PHYS-333: Problem set #2 Solutions

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Aubrey Boone
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1 PHYS-333: Problm st #2 Solutions Vrsion of March 5, Visual binary 15 points): Ovr a priod of 10 yars, two stars sparatd by an angl of 1 arcsc ar obsrvd to mov through a full circl about a point midway btwn thm on th sky. Suppos that ovr a singl yar, that midway point is obsrvd itslf to wobbl by 0.2 arcsc du to th parallax from arth s own orbit. a. How many pc is this star systm from arth? Wobbl is twic parallax angl, so p= 0.1 arcsc, which implis distanc of d = 1/0.1 = 10 pc. b. What is th physical distanc btwn th stars, in au. Angular sparation α = 1 arcsc, so s/au = d/pc) α/arcsc) = 10*1 = 10. Sparation is s = 10 au. c. What ar th masss of ach star, M 1 and M 2, in M. Kplr s 3rd law says total mass M 1 + M 2 = a 3 au/p 2 yr = 10 3 /10 2 = 10M. But sinc cntr of mass is midway btwn stars, thy hav qual masss, and thus M 1 = M 2 = 5M.

2 2 2. Doubl-lin spctroscopic binary 15 points): A doubl-lin spctroscopic binary has a spctral lin with rst wavlngth of 600 nm, which howvr is obsrvd to split into two qual pairs on both th blu and rd sids, raching a maximum wavlngth sparation of +/ nm from th rst wavlngth vry 3 months. a. What is th orbital spd in km/s) of th two stars around thir common cntr of mass? Assum a circular orbit and that w ar viwing this systm along th stars orbital plan.) By Dopplr shift formula, V 1 = V 2 = z c = c λ/λ = 0.12/600) c = km/s = 60 km/s = V 1 = V 2. b. What is th mass of ach star, in units of th sun s mass M? Hint: us th answr to part a and th vlocity form of Kplr s 3rd law, in trms of th arth s orbital spd around th sun.) First, vry 3 months is half th priod, so P = 0.5 yr. From DocOnots q. 9.8, vlocity form of Kplr s 3rd law for star 1 is M 1 M = V2 V ) 3 P yr 1 + V 1/V 2 ) 2 = ) ) 2 = 16. 1) 30 Sinc V 1 = V 2, th masss ar qual, and w hav M 1 = M 2 = 16 M. c. What is th sparation btwn th two stars, in au? Orbital spd V 1 = 2πa 1 /P so a 1 = V 1 P/2π = V 1 /V )P/2π)2πau)/yr = = 1au, and thus a = a 1 + a 2 = 2au.

3 3 3. Escap spd and nrgy 15 points): a. Rcalling that th surfac scap spd from th sun is about 620 km/s, comput V sc from stars with: 1) M = 10M and R = 10R ; 2) M = 1M and R = 100R ; 3) M = 1M and R = 0.01R? v sc = 620 km/s M/M )/R/R ). 1) v sc = 620 km/s 10/10 = 620 km/s 2) v sc = 620 km/s 1/100 = 62 km/s 3) v sc = 620 km/s 1/0.01 = 6200 km/s b. To what radius R c in km) would you hav to shrink th sun to mak its scap spd qual to th spd of light c? Rquir c = v sc = 2GM/R, solv for radius, R = 2GM/c 2 = 3km M/M ). c. Do you think it possibl for stars to hav V sc = c? Ys, thy r calld black hols.

4 4 4. Gravity and circular orbits 25 points): a. Evaluat th circular orbital vlocity V lo of an objct in Low-Earth Orbit LEO), i.. orbiting th Earth just abov th arth s atmosphr, say at a hight h 200 km abov sa lvl. Exprss your answr in km/s and in km/hr. V lo = GM / + h) GM / 7.9 km/s = 28, 000 km/hr. b. What is associatd orbital priod P lo, xprssd in sconds, hours, and days. P lo = 2π /V lo = 5000 s = 1.4 hr = 0.06 day. c. What is th ratio of V lo to th rotational vlocity of th Earth s quator? How is this rlatd to P lo valuatd in days? V rot = 2π /day = 0.46 km/s. So V rot /V lo = 0.06 = P lo day). d. At what radius from th cntr of th Earth would an objct in circular orbit hav an orbital priod of 1 day, so that it rmains fixd at a givn position abov th Earth s quator. This is th radius R g for so-calld gosynchronous orbit, or GEO. Exprss your answr in both km and in Earth radii. Lt Ω = V rot / = 2π/day. W thn hav Ω 2 R g = GM /R 2 g and so R g = GM /Ω 2 ) 1/3 = V orb /V lo ) 2/3 = 6.6 = 42, 000 km.. In trms of th arth s mass M and radius, driv an xprssion for th total nrgy E go rquird to plac an objct of mass m into GEO. Th total nrgy is th kintic nrgy of orbit mvg 2 /2, plus th work to lift from to R g. Noting that Vg 2 = GM /R g, w hav m = V g GM 1 ) R g E go = GM 1 R ) = E go 2R g m. 2) f. Evaluat this for a singl Hydrogn atom of mass m H, in both Jouls and lctron volts V). Compar this to th nrgy E sc,h rquird for a Hydrogn to compltly scap th arth. Th nrgy for Hydrogn to compltly scap is E sc,h = GM m H = J = 0.66 V, 3) Th nrgy to rach GEO is a factor 1 /2R g = 0.92 lss, or E go,h = 0.6 V = J. 4)

