Astro 162 – Planetary Astrophysics – Solutions to Set 12Problem 1. Soup Bowls and Ocean BasinsA flat bottomed bowl of radius R is filled with water to depth H.(a) Derive an approximate analytic formula for the period P of the sloshing mode.As described in lecture, the sloshing mode is a shallow water wave. That is, theentire layer of water participates in the mode, not jus t the surface. We know that forshallow water waves, the dispersion relation readsω = kpgH (1)where k = 2π/λ is the wavenumber. Now from crest to trough is λ/2. When all th ewater is heaped up on one side of the bowl, the distance from crest to trough is 2R = λ/2.Therefore λ = 4R and k = π/(2R). Since ω = 2π/P where P is the wave period, wefind thatP =4R√gH. (2)(b) Use your formula to estimate P for a soup bowl and for the Earth’s ocean basin.For a soup bowl, R ∼ 7 cm and H ∼ 5 cm. Then Psoup∼ 0.4 s .For the Earth’s ocean basin, R ∼ 6000 km and H ∼ 5 km. Then Pocean∼ 30 hr . Thefact that this ocean-sloshing period is close to the syn odic period between the Earth’srotation and the Moon’s revolu tion about the Earth means that the Moon can resonantlyexcite tides to large amplitude.Problem 2. The Storm That Launched a Thousand WavesFollowing a winter storm above the ocean, the interval between waves at Californiabeaches declined from 17–19 s on Sunday, to 16–18s on Monday, and to 15–16s onTuesday. Typical values are 10–11 s.These waves are surface g ravity waves on deep water (or somewhat misleadingly, “deepwater waves.”)(a) What was the maximum sustained wind speed during the storm?For surface gravity waves on deep water, the dispersion relation reads1ω =pgk . (3)The phase speed of such waves isvph= ω/k =rgk(4)while the group speed isvgp= dω/dk =12rgk. (5)From lecture, we know that the wind blowing across the surface of th e water generateswaves. Moreover, from lecture, the wind speed equals the PHASE speed of the wavesgenerated. So to determine the maximum wind speed during the storm, we need todetermine the maximum phase speed of the waves generated. Of the waves that weobserve at the beach, which on es h ave the maximum phase speed? By equation (4), thephase speed goes as 1/√k. S o we need to identify the waves having the smallest k tomaximize vph.What we measur e is ω, the f requency of waves crashing on the beach. From (3),k =ω2g. (6)Plug this into (4):vph=gω=gP2π(7)where ω = 2π/P and P is the wave period. So the longest period wave has th e greatestphase speed. The longest period we observed at the beach is 19 seconds. Plug in P = 19 sand g = 103[cgs] to findmax(vwind) = max(vph) = 3000 cm s−1= 30 m s−1= 100 kph . (8)(b) How distant was the storm from the beaches?Surface gravity waves are dispersive; in other words, waves of different frequenciestravel at different speeds. That is why the longest period waves arrive fir st.2Call the distance between the storm and us D. It took Sunday’s waves a time∆tSundayto travel to us:∆tSunday=Dvgp,18(9)Notice the GROUP velocity enters here: the entire group of waves travels at the groupvelocity from the storm to us. Also note the subscript “18,” which is the average waveperiod (in seconds) for Sunday.It took Tuesday’s waves a time ∆tTuesday= ∆tSunday+ 2 days to reach us:∆tTuesday= ∆tSunday+ 2 days =Dvgp,15.5(10)where 15.5 seconds is the aver age wave period for Tuesday’s waves. We can now us e (9)and (10) to solve for the desired D. Subtract th ese equations to eliminate ∆tSunday:2 days =Dvgp,15.5−Dvgp,18= D 1vgp,15,5−1vgp,18!. (11)Combine (3) and (5) to find:vgp=12gω=gP4π. (12)Use P = 18 s and P = 15.5 s to solve for vgp,18and vgp,15.5, respectively. Then use(11) to find thatD ∼ 15000 km , or roughly halfway around the world! (The PacificOcean is a big place!)Notice we have not used Monday’s data explicitly, but have simply taken Sundayand Tuesday as endpoint data (two points make a line). You could have used Monday’sdata, too, and gotten estimates for D that agree to within a factor of 2.(c) How long ago did the storm take place?We can just use (9) and our answer to (b) to calculate ∆tSunday∼ 12 days . So thestorm took place about 12 days prior to Sunday .(d) What are upper limits on the size and duration of the storm?3To estimate an upper limit on the (horizontal) size of the storm, assume firs t thatthe storm occurred instantaneously in time (that is, assu me the duration of the stormis 0 seconds). We’ll sw itch roles in this assumption later (th at is, to estimate the upperlimit on the duration of the storm, we’ll assume the storm occurred at a single point inspace.)On Sunday, waves having a variety of periods, 17–19 seconds, crashed simultaneouslyon the beach. The 19 s waves travel faster than the 17 s waves [see equation (12)]. Let’ssay the 19 s waves were born a distance D + ∆D away. And let’s say the 17 s waves wereborn a distance D − ∆D away. They arrived at the beach on the same day (Sunday).ThereforeD + ∆Dvgp,19=D − ∆Dvgp,17(13)We know every variable in this equation except for the desired ∆D (the radius of thestorm). Straightforward algebra gives∆D = D vgp,19− vgp,17vgp,19+ vgp,17!= D19 − 1719 + 17= 800 km . (14)Now we estimate th e upper limit on the duration of the storm. Assume the stormoccurred at a single point in space. Again, what we know is that on Sunday, the 19 swaves and the 17 s waves arrived at the beach at the same time. Let’s say that at thebeginning of the storm, the wind speed was such that it generated only 17 s waves. Andlet’s say that at the en d of the storm, after time δt has elapsed, the wind speed was suchthat it generated on ly 19 s waves. Even though 17 s waves have a head-start in time, the19 s waves travel faster than the 17 s waves and therefore both 17 and 19 s waves arr iveat the beach on the same day (Sunday). Both sets of waves travel the same distance:vgp,17∆tSunday= vgp,19(∆tSunday− δt) (15)which we can solve for∆t = ∆tSunday 1 −vgp,17vgp,19!= 12 days1 −1719= 1 day . (16)Now 1 day and 800 km (radius) are upper limits because the truth lies in b etween ourdual assumptions of 0-d uration/finite-size and 0-size/finite-duration.Problem 3. Tidal Disruption and the Roche Zone4This problem examines why ring systems about all the giant planets occu py planetocentricdistances that are less than ∼2 planetary radii.a) Consider a perfectly rigid, spherical satellite of