Problems:
- The sun has a mass of 2e30 kg and a mean radius of
6.4e6 m. If the sun collapsed from a very large, very diffuse cloud of
gas, how much potenital energy would have been converted into thermal energy
doring the collapse? Given that the sun is radiating now at the rate of
1340 W/m^2 at the earth (1.5e11 m from the sun), what is the total energy
radiated by the sun per second? How long would the sun "shine" if there were no
other source of energy other than that from gravitational collapse?
As shown in class, assuming a incompressible fluid for the sun (i.e. constant density all the way through), the total potential energy is
Phi = - 3/5 G M^2/R
= - 3/5 6.67e-11 2e30^2 /6.4e6 = - 2.5e43 joules
This is the amount of energy release when the sun collapsed.
The total power radiated from the sun is
P = Luminosity * Area = 1340 Watts/m * 4 Pi (1.5e11 m)^2
= 3.79e26 watts = 3.79e26 joules/sec
An esitmate of the length of time that the sun would shine is the
ratio of the two numbers.
T = Phi/P = 2.5e43 joules /3.79e26 joules/sec = 6.6e16 sec
= 2.1 billion years
- Assuming that the the temperature of the atmosphere of earth
is that same all the way up as it is on the ground and the composition is the
same, find the relationship between pressure and elevation in our atmosphere.
(Your freshman physics text may be helpful for this. Check out Pascal's
principle. Remember, however, that the density of a gas is proportional to
the pressure.) What would you predict the pressure to be outside an airplane
flying at 40,000 ft.?
From Pascal's principle, dP = - rho g dh. However, the density, rho, is a function of the pressure. Using the ideal gas law, PV = nRT, and rho = m n /V (where m in the molecular weight), rho = m P/R T. Thus,
dP/dh = - m g P/R T
The derivative of a function is the function itself imples that the function is an exponential. The solution is:
P = P_0 Exp[-m g h/R T]
The coefficient (m g/R T) = 1/h_0 = 1.11e-4/m where h_0 = 9 km is the scale height of the atmosphere. 40000 ft is 12.1 km. Thus the pressure would be
P = 1 atm Exp [12.1/9] = .26 atm
- Given equation 15.20 in your book, how many photons/m^3 are
there in intergalactic space due to the 2.7 ^oK black body radiation? Compare this with the average number of protons which is about 1 atom/m^3. Compare the
total energy density of the photons to the energy density of the protons.
Solution in a Mathematica notebook.
- The density of protons at 1 atom/m^3 is about 1/12 that needed
to close the universe. The number density of each type of neutrino should be the same as the number density of photons. How massive would one of the neutrinos have to be to "close" the universe?
Solution in a Mathematica notebook.
- SM&M 15-22
Using the Hubble constant of 17e-3 m/sec/lightyear, we get the following table:
Distance Velocity Beta lambda/lambda_0 lambda
2e6 l.y. 3.4e4 m/sec .000113 1.000113 590
2e8 l.y. 3.4e6 m/sec .0113 1.0114 597
2e9 l.y. 3.4e7 m/sec .113 1.12 660