1.3 Solutions
39
Fig. 1.27
m 1
r 1 + m 2
r 2
m 1 + m 2
= 0
m 1 r 1 − m 2 r 2 = 0
r 1 =
m 2 r 2
m 1
=
m 2 (r − r 1 )
81m 2
=
60R − r 1
81
r 1 = 0.7317R = 0.7317 × 6400 = 4683 km
along the line joining the earth and moon; thus, the centre of mass of the
earth–moon system lies within the earth.
1.48 Let the centre of mass be located at a distance r c from the carbon atom and at
r 0 from the oxygen atom along the line joining carbon and oxygen atoms. If
r is the distance between the two atoms, m c and m o the mass of carbon and
oxygen atoms, respectively
m c r c = m o r o = m o (r − r c )
r c =
m o r
m o + m o
=
16 × 1.13
12 + 16
= 0.646 Å
1.49 Let C be the centroid of the equilateral triangle formed by the three H atoms in
the xy-plane, Fig. 1.28. The N–atom lies vertically above C, along the z-axis.
The distance r CN between C and N is
r CN −
r 2
NH 3
− r 2
CH 3
r CN =
r 2
H 1 H 2
√
3
=
1.628
1.732
= 0.94 Å
r CN =
(1.014) 2 − (0.94) 2 = 0.38 Å
Now, the centre of mass of the three H atoms 3m H lies at C. The centre of
mass of the NH 3 molecule must lie along the line of symmetry joining N and
C and is located below N atom at a distance