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The core of a star collapses during a supernova, forming a neutron star. Angular momentum of the core is conserved, so the neutron star spins rapidly. If the initial core radius is $5.0\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{km}$ and it collapses to 10.0 km, find the neutron star’s angular velocity in revolutions per second, given the core’s angular velocity was originally 1 revolution per 30.0 days.
$964\phantom{\rule{0.2em}{0ex}}\text{rev/s}$
Using the solution from the previous problem, find the increase in rotational kinetic energy, given the core’s mass is 1.3 times that of our Sun. Where does this increase in kinetic energy come from?
(a) What Hubble constant corresponds to an approximate age of the universe of ${10}^{10}$ y? To get an approximate value, assume the expansion rate is constant and calculate the speed at which two galaxies must move apart to be separated by 1 Mly (present average galactic separation) in a time of ${10}^{10}$ y. (b) Similarly, what Hubble constant corresponds to a universe approximately $2\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}$ years old?
a. ${H}_{0}=\frac{\text{30km/s}}{\text{1Mly}}=30\text{km/s}\xb7\text{Mly;}$ b. ${H}_{0}=\frac{15\phantom{\rule{0.2em}{0ex}}\text{km/s}}{1\phantom{\rule{0.2em}{0ex}}\text{Mly}}=15\text{km/s}\xb7\text{Mly}$
Electrons and positrons are collided in a circular accelerator. Derive the expression for the center-of-mass energy of the particle.
The intensity of cosmic ray radiation decreases rapidly with increasing energy, but there are occasionally extremely energetic cosmic rays that create a shower of radiation from all the particles they create by striking a nucleus in the atmosphere. Suppose a cosmic ray particle having an energy of ${10}^{10}\phantom{\rule{0.2em}{0ex}}\text{GeV}$ converts its energy into particles with masses averaging $200\text{MeV/}{c}^{2}$ .
(a) How many particles are created? (b) If the particles rain down on a $1{\text{.00-km}}^{2}$ area, how many particles are there per square meter?
a. $5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}$ ; b. divide the number of particles by the area they hit: $5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\text{particles}\text{/}{\text{m}}^{2}$
(a) Calculate the relativistic quantity $\gamma =\frac{1}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}$ for 1.00-TeV protons produced at Fermilab. (b) If such a proton created a ${\pi}^{+}$ having the same speed, how long would its life be in the laboratory? (c) How far could it travel in this time?
Plans for an accelerator that produces a secondary beam of K mesons to scatter from nuclei, for the purpose of studying the strong force, call for them to have a kinetic energy of 500 MeV. (a) What would the relativistic quantity $\gamma =\frac{1}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}$ be for these particles? (b) How long would their average lifetime be in the laboratory? (c) How far could they travel in this time?
a. 2.01; b. $2.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-8}}\phantom{\rule{0.2em}{0ex}}\text{s}$ ; c. 6.50 m
In supernovae, neutrinos are produced in huge amounts. They were detected from the 1987A supernova in the Magellanic Cloud, which is about 120,000 light-years away from Earth (relatively close to our Milky Way Galaxy). If neutrinos have a mass, they cannot travel at the speed of light, but if their mass is small, their velocity would be almost that of light. (a) Suppose a neutrino with a $7\text{-eV/}{c}^{2}$ mass has a kinetic energy of 700 keV. Find the relativistic quantity $\gamma =\frac{1}{\sqrt{1-{v}^{2}\text{/}{c}^{2}}}$ for it. (b) If the neutrino leaves the 1987A supernova at the same time as a photon and both travel to Earth, how much sooner does the photon arrive? This is not a large time difference, given that it is impossible to know which neutrino left with which photon and the poor efficiency of the neutrino detectors. Thus, the fact that neutrinos were observed within hours of the brightening of the supernova only places an upper limit on the neutrino’s mass. ( Hint: You may need to use a series expansion to find v for the neutrino, since its $\gamma $ is so large.)
Assuming a circular orbit for the Sun about the center of the Milky Way Galaxy, calculate its orbital speed using the following information: The mass of the galaxy is equivalent to a single mass $1.5\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{11}$ times that of the Sun (or $3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{41}\phantom{\rule{0.2em}{0ex}}\text{kg}$ ), located 30,000 ly away.
$\begin{array}{}\\ \\ \hfill \frac{m{v}^{2}}{r}& =\hfill & \frac{GMm}{{r}^{2}}\Rightarrow \hfill \\ \\ \\ \hfill v& =\hfill & {\left(\frac{GM}{r}\right)}^{1\text{/}2}=\left[\frac{\left(6.67\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-11}}\phantom{\rule{0.2em}{0ex}}\text{N}\xb7{\text{m}}^{2}\text{/}{\text{kg}}^{2}\right)\left(3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{41}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)}{\left(\text{30,000ly}\right)\left(9.46\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\text{m/ly}\right)}\right]=2.7\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{m/s}\hfill \end{array}$
(a) What is the approximate force of gravity on a 70-kg person due to the Andromeda Galaxy, assuming its total mass is ${10}^{13}$ that of our Sun and acts like a single mass 0.613 Mpc away? (b) What is the ratio of this force to the person’s weight? Note that Andromeda is the closest large galaxy.
(a) A particle and its antiparticle are at rest relative to an observer and annihilate (completely destroying both masses), creating two $\gamma $ rays of equal energy. What is the characteristic $\gamma $ -ray energy you would look for if searching for evidence of proton-antiproton annihilation? (The fact that such radiation is rarely observed is evidence that there is very little antimatter in the universe.) (b) How does this compare with the 0.511-MeV energy associated with electron-positron annihilation?
a. 938.27 MeV; b. $1.84\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}$
The peak intensity of the CMBR occurs at a wavelength of 1.1 mm. (a) What is the energy in eV of a 1.1-mm photon? (b) There are approximately ${10}^{9}$ photons for each massive particle in deep space. Calculate the energy of ${10}^{9}$ such photons. (c) If the average massive particle in space has a mass half that of a proton, what energy would be created by converting its mass to energy? (d) Does this imply that space is “matter dominated”? Explain briefly.
(a) Use the Heisenberg uncertainty principle to calculate the uncertainty in energy for a corresponding time interval of ${10}^{\mathrm{-43}}\text{s}$ . (b) Compare this energy with the ${10}^{19}\phantom{\rule{0.2em}{0ex}}\text{GeV}$ unification-of-forces energy and discuss why they are similar.
a. $3.29\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{18}\phantom{\rule{0.2em}{0ex}}\text{GeV}\approx 3\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{18}\phantom{\rule{0.2em}{0ex}}\text{GeV}$ ; b. 0.3; Unification of the three forces breaks down shortly after the separation of gravity from the unification force (near the Planck time interval). The uncertainty in time then becomes greater. Hence the energy available becomes less than the needed unification energy.
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