1.207. M= m V2ymsrir 2 /(ri±r 2 ), where ms is the mass of
the Sun.
1.208. E = T U = —ymms12a, where ms is the mass of
the Sun.
1.209. rm = (^2) — [1. ±111 — (2 — T1) 11 Sill 2 cd, (^) where =
rov 20 /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 1 2 = p 2
r 2 — 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/r 2.
dm^2 -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 : dS 2 =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 / 2 ymM/R.
—(yMIR^3 )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 = 31^8 (1 — r2/R2)?mainR4. About 1.8.10^8 atmospheres.
()
1.214. G=
— (yM/r 3 ) r for r >R;
— 312 (1— r^213112 )TMIR for r<R,
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