5.3 Solutions
229
5.48
1
p 2 =
1
r 2 +
1
r 4
dr
dθ
2
(1)
r = a sin nθ
(2)
dr
dθ
2
= n
2 a
2
(1 − sin
2 nθ) = n
2 a
2
1 −
r 2
a 2
(3)
Using (3) in (1)
1
p 2 =
n 2 a 2
r 4 +
1 − n 2
r 2
Differentiating
−
2
p 3
d p
dr
= −
4n 2 a 2
r 5 −
2(1 − n 2 )
r 3
or
1
p 3
d p
dr
=
2n 2 a 2
r 5 +
1 − n 2
r 3
Force per unit mass
f = −
h 2
p 3
d p
dr
= −h
2
2n 2 a 2
r 5 +
1 − n 2
r 3
5.49 1
p 2 =
1
r 2 +
1
r 4
dr
dθ
2
(1)
r = a(1 − cos θ)
dr
dθ
= a sin θ
dr
dθ
2
= a
2 sin
2
θ = a
2
1 −
1 −
r
a
2
= 2ra − r
2
(2)
Using (2) in (1)
1
P 2 =
2a
r 3
or p
2
=
r 3
2a
(3)
Force per unit mass
f = −
h 2
p 3
d p
dr
(4)
Differentiating (3)