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Unformatted text preview: CHAPTER 6
GRAVITATIONAL AND CENTRAL FORCES 4 6.1 m=,0V=,0—l3—7rrs3
I
3m 3
I', = ~—
2 3
F _ Gmm _ Gm2 [47w )3 _ 2(4ﬂpj3 m;
“ (2,1)? 4 3m 4 3 ') F _ L _ Gm2 (4750);)”; W_mg 4g 3 7 in ' 2.,,72, . 73 _ 6,3“; !
5:6.67x10 N m7 kg, 4/rxll.35g cm X ll:g XIO u") X(1kg)3
W 4x9.8ms" 3 10‘ g lm‘
£=223x109
W 6.2 (a) The derivation ofthe force is identical to that in Section 6.2 except here r < R.
This means that in the last integral equation, (6.2.7), the limits on u are R — r to R + r. F : GmM [Imp + r" —R‘ )ds :1er R" 52 — Gm“? [R+r—(R—r)+ R+r R—r _ 4Rr~
GmM 4Rr2
(b) Again the derivation of the gravitational potential
energy is identical to that in Example 6.7.1, except that the limits ol'integration on s are (R —r) —> (R + r). R2 _).2 R: _r2: F: [2r+R—r—(R+r)]=0 2
¢=_GZZZ'pR R ds
rR R"
2irpR2
=—G R+— R—
,.R[ i ( d]
4 Z
¢=_G rer:_Gﬂ
R R For r < R , ¢ is independent ofr. It is constant inside the spherical shell. G2 171 t
e ‘i r I. 6.3 i=— The gravitational force on the particle is due only to the mass of the earth that is
inside the particle’s instantaneous displacement from the center of the earth, r.
The net effect of the mass of the earth outside r is zero (See Problem 6.2). 4
M = — m3 3 ,0
— 4 . .
F 2 —§ Gﬂpmre, = —kre,_ The force is a linear restoring force and induces simple
harmonic motion. a) k 4G7rp The period depends on the earth‘s density but is independent of its size.
At the surface of the earth, G'an _ Gm 4 m = — —72’R?
g R3 R3 3 ‘p
4Grrp_g
3 RL.
6 .
Tzzﬂ £2271 wx 1]” zlA/n.
g 981115" 36005
_g=—GA;ﬁnér.where M=§ﬂr3p
,. The component of the gravitational force perpendicular to the tube
is balanced by the normal force arising from the side of the tube.
The component of force along the tube is F:r = 1*; 0056’
The net force on the particle is ~ A 4
F = —i 3 Girpmr cos 0
r cos 6 = x F = —f g 6/7me = —ikr As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours. 6.6 6.7 GMm _ mv2 2 GM Z — so v =—
r r r
for a circular orbit r. v is constant.
2m
T:—
v
7 7 2
2 471‘)" 47! ’3 .3
T = , = I 0c)
v' GM’
21”
(a) T:
‘V From Example 6.5.3. the speed ofa satellite in circular orbit is l
v : gR‘: .
I. _r__ ng 3_ 242/2r2x3600252/1r'2x9.8m.\"2 3
RV 47mg 47:26.38x106m
L=6.62z7
R?
g 3
2 3 277 60R 3 3
(b) T: 1" =¥=2z 60 R?
M. gER. g _ 7” 603 x6.38x106m 3
a 9.8m as"2 x 36003::2 hr‘2 x 242/1)‘2 day 2
T = 27.27 day 2 27 day From Example 6.5.3, the speed ofa satellite in a circular orbitjust above the
earth‘s surface is Wm, 2R R
= ﬂ"=27r —" V g T This is the same expression as derived in Problem 6.3 for a particle dropped into a
hole drilled through the earth. T z 1.4 hours. 6.8 The Earth‘s orbit about the Sun is counter—clockwise as seen from. say, the north
star. It's coordinates on approach at the latus rectum are (x. y) = (£(1.—O!). . . l .
