Week 6
Day 11: Asteroids and Accretion
by Anonymous
Today in class we were discussing the formation of small bodies. To begin, a pretty plot! This is a plot of the semi-major axis vs. the inclination within the asteroid belt.
Our group looked at the worksheet we had been handed and were promptly extremely distracted by the vertical “stripes” in the data, where there appear to be no asteroids. We waved down Mike and received a cryptic answer about harmonics with Jupiter’s orbit. Not to fear though; we were promised all will be revealed in next week’s lectures.
Armed with a few basic assumptions (10 km diameter asteroids, 3 g/cm^{3} density), we were ready to tackle the problems at hand. We estimated based on the plot that the average asteroid had a semi-major axis of 2.6 AU and an inclination of 5°. To calculate the impact velocity of two of these asteroids, we determined the impact velocity would be 2 v sin θ, where v is the total velocity of the asteroid and θ is the angle of inclination. From there we balanced centripetal force with gravitational force to find that v was simply (GM/R) 0.5 where M and R are the radius and mass of the sun. Plugging in numbers gave about 3 * 10^{3} m/s, or roughly 1 km/s, the speed of a bullet.
We ran into a few snags with variable names (too many M’s! too many R’s!), but eventually we were able to solve for the cross section for collision due to gravitational focusing given by b_{2} = R_{2}(1 + (V_{esc}/V_{∞})^{2}). Do not be fooled as we were; this R is for the asteroid. V_{∞} is just the velocity we calculated earlier and V_{esc} (calculated by balancing kinetic energy and gravitational potential energy) is (2GM/R) 0.5 where M and R are for the asteroid. Using order of magnitude calculations, we determined that gravitational focusing is negligible for a body as a small as an asteroid.
Finally, we compared the impact energy to the gravitational binding energy. We calculated quickly that the impact energy wound up being 10^{6} more energetic. That means when the asteroids collided they would likely just blow apart into shards, rather than clumping into larger bodies. To reduce the energy enough to allow for coagulation, the angle of inclination needs to approach zero. This is because energy is proportional to velocity squared (which as we saw earlier is proportional to θ). Calculating gave us a value of θ = 0.001°.
We thought we were done there, but it turned out there was a back side to the page! What trickery! The group disbanded in a mixture of despondency and hunger.
by Youry Aglyamov
We started class by discussing the lectures, as usual. Professor Brown mentioned why it was so much easier to see radicals than normal molecules in comets: it’s in part because they absorb light in the visible spectrum, and in part because normal molecules are common in Earth’s atmosphere and their local absorption lines prevent us from seeing the distant ones. After a brief discussion of smog, we moved on to the main topic of the day: asteroids.
We split into groups and were given identical assignments. Today was somewhat unusual—instead of each group being assigned a TA, we worked more or less independently and asked the TAs and Professor Brown questions. This resulted in less questions being solved, but, I think, left everyone with a much better understanding of the questions that were covered. We started with a plot of asteroid inclination compared to semimajor axis, from which we decided (after, somehow, a lengthy debate) that the average asteroid had an inclination of 9 degrees and a semimajor axis of 2.7 AU. Then, we were given a list of eight questions.
Firstly, what is the average relative velocity of two typical asteroids, if their orbits are circular but inclined? You might think it’s 0, because they have identical orbital characteristics, but the orbits can be rotated relative to each other, meaning that one asteroid is highest above the plane of the solar system when the other is lowest below, and vice versa. In that case, we have:
(not to scale)
Here, the asteroids’ relative velocity at the point of impact (the only points where their orbits cross) is 2vsin(i), where v is their velocity around the Sun (√(GM/R), where M is the mass of the Sun and R is orbital radius, here 2.7 AU) and i is the inclination, here 9 degrees. (All of this is assuming the asteroids are both prograde.) If, by contrast, the relative inclinations aren’t rotated at all, the relative velocity is zero. Since this is planetary science, averaging those two values is close enough; that gives sin(i)√(GM/R). We wavered between that value and√(4GM/R)*sin(i/2), but ultimately chose the former. This is all without taking gravitational focusing into account.
