity formula: Vesc = √ 2MG/r
The Black Hole Equation
Recall the escape velocity formula:
________
Vesc = √ 2MG/r
To trap light, we must have Vesc = c
Therefore: RBlack Hole=
2
2GM/c
Black Hole’s of Various Masses
RBlack Hole= 2GM/c2:
For M = 1Msun , RBlack Hole = 3 km
For M = 10Msun , RBlack Hole = 30 km
For M = 10Mearth , RBlack Hole = 1 cm
For M = 106 Msun , RBlack Hole = 3x106 km
For M = 108 Msun , RBlack Hole = 3x108 km
Neutron Stars
• For a star spinning at the
limit of rotational stability:
V2
GMm
m
R2
R

and
P
3
G
V 
GM
R