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1.207. M= m V2ymsrir2/(ri ±r2), where ms is the mass of
the Sun.
1.208. E = T U = —ymms12a, where ms is the mass of
the Sun.
1.209. r m = 2 — [1. ±111 — (2 — T1) 11 Sill2 cd, where =
rov20/7ms, ms being the mass of the Sun.
1.210. rmin = (Trasiv0) [111 (//,02/yrns)2—11,^ where^ ms^ is the
mass of the Sun. 1.211. (a) First let us consider a thin spherical layer of radius p and mass SM. The energy of interaction of the particle with an ele- mentary belt SS of that layer is equal to (Fig. 8)
dU = —y (m8M/2/) sin 0 dO.
According to the cosine theorem in the triangle OAP 12 = p
r2— 2pr cos 0. Having determined the differential of this expres- sion, we can reduce Eq. (*) to the form that is convenient for integ- ration. After integrating over the whole layer we obtain SU =
= —ym 61111r. And finally, integrating over all layers of the sphere,
we obtain U = —ymM/r; (b) Fr= —0U/Or = —ymM/r2.
dm2^ -fag
Fig. 8. Fig. 9.
1.212. First let us consider a thin spherical layer of substance (Fig. 9). Construct a cone with a small angle of taper and the vertex
at the point A. The ratio of the areas cut out by the cone in the layer
is dSi : dS2 =71 : 71. The masses of the cut volumes are proportion-
al to their areas. Therefore these volumes will attract the particle A
with forces equal in magnitude and opposite in direction. What follows is obvious.
1.213. A = —3/2ymM/R.
—(yMIR3)r^ for^ r<R,
01)= — 1r for r> R. See Fig. 10.
1.215. G = —4/33-typ1. The field inside the cavity is uniform.
1.216. p= 318(1 —^ r2/R2)?mainR4. About 1.8.108atmospheres.
1.214. G=
— (yM/r3) r for r >R;
— 312 (1— r213112)TMIR^ for^ r<R,
19-
1.217. (a) Let us subdivide the spherical layer into small ele- ments, each of mass Sm. In this case the energy of interaction of
each element with all others is SU = —vm SmIR. Summing over all
Fig. 10.
elements and taking into account that each pair of interacting ele-
ments appears twice in the result, we obtain U = —vm2/2R;
(b) U = —3ym2/5R.
r3/2 f 4.5 days (6 = 0),
1.218. At ^...;
2x
3Ar/2r - 1 - 8 1 0.84 hour (6= 2).
1.219. w1: w2: w3= 1 : 0.0034 : 0.0006. 1.220. 32 km; 2650 km.
1.221. h = RI(2gRiv: — 1).
1.222. h = R (gR/v2— 1).
1.223. r = ryM (77231)2= 4.2.104km, where M and T are the
mass of the Earth and its period of revolution about its own axis respectively; 3.1 km/s, 0.22 m/s2.
1.224. M = (4n2R3/?T2) (1 T/'02 = 6.1024 kg, where T is
the period of revolution of the Earth about its own axis.
1.225. v' = 2n T
R V TM = 7.0 km/s, R R2^ (^1 +..x
X V = 4.9 m/s2. Here M is^ the mass of the Earth,^ T is^ its
period of revolution about its own axis. 1.226. 1.27 times.
1.227. The decrease in the total energy E of the satellite over the
time interval dt is equal to —dE = Fy dt. Representing E and v as
functions of the distance r between the satellite and the centre of the
Moon, we can reduce this equation_to the form convenient for integ-
ration. Finally, we get 't WI — 1) mlal gR
1.228. v1= 1.67 km/s, v2= 2.37 km/s.
1.229. Au= liyM/R (1 —1/- ) = — 0.70 km/s, where^ M^ and^ R
are the mass and the radius of the Moon.
1.230. Av = (1/- —1) = 3.27 km/s, where g^ is the stan-
dard free-fall acceleration, R is the radius of the Earth.
1.231. r = nR/(1 Vri). 3.8.104 km.
Fig. 11.
1.269. N = 1 /24m012sin 20.
1.270. cos 0 = 3 /2 gl 01.
1.271. Ax = 1/2 ka.
1.272. v' = 0)01/V1 + 3m/M.
1.273. F =a1232Im1= 9 N.
3m — 4M My
1.274. (a) v — 3m +4MV; (b) F — 1(1±431/3m),
1.275. (a) v = (M/m) - 112/3g/ sin (a/2);
(b) Ap = MYthig/ sin (a/2); (c) x 2/31.
1.276. (a) co = (1 -I- 2m/M) 0)0; (b) A = 12mco:R2(1 + 2m1 M).
1.277. (a) cp — 2m12+m1m2q)'; (b) Nz=^2 mmlimi!Rin2^ ddv;
1.278. (a) w — h1 ; (b) A-
2(/111+12/2) W2)2" 1.279. v' = v (4 — 11)/(4 -I- II), (^) = 12v// (4 + TO. For it = 4 and 11>4.
