Blog #32 - Types of Supernovae

Types of Supernovae

Supernovae are the most powerful known objects in the universe. Formed from old stars that collapse and explode, a supernova can emit energy on the scale of \(10^{44} \) joules, or as much energy as the Sun emits in its entire lifetime. Within a few moments of the star's collapse, it's apparent magnitude can increase by orders of 11 magnitude, often times making it the brightest object in its respective galaxy for months on end.

Supernovae are very rare occurrences and happen only once or twice per century. However in the Milky Way, only 6 have ever been recorded, the most recent being noted in 1604, before the invention of telescopes! The lack of supernovae imaging in our galaxy is due to our solar system's position in an arm of the Milky Way where most of the stars exist in the inner core. The obfuscation of light from dust and the ISM makes it nearly impossible to detect supernovae towards the center of our galaxy.

However, data from other galaxies has allowed astronomers to characterize two main types of supernovae distinguished in their light curves and process of formation. Type I supernova have a much sharper maximum peak before their light intensity gradually drops. It is believed that these types of supernova are caused by White Dwarf stars in binary systems with much larger Red Giant. The larger star gradually pulls mass from the white dwarf until it goes below a stable limit and begins to collapse on itself, causing the supernova explosion.

Type II supernova have less dramatic light curves and occur over a longer period. The progenitor star for a type II supernova is Red or Blue Supergiant. This is the more typical type of supernova and occurs when the fusion of the star's core dies out causing the gravitation equivalence to collapse and the star implodes. This explosion is typically linked to hydrogen emission lines which helps identify the type of supernova.

Blog #31 - Supernovae

Supernovae

(a) Let's estimate the energy output of a supernova. In a core-collapse supernova, the star collapses to a neutron star. A neutron star is a star/stellar remnant supported by neutron

degeneracy pressure (in contrast with white dwarfs, which are held by electron degeneracy pressure) with a diameter of ∼ 10 km and mass of ∼ 1.4 M. Using conservation of energy, we know that the energy in a supernova explosion must come from the liberation of gravitational potential energy. Estimate the energy available in a supernova explosion. Assume that the star's radius prior to collapse is 10 km.

For the first part we assume all of the kinetic energy is converted from potential energy upon explosion: \[KE = PE \rightarrow KE = \frac{3GM^2}{5R}\] Plugging in the numbers given in the problem we get \[KE = \frac{3*6.67*10^{-11}*(1.4*2*10^{30})^2}{5*10^3} = 2.1*10^{46} \space joules \]

(b) Most of the energy from part (a) is carried away by neutrinos, tiny charge less particles that rarely interact with other matter. A portion of it, however, is injected directly into the gas of the interstellar medium. When an energy E is suddenly injected to a background gas of density ρ, a strong blast wave that travels outwards is created. Using an order of magnitude estimate,show that the radius of the blast wave R scales with time t and density as R ∼ t^2/5 ρ^-1/5.
Hint: Assume that the Kinetic Energy of the blast wave is constant. Why might this be a good assumption?

Now we convert the kinetic energy of the blast into the radius that it scales to. \[KE = \frac{1}{2} m v^2 \approx mv^2\] The kinetic energy can be extrapolated by order of magnitude approximation into its components of density and radius as \[mv^2 \approx R^3 \rho (\frac{R}{t})^2 \] Which simplifies into \[KE \approx \frac{R^5 \rho}{t^2}\]Assuming the kinetic energy is constant considering that the losses in free space are negligible, we can rearrange to get [R^5 = \frac{t^2}{rho} \] or simplified to the scaling relation we expected \[R \approx \frac{t^{\frac{2}{5}}}{\rho^{\frac{1}{5}}} \]

(c) The blast wave from part (b) is known as the Sedov-Taylor blast wave, and was studied during the Second World War in the context of trying to estimate the destructive potential of the (then theoretical) atomic bomb. Use your order-of-magnitude estimate from (b) to estimate the velocity v(t) of the blast wave. Hint: Write the velocity v(t) in terms of the derivative of the radius R(t).