5 5 5. Challng Problm: Spac Elvator 30 points): Thr ar srious nginring proposals to build a spac lvator cntrd on a cabl that xtnds from th arth s quatorial surfac radius to th gosynchronous orbit radius R g computd in th prvious problm. a. For a cabl with mass-pr-lngth µ, th gravitational wight of th cabl sction blow any givn radius r must b supportd by th cabl s local tnsion forc T g r). Ignoring cntrifugal forcs, but accounting for th radial variation of arth s gravitational acclration, driv an xprssion for T g r) for r R g. A cabl lmnt of lngth dr at radius r has mass dm = µdr and wight dw = dm gr) = GM µdr/r 2, which incrass th cabl tnsion by dt g = dw. Intgration from th arth s radius to r thus givs th gravitational tnsion at r, r GM µ T g r) = dr = GM µ 1 R ). 5) r 2 r b. Th strngth of cabl matrial can b charactrizd by its braking lngth l b, dfind as th lngth that an untaprd cabl will brak undr its own wight in arth s constant surfac gravity g = 9.8 m/s 2. Comput th braking lngth rquird to support th gravitational tnsion at GEO, T g R go ), from part a. Writ this in trms of arth s radius. Th minimum rquird braking lngth must allow support of th tnsion at GEO, so l b µg = T g R g ). Using th rsult from part a), w can solv for th braking lngth as l b = T gr g ) µg ) = GM µ/ µgm /R 2 1 R ) = 1 R ) = 0.85 = 5400 km. 6) R g R g c. Considr thn a strand of cabl that consists of a singl chain of atoms of mass m a sparatd by a distanc s, for which thus µ = m a /s. For binding nrgy pr atom E b, th associatd forc binding thm can b approximatd by F E b /s sinc Enrgy = Forc x Distanc). Driv thn an xprssion for th strand s braking lngth l b in trms of E b, m a, and g. Stting th forc qual to th cabl wight at braking lngth undr arth s gravity, w hav F = E b /s = µl b g, which upon using µ = m a /s, solvs to l b = E b m a g. 7)

6 6 d. Now us this to driv th binding nrgy pr unit mass, E b /m a in V/m H ), rquird to support th rquird tnsion at GEO, T g R g ), as drivd in b). For m a = m H, compar this rquird binding nrgy E b to th Hydrogn scap nrgy from th arth, E sc,h, as computd in problm 4f. E b m a = l b g = 0.85 GM R 2 = 0.85 GM = 0.56 V m a. 8) If m a = m H, thn E b = 0.85 E sc,h.. Finally, rfrring to th Wiki pag strngth What matrial has a braking lngth that might qualify it to build a cabl to GEO? Carbon nanotubs hav a braking lngth l c = 4716 km. If taprd to incrasing thicknss upward, thy could in principal provid th basis for building a spac lvator to GEO. h. EXTRA CREDIT: A cabl to GEO would hav a 1-day rotation priod. By what fraction is th rquird braking lngth rducd if on accounts for cntrifugal acclration in rducing th gravitational tnsion from part b)? Th cntrifugal acclration at radius r is givn for rigid rotation of priod P by g c = 4π 2 r/p 2. Intgrating µg c from th arth s surfac to GEO givs for th cntrifugal tnsion rduction, Rg [ R 2 g T c R g ) = 4π2 µ P 2 rot rdr = µv 2 2 R 2 ] 1, 9) whr V rot = 2π /P is th arth s surfac rotation spd. Noting that V rot / ) 2 R g = GM /Rg, 2 w find that th ratio of th cntrifugal to gravitational tnsion at GEO is T c R g ) T g R g ) = 2R g 1 + R ) = 0.09, 10) R g which implis that cntrifugal acclration rducs th nt cabl tnsion at GEO by just 9% rlativ to th purly gravitational rsult computd in b). This maks th rquird braking lngth km, or just a littl abov th braking lngth 4700 km) of Carbon nanotubs. Again, tapring th cabl could in principl mak it all work.

Hydrogen Atom and One Electron Ions

Hydrogen Atom and One Electron Ions Hydrogn Atom and On Elctron Ions Th Schrödingr quation for this two-body problm starts out th sam as th gnral two-body Schrödingr quation. First w sparat out th motion of th cntr of mass. Th intrnal potntial