Transcript Week 8

Announcements
• Homework: Chapter 14 # 47 & 48 +
Supplemental Problems.
• Exam 3 will be during the final exam
period: Monday May 2 @ 4:00pm. Will
cover everything since the last exam.
The energy source of stars
A young Albert Einstein
proposed that mass and
energy were two sides of
the same coin and could
be interchanged. His
famous equation from
special relativity is
E = mc2
Fusion: the means of
converting mass into energy
The first person to propose that
stars fused hydrogen into
helium was Sir Arthur
Eddington in the 1920’s. His
critics said that stars were not
hot enough. His retort: “I am
aware that many critics
consider the stars are not hot
enough. The critics lay
themselves open to an obvious
retort; we tell them to go and
find a hotter place.”
The Proton-Proton Cycle: The
Mechanism of Solar Fusion
Hans Bethe worked out the details
of how the Sun fused hydrogen into
helium in the 1930’s. He won the
Nobel Prize in 1967 for his work
Example
Determine the energy
released (or required) in each
step of the CNO Cycle
The mass of the neutrino can
be ignored but the mass of
an electron or positron
cannot. Gamma rays have no
mass
Masses can be found at
Wikipedia. Just google the
isotope.
Example Solution
Look up masses on Wikipedia
1u = 1.660538921×10−27 kg
Example Solution 2
To find the energy from each step add up the masses
of all the particles going into a step then subtract all
the mass coming out of that step. Multiply the result by
the speed of light squared to get the energy in Joules.
Gamma rays have no mass and the mass of the
neutrino can be ignored
Step 1: 12C + 1H
13N
2.000000 1026 kg  1.67372 1027 kg  2.167372  1026 kg
 2.159653 1027 kg
 1.951411026 kg

 2.997925  108 m s
 1.753837 109 J

2
Example Solution 3
Step 2: 13N
13C

+ e+
2.159653 1026 kg  2.159258 1026 kg  9.102176 1031 kg

 3.039782 1030 kg

 2.997925 108 m s

2
 2.7320211013 J
Step 3: 13C + 1H
 2.159258 10
14N
26
 1.365 1029 kg

 2.997925 10

kg  1.67372 1027 kg  2.325265 1026 kg
8 m
2
 1.2268011012 J

2
Example Solution 4
Step 4: 14N + 1H
 2.325265 10
26
15O

kg  1.67372 1027 kg  2.491317 1026 kg
 1.32 1029 kg

 2.997925 108 m 2

2
 1.186357 1012 J
Step 5: 15O
15N

+ e+
2.491317 1026 kg  2.490826 1026 kg  9.102176  1031 kg
 3.999782 1030 kg

 2.997925 10
8 m
s
 3.594826 1013 J

2

Example Solution 4
Final Step: 15N + 1H
 2.490826 10
26
12C
+ 4He
 
kg  1.67372 1027 kg  2.491317 1026 kg  6.6465 10 27 kg
 4.97769 1027 kg

 2.997925 108 m 2

2
 4.473726 1010 J
This final step actually consumes energy rather than
releasing it!
Overall Energy produced
1.753837  109 J  2.732021 1013 J  1.226801 1012 J
1.186357  1012 J  3.594826  1013 J  4.473726  1010 J
 1.309510 109 J

Angular Momentum
Objects in orbit have orbital
angular momentum
For a point mass moving in
a circular orbit of radius r
and at a speed v, the orbital
angular momentum is just
L = mvr
Rotating objects have
rotational angular momentum
For solid objects we
express the rotational
angular momentum as
the product of their
moment of inertia (I) and
the angular velocity they
rotate at (w).
L = Iw
I will depend on the
shape of the body and
the axis of rotation.
Moments
of Inertia
The moment of
inertia (I) depends
on both the shape of
the body and the
axis of rotation. For
a point mass m,
moving in a circular
orbit with a radius R,
I = mR2
If gravity is the only force
acting on a system, angular
momentum is conserved
Internal forces cannot change the total angular momentum
Example
When a massive star explodes in a Type II supernova
the core of the star collapses from an iron core with a
diameter of 12,500 km to a neutron star with a diameter
of 18.0 km. If the total mass of the core remains the
same and the original rotational period before the
collapse was 31.5 days, what is the rotational period of
the resulting neutron star?
Example Solution
A conserved quantity is simply one that stays the same so
Lbefore = Lafter
In this case, the object is a sphere before and after the
supernova so I = 2/5MR2 and M remains the same
Lbefore  Lafter  I beforewbefore  I afterwafter

2
5
2
before
MR
2
Rbefore
Tbefore


 2 

 
 Tbefore 
2
Rafter
Tafter

2
5
2
after
MR
 Tafter  Tbefore

 2

 Tafter



2
Rafter
 18km 
 31.5days 

12,500
km


2
Rbefore
Tafter  6.53 105 days  86400 s day  5.64s
2