It is interesting to examine the y-velocity at the extremes:
0
0
0
0
g
For 2.04 m/s,
42.049.81
c
vt
v
==
−
()
,
0.385s 3.31 m/s
cy
tv==!
2
d
0
0
3.11m/s :2.81 m
v
=
C
A
x
d/2
d
√
2g
2d
Solution 3.219
823
−
Solution 3.220
Solution 3.221
2
2
m/d
ssms
FGm
−
=
Moon m
Solution 3.222
Speed in circular orbit:
3
g9.825
p
max
g(1)
r
Solution 3.223
From Appendix D, for the moon
2
max
min
080 mi,g 5.32 ft/sec
1080 180 1260 mi
1080 60 1140 m
1
so
i
R
r
r
=
=+=
=
+=
=
min
5.32 1 1260
p
p
ar
Solution 3.224
408m/s
=
9
6.67(10 ) J
=
2
33
2
23
2
22
12
2
12
22 11
12
2.39(10 )J
9.825
6371(10 ) 7.73(10 ) m/s
(6371 300)(1000)
11
80 000 7.73(10 )
22
mg 80 000 (9.825)(6371 1000) 4.78(10 )J
(6371 300) 1000
( ) 2.61(10
r
Tmv
R
Vr
ETV TV
==
+
−⋅
=− =−
+⋅
Δ= + − + =
== =
=
)J
Solution 3.225
min
2, 3
2.5 2 5
p
rRrR
aR
==
The velocity in the original circular orbit is
gg1
g
cRR R
V==
=
28.5°
ω
R
Solution 3.226
min
g 9.825
(
(
g 1 9.825 1 0.1
1.1(6371)
(
c) 70 081 km
6
p
e
r
ae
v
++
== =
=
T
9.825 1 0.9
371(1000) 70 081(1000) 1 0.9
10398 m/s
(
v
+
−
==
Solution 3.227
()
22
w
(3 / 47),
S
peed at in parabolic orb
it
c
a
P
=∞
f
1cos
p
d
θ
Solution 3.228
g 32.23
(3959)(5280) 23,676 ft/sec
A
vR
== =
Momentum conservation during impact:
().But ,so
123,796 ft/sec
AA BB A B C A B
mv mv m m v m m
+=+ =
Solution 3.229
3/2 3/2
93/2
22
E
q. 3.44 : g
f
A
aa
RGm
ππ
τ
==
Solution 3.230
232
C
g
gives
E
q.
3.47with
v
arRH
RRH
=+
== +
R
β/2
υ
υo
E
Solution 3.231
ω
=
=+
Let angular velocity of
s
A
V
OA R
H
H
Solution 3.232
The orbiter speed in the initial elliptical orbit, at A, is
min
max
g
A
r
v
Rar
=
Solution 3.233
Circular orbit speed
gg1
g
vR R R
== =
=
2
284 m s(directed rearward)
r
Call the circular orbit period
τ
0 and the elliptical orbit period
τ
AB
3/2 3/2
3/2
0
23 98.0
τ
°
Solution 3.234
cos 2000 cos 30 1732 m/s
vv
α
== °=
7
7
6.371 10
6.2595 10 m
=− ×
24
mg
R
=×
MAX
6
6.4249 10 m
max max 53 900 m or 53.9 kmhrR
υ
B
Solution 3.235
Speed in circular orbit is
g 32.23
Time required for B to return to Cs
2
11
A
t apogee, 2g 24,156 ft/sec
2
C
vR
ra
=−=
Solution 3.236
g
C
irculer orbit vR
=
()
2 6000 ft
F
Substitute conditions at B to find a = 2.1403 (107) ft
A
g1
Use to obtain 0.02599
1
e
vR e
ae
−
==
+
Solution 3.237
3/2
2
a
π
Solution 3.238
2
From Eq. 3.43,
11 cos
e
θ
+
=−
()
()
2
1
RH a e
ae
++
−
From Eq. 3.48
()
g1 g
1
eRR
R
−
++
Solution 3.239
Eq. 3.47 at perigee P:
2
3
11
a
() ()
ωω
−
−−
=−= −
=
4
3
8179 465 0.7292 10
150 10
0.0514 rad/s
vR
PH
A
N
H
υ
A
RH
P
ω
Solution 3.240
Fs: force exerted on spacecraft by sun
Fe: force exerted on spacecraft by earth
2
() ()
()
23 6
me 4.095 10 slugs, 92.96 10 5280 ft,
D
==
Solution 3.241
For 1, g ; for 2, gvR r vR r==
F
12
121
F
or transfer ellipse at , g 2
rr
Av R a r r a
+
′==
+
42171 6871 42171
B
m
F
s
F
e
n
Solution 3.242
E
q. 3.47 : 2g 2
(7791) 2(9.825)(6371 1000) 6571(1000) 2
vR
a
=−
=⋅ −
(Semi major axis is parallel to x-axis.)
6
min
6
max
max
min
min
max
(1 ) 6.49(10 ) m
(1 ) 6.66(10 ) m
g7890 m/s
g7690 m/s
p
a
rae
rae
r
vR
ar
r
vR
ar
−=
+=
=
=
=
=
=
=
Solution 3.243
θ
= ±72.8°
2
2
11
ra
a
Solution 3.244
A
t, 29
2g 2
Br R
vR
ra
=
=−
6
7
29 6.371(10 )
2
0.9295
=
111.8
θ
θ
=°
111.8 2.7 109.1
β
=−=°
α
β
158.2
D
α = 21.8°
Solution 3.245
Truck bed is a constant-velocity frame of reference
1/2 rel
12
mm
Solution 3.247
rel
rel rel
2
0.5(2) 1m/s
(3)(1) 1.5
22
kg m
vl
Tmv J
θ
=
==→
==
⋅
=
y
100(9.81) N 100 15
Solution 3.248
=Δ
rel rel
Rel. to carrier
UT
Solution 3.249
Solution 3.250
sin165 sin
90 16
α
°=
4000 lb
Solution 3.251
stop 2
/
15 m/s
For truck, 0.9g, 1.699s
T
CT C T
at
=
−=
(as long as truck is moving)
2
stop
At ,
1
tt
=
Solution 3.252
22
11
s
F = Constant
x
o
xx
. = υ