We know from (b) that \[R \approx \frac{t^{\frac{2}{5}}}{\rho^{\frac{1}{5}}} \] from this we can use the differential form of \(\frac{dR}{dt} = v(t) \) to solve for v(t). \[\frac{dR}{dt} = \frac{d}{dt}(t^{\frac{2}{5}}\rho^{-\frac{1}{5}}) \] Taking the derivative of the right side we get \[v(t) = \frac{2}{5} t^{-\frac{3}{5}}\rho^{-\frac{1}{5}} \]

Blog #30 - Death of a Star

Death of a Star 

1. Stellar death. We’ve now covered the main portions of a star’s life: it’s“birth” from the collapse of molecular clouds as well as the majority of it’s“life” on the Main Sequence in hydrostatic equilibrium. Now we’re going to take a look at what happens when a star like the Sun“dies” after it can no longer support itself via nuclear fusion.

(a) Life. The sun’s luminosity is L = 1033 erg/s. If fusion converts matter into energy with a 0.7%
efficiency and the Sun has 10% of its mass (the mass at the center/core) available for fusion, how long does it take to use up its fuel supply?

We start with the relationship between mass, luminosity and energy from fusion \[L \approx M^4 \approx \frac{E}{t} \] From this we can use the mass-energy equivalence to convert the energy relation to the mass of the core \[E = m*c^2 \rightarrow L \approx \frac{mc^2}{t} \] From here we can use the 0.7% conversion relation to find the time until all of the fusion energy is used. \[t = (0.007) \frac{.1 M_{\odot} c^2}{L_{\odot}} = 1.26*10^{18} seconds \approx 40 billion years\]

(b) Death. We previously derived that the time it takes for a cloud of gas to collapse under its own weight with no opposing force of pressure (i.e. the free-fall time tf f) is: \[t_{ff} = \sqrt{\frac{3 \pi}{32 G \rho}} \]

How long does it take the core of the Sun (again containing 10% of its mass) to collapse under its own weight once it runs out of fuel and there is no longer any opposing pressure?

First we solve for the density of the core \[\rho_c = \frac{M_{\odot}}{\frac{4}{3} \pi R_{\odot}^3} = \frac{3 M_{\odot}}{4 \pi R_{\odot}^3} \] Plugging this density into the equation for collapse time we get \[t_{ff} = \sqrt{\frac{3 \pi}{32 G \rho_c}} = \sqrt{\frac{\pi^2 R_{\odot}^3}{8 G M_{\odot}}} \] Solving for this value we get the time of collapse to be \[t_{ff} = 10^4 \space seconds \]This result is pretty shocking but thankfully we have fusion energy to keep our Sun alive! 

Blog #29 - Alien Megastructure

Alien Megastructure 

In 2015, a team of Yale researchers led by astronomer Tabetha Boyajian found an incredibly strange transit event occurring on the star KIC 8462852, or "Tabby's Star". Observed by the Kepler Space Telescope, KIC 8462852 showed multiple massive dips of the incoming stellar flux. Typically these types of drops signify a transiting planet but in this case the light curve peaks showed incredibly large drops of almost 20% of the incoming light. For reference, when Jupiter transits the Sun, the light curve drops only around 1% of the total flux.

Further contributing to the strangeness of the star was the dimming over time. After hearing of "Tabby's Star", researchers at Cal Tech went back to old observations of the star and found that from 2009-2013, the star displayed a drop of nearly 3% in the short time span. Of the 200 nearby comparison stars they observed, not a single one displayed such drastic results. 

These massive drops were initially contributed to a group of passive comets or planetary building blocks but as the drops showed repetitive transits, these theories were dispelled. Soon, astronomers from around the world were attempting to solve the mystery of KIC 8462852. The Search for Extraterrestrial Intelligence (SETI) is a group attempting to find life beyond our planet and currently leads research on KIC 8462852 in hopes of finding what is causing the strange light curves.

Currently the most exciting theory surrounding Tabby's Star is that of the alien megastructure. Though mostly speculation, SETI and other alien optimists believe that a large intelligently made structure surrounds the star, rotating to collect the energy from the stellar structure. They believe that the "megastructure" blocks the outgoing light of the star for large portions of the orbit, causing the severe drops in the incoming light. 