1.280. (a) Age = '/2/:(0:/(I /07 A180. = 2/:co://; (b) N =
1.281. co = V 6.0 rad/s; F =mg1011= 25 N.
1.282. (a) M = I/12 mw/2 sin 0,^ M, = M sin 0. (b) AM I
1 112m0)12 sin 20; (c)^ N = 1l24me)212 X
X sin 20.
1.283. (a) w' = mg11.1w = 0.7 rad/s;
(b) F = mo.)121 sin 0 = 10 mN. See Fig. 11.
1.284. w = (g w) IlanR2 = 3 x
x 102rad/s.
1.285. co' = ml Vg2 w2/ =
= 0.8 rad/s. The vector w' forms the
angle 0 = arctan (w/ g) = 6° with the ver-
tical.
1.286. F' = 215 mR20)co'// = 0.30 kN.
1.287. Fmax = nmr2q)m0)//T=0.09 kN.
1.288. N = 2nnIvIR = 6 kN•m.
1.289. Fodd = 2nnIvl RI = 1.4 kN. The force exerted on the
outside rail increases by this value while that exerted on the inside one decreases by the same value.
1.290. p = aE AT =^ 2.2.102atm, where a is the thermal expan-
sion coefficient.
1.291. (a) p amAr/r = 20 atm; (b)^ p^ 2amAr/r = 40 atm.
Here amis the glass strength.
1.292. rt=j12amlpInl= 0.8.102rps, where amis the tensile
strength, and p is the density of copper.
1.293. n = Vam/p/2nR = 23 rps, where am^ is the tensile
strength, and p is the density of lead.
1.294. x l Vmg/23-c^ (PE=^ 2.5 cm
1.295. s = 2 /2F0/ES.
1.296. T = 11277-14o21 (1 -.- r2/12), Al^ = 1/3p.o2l3IE, where p is the density of copper. 1.297. AV'--= (1 — 2μl FlIE = 1.6 mm3, where p. is Poisson's ratio for copper.
is the density, and p, is Poisson's ratio for copper.
1.298. (a) Al = 1/2pg/2/E; (b) AVIV = (1 — 2p,) 6.111, where p
1.299. (a) ATI/V = —3 (1 — 20 pIE; (b) 13 = 3 (1 — 20/E. 1.300. R = 116 Eh2/pgl2=-- 0.12 km, where p is the density of steel. 1.301. (a) Here N is independent of^ x^ and equal to No. Integrat- ing twice the initial equation with regard to the boundary condi- tions dyldx (0) = 0 and^ y^ (0) = 0, we obtain y^ 0 12E1)^ x2. This is the equation of a parabola. The bending deflection is X = =N 0/2/2E/, where I = a4/12. (b) In this case N (x) = F (1 — x) and y = (F12EI) (1 — x13) x2; = F1313EI, where I is of the same magnitude as in (a). 1.302. 1. = F13148EI. 1.303. (a) X = 3/2 pg141Eh2; (b) = 5/2 pg141Eh2. Here p is the density of steel. 1.304. X, = 2/513p15/Eh2, where p is the density of steel. 1.305. (a) q=- (l/2nr3 ArG)- N;^ (b)^ (211 a-v.4G) • N. 1.306. N = n (d: — (4) G(p/32/ = 0.5 kN•m. 1.307. P = i/2 ar4G(pco = 17 kW. 1.308. N = 1/2 pm (r: - (^) r4)I(r22— 1.309. U = 1 /2mE82/p = 0.04 kJ, where p is the density of steel. 1.310. (a) U =^ 1/onr2 /3o2g2/E; (b) U = 2/3 nr2lE (A111)2.^ Here p is the density of steel. 1.311. A (^) 116 n211(53E11 = 0.08 kJ. 1.312. U =^1 /4 ar4Gy2// = 7 J. 1.313. u = 1/2GT2r2//2. 1.314. u = 1/213 (pgh)2 = 23.5 kJ/m3, where 13 is the compressi- bility. 1.315. pi > p2, v1< v2. The density of streamlines grows on transition from point 1 to point^ 2. 1.316. Q = S1S2 1/^20111 (S22 — Si). 1.317. Q = Si/ 2gAhp0Ip. 1.318. v= 2g (hi+ hzPziPi) = 3 m/s, where piand p2are the densities of water and kerosene. 1.319. h = 25 cm; lmax =50 cm. 1.320. h 1 /2 v2/g — ho =^ 20 cm. 1.321. p = Po pgh (1. — R:/r2), where R1 <r <R2, (^) Pois the atmospheric pressure. 1.322. A =^1 /2073/s2t2, where p is the density of water. 1.323. r = V2h/g S/s. 1.324. v = colt 17211h-1. 1.326. F = 2pgS c1 h = 0.50 N. 1.327. F= pgbl (2h — 1) = 5 N.