Then the second question: how important is gravitational focusing? The radius of the cross-section from gravitational focusing is b. We fumbled around, trying to either find or re-derive the equation of gravitational focusing from the lectures; unfortunately, both our notes and our attempts at derivation gave incorrect equations. Eventually, we managed to come up with a derivation from conservation of energy and angular momentum that b=r√(1+mR/(Mrsin^{2}(i))), where m is asteroid mass, r is asteroid radius, and b is the effective asteroid radius once gravitational focusing was taken into account. Since the parameters of asteroid size were given to us, we could calculate the ratio of the cross-sectional areas with (πb^{2}) or without (πr^{2}) gravitational focusing. It ended up close to 1.0001, implying focusing was insignificant.
By this point the chaos of the end of class was beginning. We started work on question #3, which asked to compare the impact energy to the binding energy of the asteroid. The former was on the order of (GMmsin^{2}(i))/R without taking gravitational focusing into account, and the latter on the order of (Gm^{2})/r. The former was a lot bigger, implying (as the solution to question 4) the asteroids weren’t going to coagulate unless the inclination was tiny indeed—a correct result. We didn’t get any more precise results for the energies because time ran out.
by Christine
From the lectures, we’ve learned the general process by which small bodies and oligarchs are formed. However, today, we took a closer look at exactly what conditions govern the formation of these small bodies. We worked through a set of guided questions that initially started with the computation of asteroid impact velocity. But first, there were some assumptions to make. In order to go about solving these questions, we first needed to determine some intrinsic values for our asteroids. For simplicity’s sake, we can assume that all the asteroids we are dealing with have the same “typical” inclinations and semimajor axes, all with the same densities and radii (we took these values to be 5 degrees, 2.25 AU, 3 grams per centimeter cubed, and 5 km, respectively. The actual values are not particularly necessary for understanding the process, but I’ve provided them anyway for the home viewers that would like to follow along).
In calculating the impact velocities of two asteroids, we are calculating the velocity of one asteroid relative to another. Using the formula for velocity of an orbiting body that we had determined in class last week (v = (GM/a)^{1/2}), we can say that the impact of the two asteroids traveling at the same inclination yields an impact velocity, v_{r} = 2 v sin(θ).
Then, we can use this velocity when calculating the impact velocity due to gravitational focusing. Using the conservation of energy and angular momentum, we can easily find an expression for v_{I}, the new impact velocity, taking v_{r} to be the velocity of the asteroid at “infinity.”
If we look at the gravitational binding energy, U_{g} = 3GM^{2} / 5r, we can compare it to the impact energy to determine asteroids’ behavior. Simply, if the impact energy is greater than the gravitational binding energy, then they will shatter. Otherwise, if the impact energy is less than or equal to the binding energy, the asteroids will be expected to coagulate.
Then, we can determine the inclination at which we would expect the asteroids to coagulate. This leads us into a better understanding in the formation of oligarchs. As we worked through these problems, we were able to more deeply understand the process by which we would expect oligarch formation, but once again, we were out of time before we could finish this fascinating exercise.
Day 12: Comets and Isotopes
by Alice Michel
Today we spent a good deal of time talking about the Oort Cloud. The class and Mike Brown brought up some interesting questions. We answered some of them; some of them are more opinion-dependent. Here’s a sampling:
Do all planetary systems have an Oort Cloud? Well, first we said we would need a large planet and material to scatter, at the minimum. Since Kepler has found some planetary systems with near-Earth size planets within the orbit of Mercury, clearly not all planetary systems would have one. What about all Hot Jupiter systems? Since they’re so close to their star, the objects they would be trying to scatter would be too tightly bound to the star, thus modifying our original minimum conditions. Then we realized that since stellar formation ejects lots of comets, we don’t even need any planets to have an Oort Cloud! There is even the loose possibility that our Oort Cloud was captured from other stars and planets didn’t play a role. So no easy answer to this one.
We continued talking about extrasolar-system Oort Clouds. Another interesting thought was that, though our Oort Cloud is made up of icy objects, there is nothing preventing one from being made up of other things, like a rocky Oort Cloud. Now what about binary star systems? It depends - some binary star systems can be modeled like uni-star systems, with all the planets going around both. But what if they only orbit one of the stars. It gets complicated fast!
Lastly, Mike Brown asked us whether we thought the Oort Cloud was real or just a hypothesis. It’s so far away that we’ll never (?) be able to see it. We do see things as a consequence of it. But there is an alternate theory that those are just really eccentric objects, which really can’t be right because of dynamic disappearance. So is it real?...