Unfortunately further research into the megastructure has come up empty. Both radio and infrared telescopes were pointed at the structure in hopes of receiving outgoing signals or "waste heat" if the structure was in fact absorbing heat from the central star. Neither was found and astronomers more or less gave up on the idea that there was a circumstellar structure around Tabby's Star. 

Still the hunt continues to try and identify what was happening in 2015 at Yale. Most astronomers now agree that the curves were probably distorted by some phenomena of the interstellar medium, though no evidence has proven this hypothesis either. 

Blog #28 - Habitable Zone pt. 2

Habitable Zone pt.2

2. In this problem we’ll figure out how the habitable zone distance, Ahz, depends on stellar mass. Recall the average mass-luminosity relation we derived earlier, as well as the mass-radius relation for stars on the main sequence. If you don’t recall, this is a good time to practice something that will very likely show up on the final!

(a) Express aHZ in terms of stellar properties as a scaling relationship, using squiggles instead of equal signs and ditching constants.

Simplifying the answer from part one, we get the relation \[T_p \approx C*L_s^{\frac{1}{4}}*a_{hz}^{\frac{1}{2}} \]Rearranging for a we get \[a_{hz}^{\frac{1}{2}} \approx C*L_s^{\frac{1}{4}}*T_p^{-1}\]Which gives us \[a_hz \approx C*L_s^{\frac{1}{2}}*T_p^{-2} \]

(b) Replace the stellar parameters with their dependence on stellar mass, such that aHZ ∼ Mα. Find α.

From an earlier worksheet we know the relation that \[L_s \approx M_s^4 \] This relation can help turn the semi major axis into a number based solely on the physical makeup of the star and planet. From part (a) we have \[a_hz \approx \frac{L^{\frac{1}{2}}}{T_{hz}^{2}} \approx \frac{M_s^{2}}{T_{hz}^{2}}\]

(c) If the Sun were half as massive and the Earth had the same equilibrium temperature, how many days would our year contain?

Converting this to a relationship for the orbital period, we have \[P^2 \approx \frac{a^3}{M_s}\] Substituting in our term for \(a_{hz}\) we get \[P^2 \approx \frac{(\frac{M_s^2}{T_{hz}^2})^3}{M_s} \]Using the information that \(T_{hz}\) is the same and \(M = .5 M_s \) we get \[P^2 \approx \frac{(\frac{\frac{1}{2}M_s^2}{T_{hz}^2})^3}{\frac{1}{2}M_s} \approx \frac{\frac{1}{64}M_s}{\frac{1}{2} M_s} \rightarrow P \approx \sqrt{\frac{1}{32}}*P_E \]Giving us a final answer of \[P \approx \sqrt{\frac{1}{32}}*365 \space days = 64.52 \space days \]

Blog #27 - Habitable Zone pt.1

Habitable Zone Pt. 1

The Earth resides in a “Goldilocks Zone” or habitable zone (HZ) around the Sun. At our semi major axis we receive just enough Sunlight to prevent the planet from freezing over and not too much to boil off our oceans. Not too cold, not too hot. Just right. In this problem we’ll calculate how the temperature of a planet, Tp, depends on the properties of the central star and the orbital properties of the planet.

(a) Draw the Sun on the left, and a planet on the right, separated by a distance a. The planet has a radius Rp and temperature Tp. The star has a radius R and a luminosity L and a temperature Teff.

We begin by putting together a picture of the setup, seen below: 

(b) Due to energy conservation, the amount of energy received per unit time by the planet is equal to the energy emitted isotropically under the assumption that it is a blackbody. How much energy per time does the planet receive from the star?How much energy per time does the Earth radiate as a blackbody?

We start by finding the amount of energy incident on the planet. This number comes from the amount of flux that hits the planet's surface. \[F_s = \frac{L_s}{4 \pi a^2}\] \[E_{in} = F_s * \pi R_p^2 = \frac{L_s R_p^2}{4 a^2} \]We then solve for energy coming out of the planet as simply the luminosity out of the surface of the Earth. \[E_{out} = L_p = \sigma T_p^4 (4 \pi R_p^2)\]

(c) Set these two quantities equal to each other and solve for Tp.

