2.3 Solutions
97
Using (5) in (4)
dv
dt
= −
α v r
m o − αt
dv =
(v r α/m 0 )dt
1 −
α
m 0
t
Integrating between v = 0 and v
v = −v r ln
1 −
αt
m 0
(6)
Writing
1 −
αt
m 0
=
m
m 0
in (6) with the aid of (5), the rocket equation
simplifies to
v = −v r ln
m
m 0
or m = m 0 e
−v/v r
(7)
(d) Time taken for the rocket to reach the burn-out velocity is given by (5):
t = t 0 =
m 0 − m
α
(8)
2.61 a = 0.5 g =
v r
m
dm
dt
− g
dm
dt
= 1.5
mg
v r
=
1.5 × 10 6 × 9.8
2000
= 7350 kg/s
2.62 (a) Rocket thrust = v r
dm
dt
= 55 × 10 3 × 1290 = 71 × 10 6 N.
(b) Net acceleration a =
v r
m
dm
dt
− g =
71 × 10 6
2.72 × 10 6 − 9.8 = 16.3 m/s 2 .
(c) Time to reach the burn-out velocity t =
m 0 − m B
α
=
2.72 × 10 6 − 2.52 × 10 6
1290
= 155 s.
(d) Burn-out velocity v B = v i + v r ln
m 0
m β
− gt
= 0 + 55, 000 ln
2.72 × 10 6
2.52 × 10 6 − (9.8 × 155) = 2714 m/s = 2.7 km/s
.
2.63 (a) Weight of the rocket
M 0 g = 5000 × 9.8 = 49, 000 N