Then we moved on to radioactive decay and some math and physics. We can use radioactive decay to figure out the ages of chondritic meteorites, which would give up the age of our solar system. U-235 decays to Pb-207 and U-238 decays to Pb-206. The stable isotope of Pb is 204, so we crafted the following equations:
We then divided them by each other, which gives us an equation that we could theoretically solve for t. But it is super ugly. We then make the assumption that the starting ratios are the same. With that, class time was up—leaving the rest of the problem as this week’s homework set.
by Allison Maker
Today, our main topic of discussion for the first part of class was the Oort Cloud! This funny-named group of mostly icy objects lies in the far reaches of the solar system. It contains a gigantic number of objects, being far larger than the Kuiper Belt, which lies just outside Neptune’s orbit (its most famous object is ex-planet Pluto (thanks Mike Brown)). It is the source of many of the comets we see fly past Earth. Oort’s process involves objects in the sun’s orbit, but at such a large distance (up to 50,000 AU!) that nearby stars can interfere with their orbit. When this happens, the orbit can change so dramatically that the comet can pass close to the Earth. Halley’s Comet is a well-known comet thought to originate in the Oort Cloud. It is interesting because after being kicked out of the Oort Cloud and into the inner solar system, it has been pulled into a short period orbit, allowing us to see it about every 75 years.
(Wikimedia Commons)
These Oort Cloud comets are also super cool because when we see them, unless they have been pulled into a short orbit like Halley, it is likely the first time they have ever been warmed up since the formation of the solar system.
We also discussed the possibility of Oort-like clouds around other stars. This would be evidence for planets around these stars – as a matter of fact, certain types of planets must exist. A star needs bit, far out planets so that the orbit of the outer objects is capable of being influenced by other stars. So, a star with planets too close by could not have an Oort Cloud. Even hot Jupiters, though they are large, orbit too closely to the stars to allow for the cloud to form.
It’s probably important to note before I conclude that we actually haven’t seen the Oort Cloud visibly, in that it is too far out and lacks illumination. Our only evidence is the comets it sends our way. However, the cloud is the only real viable explanation for the presence of these bodies that light up our sky (and hopefully don’t crash into us anytime soon!)
by Jingyuan Li
We began Thursday's class by discussing some general questions, starting with the motion of stars. While the exact motion of stars is unknown, we do know that they move extremely quickly - 30 to 40 m/s, which is why they don't destroy the Oort Cloud going through it. As a star moves close to a planet, however, the gravitational force of the star will change the orbit of the planet. If, in the 1% chance it reaches within 400AU of a planet, the star becomes so close that its gravitational field will completely dominate whatever other force is acting on the planet.
Next, we discussed the Oort cloud specifically. It is possible to have Oort clouds without any planets, but the particles composing our Oort cloud are believed to have been scattered there by the planets. Objects are composed of ice in the Oort cloud around our sun, but they do not have to be (can be rocky materials as well).
For the second half of class, we split up into three groups and worked on some isotope dating equations, in an attempt to replicate Clair Patterson's results. We start off with two decay equations:
^{235}U → ^{207}Pb, with λ_{235}
^{238}U → ^{206}Pb, with λ_{238}
We next write the equations for the ratios of ^{207}Pb and ^{206}Pb to the stable isotopes of lead (^{204}Pb), which end up being of the form
^{207}Pb(t)/^{204}Pb = ^{207}Pb(0)/^{204}Pb + (^{235}U(t)/^{204}Pb )*(e^{λt} – 1).
From this, we can simply rearrange terms to write an equation in the form y = mx + b, where y is ^{207}Pb(t)/^{204}Pb and x is 206Pb(t)/^{204}Pb. While the equation looks messy, it is quite straightforward. Clair Patterson's data values of ^{207}Pb/^{204}Pb and ^{206}Pb/^{204}Pb from a set of meteorites gives us a linear line. From this line we can find the age of these meteorites using the slope m, which is
(^{238}U/^{235}U)*( e^{λ}^{238}^{t} – 1/e^{λ}^{235}^{t} – 1).
Of course, this requires several big assumptions, such as that the initial lead ratios have to be equal. However, despite this isotope dating can still provide plenty of insight into ages of objects in the universe.