The easiest way to solve this problem IS to note that 8 = — 18 small. The 60
orbit is almost circular! . Gil/{Sm = mv2 and v2 : GMS
)" r r
with r = a z a z b when 5 z 0
vz[Gll‘Is]=3.104:71
a s
More araclly
;xr=avcosﬂ=1. but a=% (equation 6.5.19)
2 I
Since k = GMgm a = l i hence [2 avcosﬂ = (aGMs )5
‘ GMS

Or v =(GMS )2 l
a cosﬂ
The angle ,6 can be calculated as follows:
.y_, + x3 = 1 (see appendix C)
b” a‘
cit“ [)2 .t' 4 = ——,— and at x, = 6a,—a
dx (1‘ y ( y) ( )
dy 1)2 ga b2 E . 2 b2
so —= ,—=—, =851ncel—g =—,
cit a" a a’(l—gz) a~
dv
here A = tan ﬂ = a or ,6 z .9 (small a)
(it
 
GM 5 l 1’ 5
and v =( S] 2 [GA 5] as before.
a 0056 a 6.9 Pp) = 11 + Fd F; = _ Gilz'm
r
Fd : _ GM/n The net effect ofthe dust outside the planet‘s radius is zero (Problem 6.2). The mass of
the dust inside the planet‘s radius is: 4 1W =—71rr3 )
(I 3 It
A/ 4
F(r) = — G {In ——7rpmGr
r” 3
6.10 It =1=—1—e'm
r );
§:=_£ew
d6 1:
2 ’I
d 1: =A—e'w :11. u
d6” Iv
From equation 6.5.10 dzu ,  _ \
+u=k'u+u=—T7 u'
duz ml”u" ‘ ( f(u_' ) = —n112(k2 + l)“3 ml2 1k2+l
f(")= (3 ) r
The force varies as the inverse cube of r. From equation 6.5.4, r29 =1 :13 z 4
dt rj
1“) l
e“ = ﬁg]!
1;”
_l_62k0 2k r; ’) y
0 =—1—ln —J\2h +C'
2k I;
6? varies logarithmically with t. 6.11 f(r)=—k3=ku3
,.
From equation 6.5.10 d2 1 '
1: +u=— z , ku3 =— A”,
(10‘ ml 11‘ ml“ 2
d 1: +£1+ [(1)1120
(19‘ ml' 7 (119' , ,2
If (1+ [‘2]=0, ‘1” =0 e k dill . . .
It (I +—— < 0. —cu = 0 , c > O , for which 11 = uew IS a solution. dt92 u=q6+g
l
q0+q r: 1f (1+ k,)>0‘ dblz+cu=0. c>0
ml“ (10‘ u=x1eos(\/Zt9+5) 7
I'=:ACOS[ 1+ k,0+6]]
ml‘ 1
6.12 u=—= l
r racost‘) ﬂ_ sinO
d0 i;cosz() d‘iu__l_ I +231112t9 _ I l+2—2cr)526\_ l [ 2 1]
£103 ;; cosO 00530 500519 (30520 i rbcosO c0536 (121/ 1
a = 1((2r;21l2 — 1) = 21;"113 —u
(1'6"
Substituting into equation 6.5.10 1
2751/3 —u +11 = — f u"
mlzu2 ( )
f(u" ) = 2rozmlzu5
lezml2 = _ 5 r 6.13 From Chapter I. the transverse component of the acceleration is a” = r61 + 2ft?
If this term is nonzero, then there must be a transverse force given by f((}) = m(rt§ + 21‘?)
