Revisiting the Astro 16 Midterm

On March 11, I got to spend my afternoon tackling some pretty interesting problems (terrifyingly, without the aid of the Internet or even my ancient graphing calculator which, for some reason, is still kicking around). It generally went pretty well, but there were a few hiccups which I'll go over here.

To start with, problem four, and a return to Io. I started by calculating the semi-major axis of its orbit around Jupiter which, given its orbital period and the mass of Jupiter, could be found using Kepler's Third Law--it came out to $8 \times 10^{10}$ cm. Next, I did some work using the mathematics behind tides that we worked out here in order to determine the tidal force exercised by Jupiter on Io: $\pm 1.1 \times 10^{-2}$ g $\times$ cm $\times$ s$^{-2}$.

Finally, on the third part, I (predictably) got into some math trouble. In order to compare the tidal forces on Io due to Jupiter to those on Earth due to the Moon, I was trying to calculate the tidal force we experience here, just by fiddling with the constants in the tidal force equation. I got as far as \[\Delta F =  \frac{6 \times 10^2 \times 10^{-8} \times 10^8 \times 10^{25}}{6.4 \times 10^31}\] But, for once having to rely on my own brain rather than Stephen Wolfram's to do math, I mistakenly simplified it to $10^{-7}$ g $\times$ cm $\times$ s$^{-2}$, putting it a whopping five orders of magnitude below what Io feels due to Jupiter. In reality, we experience tidal forces of around $10^{-4}$ $ g $\times$ cm $\times$ s$^{-2}$--closer, but still a couple of orders of magnitude less than what Io's up against. Even moving Earth's and Io's predicaments that much closer, it still makes a great deal of sense that Io's extreme geological activity is thanks to the nearby tug of Jupiter.

Another image of Io, this one taken by the Galileo spacecraft. It is one crazy-looking planet (source).

The next issue, at the start of problem five, concerned converting keV (that'd be kiloelectronvolts) in to ergs--right off the bat, I dropped a factor of ten giving me $2 /times 10^{-9}$ erg where I should have had $2 \times 10^{-8}$ erg. I also made an arithmetic error when dividing this by $h$ in order to get frequency, $\nu$. As a result, I ended up with a photon energy of $3.5 \times 10^{17}$ Hz rather than what it should have been--$3 \times 10^18$. Thus, my final photon wavelength was $10^{-7}$ cm, rather than the correct answer of $10^{-8}$ cm, and an angular resolution of $2.5 \times 10^{-9}$ radians instead of the actual resolution of $2.5 \times 10^{-10}$ radians. However, this doesn't change the overall ranking of the three telescopes being compared in the problem (I calculated the resolution of a 5-meter telescope observing photons at 0.5 micron wavelength as $10^{-7}$ radians, and that of a kilometer-wide interferometer observing at 30 GHz as $10^{-5}$ radians. So A is still on top, resolution-wise. 

As a sidenote, A likely involves the most interesting engineering as well. X-rays, you might be surprised to find, tend to pass straight through most things, making conventional telescopes useless for such high-energy photons. As a result, when designing the Chandra X-Ray Observatory, NASA had to come up with a pretty nifty setup where the photons, rather than being reflected straight off a mirror, graze a series of angled mirrors in order to gradually alter their course into the detector. Chandra, one of NASA's Great Observatories, has been trucking along since 1999--despite only being planned to last until 2004. As an added bonus, it's operated from just across the street from where I'm sitting right now, at our very own Harvard-Smithsonian Center for Astrophysics.

Cutaway of the Chandra design (source).