361
13–117.
SOLUTION
For and ,
Eliminating C, from Eqs. 13–28 and 13–29,
From Eq. 13–31,
Thus,
Q.E.D.T2=a4p2
GMsba3
4p2a3
T2h2=GMs
h2
b2=T2h2
4p2a2
T=
p
h(2a)(b)
2a
b2=2GMs
h2
1
ra
=-C+GMs
h2
1
rp
=C+GMs
h2
u=180°u=0°
Prove Kep
l
er’s t
hi
r
d
l
aw of mot
i
on. Hint: Use Eqs. 13–19,
13–28, 13–29, and 13–31.
GMsb
13–118.
The satellite is moving in an elliptical orbit with an eccentricity
.Determine its speed when it is at its maximum
distance Aand minimum distance Bfrom the earth.
e=0.25
SOLUTION
.
where .
Ans.
Ans.vA=
rp
ra
vB=8.378(106)
13.96(106)(7713) =4628 m>s=4.63 km>s
ra=r0
2GMe
r0v0
–1
=8.378(106)
2(66.73)(10–12)(5.976)(1024)
8.378(106)(7713)2–1
=13.96
A
106
B
m
vB=v0=
C
66.73(10–12)(5.976)(1024)(0.25 +1)
8.378(106)
=7713 m>s=7.71 km>s
r0=rp=2
A
106
B
+6378
A
103
B
=8.378
A
106
B
m
r0v2
0
GMe
=e+1v0=BGMe(e+1)
r0
e=
¢
r0v2
0
GMe
–1
≤
e=1
GMer0
¢
1–GMe
r0v2
0
≤
(r0v0)2
where C=1
r0
¢
1–GMe
r0v2
0
≤
and h=r0v0
e=Ch2
GMe
B
A2Mm
363
13–119.
The rocket is traveling in free flight along the elliptical orbit.
The planet has no atmosphere, and its mass is 0.60 times that
of the earth. If the rocket has the orbit shown, determine the
rocket’s speed when it is at A and at B.
SOLUTION
18.3 Mm 7.60 Mm
BA
O
*13–120.
D
etermine the constant speed of satellite Sso that it
circles the earth with an orbit of radius . Hint:
Use Eq. 13–1.
r=15 Mm
SOLUTION
Ans.y=AGme
r=B66.73(10–12)a5.976(1024)
15(106)b=5156 m>s=5.16 km>s
msay2
0
rb=Gmsme
r2
F=Gmsme
r2Also
F=msay2
s
rbHence
S
r15 Mm
13–121.
The rocket is in free flight along an elliptical trajectory .
The planet has no atmosphere,and its mass is 0.70 times that
of the earth.If the rocket has an apoapsis and periapsis as
shown in the figure,determine the speed of the rocket when
it is at point A.
A
¿
A
SOLUTION
, we have
Ans.y
A=7471.89 m>s=7.47 km>s
9
A
106
B
=6(10)6
¢
2(66.73) (10–12)(0.7) [5.976(1024)]
6(106)y2
P
≤
–1
M=0.70Me
6M
m9
Mm
BA A¿
r3Mm
O
13–122.
The Viking Explorer approaches the planet Mars on a
parabolic trajectory as shown. When it reaches point Aits
velocity is 10 Mm h. Determine and the required velocity
at Aso that it can then maintain a circular orbit as shown.
The mass of Mars is 0.1074 times the mass of the earth.
r0
>
SOLUTION
A
13–123.
The rocket is initially in free-flight circular orbit around the
earth. Determine the speed of the rocket at A. What change
in the speed at A is required so that it can move in an
elliptical orbit to reach point
A′
?
A¿
A
O
8 Mm
19 Mm
SOLUTION
SOLUTION
A′
A′
2
2
*13–124.
The rocket is in free-flight circular orbit around the earth.
Determine the time needed for the rocket to travel from the
inner orbit at A to the outer orbit at
A′
.
A
¿
A
O
8 Mm
19 Mm
369
13–125.
SOLUTION
(a)
Ans.
(b)
Ans.
(c)
(d)
Ans.
Ans.r7392(103)mi
e71
194(103)mi6r6392(103)mi
e61
r=396(103)–3960 =392(103)mi
r0=2GMe
v2
0
=2(14.07)(1015)
[3.67(103)]2=2.09(109)ft=396(103)mi
1
GMe
(r2
0v2
0)a1
r0b
¢
1–GMe
r0v2
0
≤
=1
e=C2h
GMe
=1
r=1.047(109)
5280 –3960 =194(103)mi
r0=GMe
v2
0
=14.07(1015)
[3.67(1013)]2=1.046(109)ft
=14.07(1015)
GMe=34.4(10–9)(409)(1021)
1=GMe
r0v2
0
e=C2h
GMe
=0 or C=0
v0=2500 mi>h=3.67(103)ft>s
A satellite is launched with an initial velocity
parallel to the surface of the earth.
Determine the required altitude (or range of altitudes)
above the earth’s surface for launching if the free-flight
trajectory is to be (a) circular, (b) parabolic, (c) elliptical,
and (d) hyperbolic. Take
the earth’s radius and
1mi=5280 ft.
re=3960 mi,Me=409110212slug,
G=34.4110–921lb #ft22>slug2,
v0=2500 mi>h
370
SOLUTION
6
6
13–126.
The rocket is traveling around the earth in free flight along
the elliptical orbit. If the rocket has the orbit shown,
determine the speed of the rocket when it is at A and at B.
B
A
13–127.
An elliptical path of a satellite has an eccentricity
If it has a speed of when it is at perigee,
P, determine its speed when it arrives at apogee, A.Also,
how far is it from the earth’s surface when it is at A?
15 Mm>he=0.130.
SOLUTION
Ans.
Ans. =27.3 Mm
d=33.7111062 – 6.37811062
=11.5 Mm>h
=15125.96211062
33.7111062
nA=
n0r0
rA
=33.7111062 m=33.7 Mm
=25.961106211.1302
0.870
rA=r01e+12
1–e
rA=r0
2GMe
r0n2
0
–1
= r0
A
2
e+1
B
–1
GMe
r0 n2
0
=1
e+1
=25.96 Mm
=1.130166.732110–12215.9762110242
C
4.16711032
D
2
r0=(e+1)GMe
n2
0
r0 n2
0
GMe
=e+1
e=
¢
r0 n2
0
GMe
–1
≤
e=Ch2
GMe
=1
r0
¢
1–GMe
r0v2
0
≤
ar2
0 v2
0
GMeb
np=n0=15 Mm>h=4.167 km>s
e=0.130
P
A
*13–128.
SOLUTION
A rocket is in free-flight elliptical orbit around the planet
Venus. Knowing that the periapsis and apoapsis of the orbit
are 8 Mm and 26 Mm, respectively, determine (a) the speed
of the rocket at point (b) the required speed it must
attain at Ajust after braking so that it undergoes an 8-Mm
free-flight circular orbit around Venus, and (c) the periods
of both the circular and elliptical orbits.The mass of Venus
is 0.816 times the mass of the earth.
A¿,
A¿A
O
8 Mm
373
SOLUTION
A′B
2
13–130.
If the rocket is to land on the surface of the planet,
determine the required free-flight speed it must have at
A′
so that the landing occurs at B. How long does it take for
the rocket to land, going from
A′
to B? The planet has no
atmosphere, and its mass is 0.6 times that of the earth.
BAA¿
r 6 Mm
O
375
SOLUTION
*13–132.
The rocket is traveling around the earth in free flight along
the elliptical orbit AC. Determine its change in speed when
it reaches A so that it travels along the elliptical orbit AB.
A
CB