For r = (10 , and 6 = bf _/'(()) = 2mab2 ¢ 0 Since f(()) :t 0 . the force is not a central field. For 2' = (16 , and the force to be central, try 6 = bl”
f(6)=I72[2£1l)2r1212”_2 + ab2n(n — l)t2""2:
For a central ﬁeld f(l9) = O
2n+(n—l) = 0 n:—
3 6 = [#3 6.14
Calculating the potential energy ﬂ= '(r)=—k[is+"—:] I‘ I‘ 2
Thus, V = —k a 4) The total energy is E=]1+I{=lv:_k[_27+ 1 ]:i[9k] 9k :0 2 a" 402 2 2a2 4a2
Its angular momentum is 9k ‘1
l2 = agi;2 = 7 = constant = r46'
Its KE is [ﬂ] 92:1 [a L
2 2 (16? 2 d9 r4 The energy equation of the orbit is T+V=0=l +r2 [—;—k a“
2 d6 r 7" 4r4
dr 2 2 9k 2 a2
= — +r —4—k —2+—4
d0 4r r 41‘
dr 2 l 2 2
01" =5(CI —I' )
dr d¢ Letting r = (1 cos {15 then — = —a Sin 45% 1
Thus r= mos—(9 r=a 6=0°
( 3 < @ )
37: . . .
(b) at 6 = 7 r —> O the orlgln of the lorce.
6. = ‘% = mg 1 : v” 1
’ 0" cos2 —0 aces2 —6
3
dt = ices2 10m)
V0 3
L” 5
T = Iﬂcos2 l49(4’6’ =22 Icosz ¢d¢ =21]
0 v: 3 vb 0 41g 9k Since v°=( ,]2
2a" 3 ,[2 '
Tzzra’ —
4 9k 4 (c) Since the particle falls into the center of the force NI—
H
m
Q4
/—\
>.~N
w
NI— V —> 00 (since I = wiL = const )
i
6.15 From Example 6.5.4 i = 2" vc rI +1;
1
. v 2
Letting V = —°— we have V = —
Vt 1+ 5
’1
So: ﬂ. = _l_ _2 1+ E _%
‘1": 2V I} rl’
dV
_ l . ‘
Thus _V—=___1__2L:l 1 I;
— 1+—" 1+—
rl 2 ’3 rl
1+ '—° To ﬁnd how long it takes ld—VJ
(a) V ~15 (b) [ﬂ]=ﬂ[ﬂ]=2(6o)1%=120%! [girl] 2 rl rl I; V
7i The approximation of a differential has broken down — a correct result can be obtained by
calculating ﬁnite differences, but the implication is clear — a 1% error in boost causes rocket to miss the moon by a huge factor  ~ 2! 6.16 From section 6.5, 520.967 and I; =55x10°mi. From equations 6.5.213&b, rl = i;— l I; 55X106mi lAU
a:—(z;+r,)= =———><———6 .
2 lE 1—0.967 93x10 Im
a = 17.92AU
3
From equation 6.6.5, 2' = ca2
1 Z 3
r=1yrAU 2 x 17.922 AU2
r = 75.9yr
From equation 6.5213 and 6.5.19 (1 ml2
5 = — — 1: ——
1;, [Oh
_ migvf 6 — —1 and k = GMm k 1
v3 = [6]” (6+ 1)]
I; From Example 6.5.3 we can translate the factor GM into the more convenient GM = aeve2 with ae the radius of a circular orbit and ve the orbital speed I l
a v2 3 93x106mi 5
vb: e “ 5+1 = — 1.967 v
i I; ( [55x106mi( 2
v0 = 1.824%
Since I is constant rlvl : r0120
l—— —.
v, = Eva = 6 v? =1 9‘67 x 1 .824ve = 0.030612,
r] 1+ 8 1.967 '
27mg _ 271x 93 x106 mi .~
~
l/e _ F = 66, 705 /
T lyr X da)’ ' yr X hr  (fa)?! "1p 1 VO = 1.22x 105 mph and VI = 2.04x 103 mph 6.17 From Example 6.10.1 V 1' l
(an[1+0]? —§)(quin¢)2] where q=——— and d:— V a E f are dimensionless ratios of the comet‘s speed and distance from the Sun in terms of the
Earth‘s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1). (15 is the angle between the comet‘s orbital velocity and direction
vector towards the Sun (see Figure 6.10.1).
