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 allFig. 10.elements and taking into account that each pair of interacting ele-
ments appears twice in the result, we obtain U = —vm 2 /2R;
(b) U = —3ym 2 /5R.
r3/2^ f 4.5 days (6 = 0),
1.218. At ^...;2x3Ar/2r- 1 - (^8 1) 0.84 hour (6= 2).
1.219. w 1 : w 2 : w 3 = 1 : 0.0034 : 0.0006.
1.220. 32 km; 2650 km.
1.221. h = RI(2gRiv: — 1).
1.222. h = R (gR/v^2 — 1).
1.223. r = ryM (77231) 2 = 4.2.10 4 km, 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/s^2.
1.224. M = (4n 2 R 3 /?T 2 ) (1 T/'0 2 = 6.10 24 kg, where T is
the period of revolution of the Earth about its own axis.
1.225. v' = 2nT
R V TM = 7.0 km/s,
R^ R2 (^1 + ..x
X V = 4.9 m/s 2. 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. v 1 = 1.67 km/s, v^2 = 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). 3.8.10 4 km.