5.3 Solutions
221
ε =
1 −
R
a
+
1
2
R 2
a 2
(5)
Now
R
a
=
6400
80000
= 0.08
∴ ε = 0.96
5.37 The resultant velocity v of each fragment is obtained by combining the velocities
1
2
v 0 and v 0 vectorially, Fig. 5.17.
Fig. 5.17
v =
1
2
v 0
2
+ v 2
0 =
1
2
√
5v 0
Kinetic energy of each fragment
K =
1
2
m
2
√
5
2
v 0
2
=
5
16
mv
2
0 =
5
16
m
G M
r
Potential energy of each fragment U = −
G M
1
2
m
r
∴ Total energy E = K + U =
5G Mm
16r
−
1
2
G Mm
r
= −
3
16
G Mm
r
If v makes on angle θ with the radius vector r, then v sin θ = v 0 . The angular
momentum of either fragment about the centre of the earth is
J =
1
2
mv 0 r =
mr
2
G M
r
=
1
2
m
√
G Mr