The orbit is hyperbolic, parabolic, or elliptic as e is > , = , or < l 2 i.e., as [(12 —E)is > , = . or < O. [q3—%) is>,=.or<0as (12d is>,=,or<2. 6.18 Since I is constant, vmx occurs at I; and vmin occurs at rI , i.e. vm = v, and v in = VI and form the constancy of l vlrl = var, m ,. vrnin vinax : vlvn = ’i
2 k
Igig = —(5+ 1) (See Example 6.5.4)
m
. k 2
From equatlon 6.6.5 —: GMO =(gﬂ) a
m r
Vminvml‘ :[27za] . T ’i
. , 1+8 .
From equation 6.5.213&b r1 = r; 1 . With 2a = 1; +7}:
6
1 6+1 1 );+r 5+1 1 i _
(( )=—£—')(—)=— ’—’+l (8+l)=1—(1—g]+l (5+1)=1
r. 2 rl 2 rl 2 1+8 2
27m
vmin vinax =
T 6.19 As a result of the impulse, the speed of the planet instantaneously changes; its
orbital radius does not, so there is no change in its potential energy V. The instantaneous
change in its total orbital energy E is due to the change in its kinetic energy, T, only, so 5E = 5T = 5(lmv1]= mw5v = 17sz ﬂ = 27"2
2 v v T v 10 But the total orbital energy is E = —_/_c_ so : kw 60
2a 2a“
Since planetary orbits are nearly circular
V ~ —£ and T ~ ji
0 2a
‘E (5
Thus. (5E ; TE and ‘5 = _a
a T a
3.
We obtain (—6 = 2ﬂ
(I V 6.20 (a) V = 1 £de
T V(r) = —£
,.
From equation 6.5.4, I = r26? (:16 l rde —=—., 01' a't=
" 1 dt I
I: m =— fng From equation 6.5.18a _2
r_a(16) — l+gcoSH
' ka 182 25'
“dbl—l L i
’ l 1+80056
From equation 6.6.4 r = Zillw 152 I?” 27m " l+£c036
fir—d6 = 2” .52 <1 .'.l7=—£
1+50059 1_53 a (b) This problem is an example of the virial theorem which. for a bounded, periodic system, relates the time average of the quantity to its kinetic energy T. We will
0 i derive it for planetary motion as follows: 1 r T _ l r — _ — Imr rdt = — IF  rd!
z 0 0 er __
—Iprdt=
To 2' ll Integrate LHS by parts
1 T _
— mrr
Tl ] The ﬁrst term is zero — since the quantity has the same value at 0 and r . l' —') lr — _
'—— mi"dt=— Frdl Thus 2(T) = —(l?F) where ( ) denote time average of the quantity within brackets. <FF>=<rW>=<ﬂ>=<i>=~<v>
hence 2(T)=_<V) but (E)=(T)+(V)=_<LQ+<V>=(V) hence (V): 2(E) but E = —21 = constant
a l r k k
an < ) T6) (1 20 so a
Thus: (V) = —S as before and therefore (T) = —%(V) : 6.21 The energy of the initial orbit is
l 2 k k —mv —— = E =——
2 r 20  7
(1) v3=i[:__l_]
m r 0 Since 1;, = u(l +6) at apogee, the speed 1". at apogee is v25 2 1 _i(“5)
I m a(l+8) a ma(l+£) To place satellite in circular orbit, we need to boost its speed to vc such that 7 k . . . .