Setting the two energies equal to each other gives us: \[E_{in} = E_{out} \rightarrow \sigma T_p^4 (4 \pi R_p^2) = \frac{L_s R_p^2}{4 a^2}\] Rearranging for \(T_p\) we get \[T_p = (\frac{L_s}{16 \pi a^2 \sigma})^{\frac{1}{4}} \]

(d) How does the temperature change if the planet were much larger or much smaller?

Notice that this equation does not depend on the size of the planet so the distance is inherently based on the distance from and luminosity of the star. 

(e) Not all of the energy incident on the planet will be absorbed. Some fraction, A,will be reflected
back out into space. How does this affect the amount of energy received per unit time, and thus how does this affect Tp?

By adding a term for reflectance the \(E_{in}\) becomes \[E_{in} = \frac{L_s R_p^2}{4 a^2}*(1-A)\] Meaning that our \(E_{in}\) will not reach the full flux of the star incident to the planet. This means that the energy out of the planet will also have to decrease meaning the overall temperature of the planet will decrease as well.

Blog #26 - Angular Momentum

Angular Momentum




1. Angular momentum. In this problem we will obtain some intuition on why a disk must form during star formation if angular momentum is to be preserved.

(a) Cloud angular momentum. Consider a typical interstellar cloud core that forms a single star. You can assume it has a mass of 1 M and a diameter of 0.1 pc. A typical cloud rotational velocity is 1 m/s at the cloud edge. Calculate the angular momentum of the cloud assuming constant density. If the core collapses to form a sunlike star, what would the velocity at the surface of the star be if angular momentum is conserved? How does this compare with the break-up velocity of the Sun which is ∼300 km/s?

We begin by finding the equation for the angular momentum, L. We know we can relate the moment of inertia for a sphere and angular velocity to their linear equivalents to turn rotational movement into instantaneous linear movement. \[L = I \omega = M R v \] Using the information in the problem we can equate the rotational speed of the gas cloud and the speed of the rotating star to find the speed at the surface of the star. \[M_{\star} R_{\star} v_{\star} = M_{cloud} R_{cloud} v_{cloud} \] We know the cloud and the sun-like collapsed star have roughly the same mass so our equation for the speed of the star surface becomes \[v_{\star} = \frac{R_{cloud} v_{cloud}}{R_{\star}} \] Plugging in the given values we get \[v_{\star} = \frac{.05 pc*1 \frac{m}{s}}{2.3 * 10^{-8} pc} = 2170 \frac{km}{s} \] This value is much higher than the given \(300 \frac{km}{s}\) for the Sun. 

(b) Disk angular momentum. Assume that all the angular momentum is instead transferred to a disk of size 10 AU and negligible height. How massive must the disk be? You can assume constant density. (Hint: You can assume that the disk rotates with a Keplerian velocity given by v = GM/r where M is the mass and r is the radius.)

We again begin by solving for the rotational inertia by equating the moment of inertia and velocity of a sphere (star) to the inertia of the disk. \[L_s = \frac{2}{5} M_s R_s v_s  = \frac{1}{2} M_d R_d v_d = L_d \] We also know that the speed of the disk can be described by Keplerian velocity so we substitute \[L_d = \frac{1}{2} M_d R_d * (\sqrt{\frac{G M_s}{R_d}}) \] Notice the mass of the velocity term is dictated by the central star as it is assumed to be much more massive than the surrounding disk material. Simplifying and solving we get \[M_d = \frac{4 (M_s)^{\frac{1}{2}} R_s v_s}{5 (R_d)^{\frac{1}{2}} G^{\frac{1}{2}}} \]Solving for our constants, we get \[M_d = 1.71*10^{29} \space kg \]

(c) Solar System. The Sun has a surface rotational velocity of ∼2 km/s at the equator. How do the angular momenta of the Sun and Jupiter compare?

Comparing the angular momentums \[\frac{L_s}{L_j} = \frac{M_s R_s v_s}{M_j R_j v_j}\] which gives us a ratio of around 18 times the angular momentum for the Sun given the sun moves at around 2 km/s and Jupiter moves around 12 km/s.