5.3 Solutions
227
Fig. 5.19
dt
d θ
=
r 2
h
Time taken for the object to move from P 1 to P 2 (Fig. 5.19) is given by
t =
dt =
r 2 d θ
h
=
4a 2
h
θ
0
d θ
(1 + cos θ) 2
=
a 2
h
θ
0
sec
4
θ
2
d θ =
2a 2
h
θ
0
1 + tan
2 θ
2
d
tan
1
2
θ
=
2a 2
h
tan
1
2
θ +
1
3
tan
3 1
2
θ
But h =
√
G M × semi - latus rectum =
√
2aG M
∴ t =
2a 3
G M
tan
1
2
θ +
1
3
tan
3 1
2
θ
5.47 Required time for traversing the arc PQT is obtained by the formula derived
in problem (5.46), Fig. 5.20
t 0 = 2t = 2
2a 3
G M
tan
1
2
θ +
1
3
tan
3 1
2
θ
(1)