inn); —£ = —— smee the radius ofthe orbit IS I;
2 r” 21'” k k v" = — C mru ma(l+£) Thus. the boost in speed Avl = vL, — VI
1 (2) Now we solve for the semimajor axis a and the eccentricity e ofthe ﬁrst orbit. From (1) above. at launch v = V“ at r = R5 , so 2 k 2 1
v5 =— ———
m RE 0 and solving for a
RE a =—R— noting that 2 — quz —E
k
k GM
(3) = E = gRE
mRE RE
RE R E =4.49103km [2_ v; J 1.426
gRE The eccentricity s can be found from the angular momentum per unit mass, 1, equation 6.5.19. and the data on ellipses deﬁned in figure 6.5.1 I '3
 3 In 1—1;“
I: r26=vD(RE sim93):[k_a] tul— m I}? where va, 09 are the launch velocity, angle Solving for 5 (using (3) above) 52:1— V; [2— Vi jsinzaﬁows
gRE 8R5
.'.g=0.892 Inserting these values for a, 6 into (2) and using (3) gives
1 (21) Av, =[gRELRE/aﬂz[1—(1—£)%]=4.61~103 kms" 1+5 (b) h = a(l + 8)— R5 = 2.09 . 103 km {altitude above the Earth at perigee} _ br —br br
6.22 mar/(E bi 283 J46? (M2) I r I I —1 a
fl“) — b+a
From equation 6.14.3, 1,1/=7r[3+a/;((:))] 2 =”[3_(ab+2)]%
= 71
W l—ab 13 _ GM»: 4 6.23 From Problem 6.9. = 2 —§7rpmGr
r
f’(r) = M—ﬂﬂpmG
r 3 3 f.((’) = 2GMma‘3 —§rrpmG = —2+4:’j:; 3 f(a) —GMma'237rpmGa a 1+4ﬂpa 3 3M Ha)?
f (a) __ ‘ 73 __
_2+4”""'3 2 1+4[4”p” ] From equation 6.14.3, V = ir[3 + a 3M
(/1 = 7r 3+————4 3M3 = if 4 3
1+ Epa 1+ ﬂpa
3M 3M
l
E 4 3
W = ” H c ’ C : ﬂpa
1+ 46 3M
1U '
6.24 We differentiate equation 6.1 1.1b to obtain mi" = — C l0 )
c r For a circular orbit at r = a. 2" = 0 so dU
W 0 For small displacements x from r = a .
r = x + a and 5" = From Appendix D 2 f(x+a)=f(a)+.#'(a)+if'(a)+... 2
dU d2U
Takin r tobe—. ' ' =
gm dr f(') drz
Near r=u a’U dU dzU
—=—r:u+x , ,=u+..
dr dr dr“
.. dZU
I?L\'=—.\' 7.
drz “a c/2L/ , >0 at r=a .
c/r“ This represents a “restoring force,” i.e., stable motion, so long as 14 6.25 f’(l‘)=r—3+—5" From equation 6.13.7, the condition for stability is f(a)+§f’(a) < 0 k a a 2k 45
——2——+— —3+—— <0 5
(I a fir)=—ke‘b*[—%—%]=ke‘:' [123.] r I' r I From equation 6.13.7, the condition for stability is f(a)+§f’(a)< 0 bu » —ba
—ke—:—+ﬂke (b+—2—]<0 2 a 3 a a
—bu bu
—ke , +kbe <0
30' 3a
b<l a<l
a b
k
(b) f(")=r—3
I . _E
f(') a , k a 3k
f(a)+§f (a)=—?+§[—a—4]=O . . a ., . . .
Since f (a)+§ ] (a) is not less than zero. the orb1t 18 not stable. 15 6.27 (See Figure 6.10.1) From equation 6.5.183 r = ﬂ
1 + 8 cos 6’
and the data on ellipses in Figure 6.5.1 p = (10—8) so
(1+ 5)
1 + acos 6
For a parabolic orbit, 8 = l
The comet intersects earth‘s orbit at r = a. r: 6.28 (See Figure 6.10.1) T = Id! along the comet’s trajectory inside earth’s orbit .  7 ' 7 '2 From equation 6.5.4. 2"6 = r‘ (—— =l so dt = I C
d! 1
rde
T :
l 1
. 1— ‘2
From equation 6.5.1821 r = M
1+ 50030
and the data on ellipses in Figure 6.5.1 p = a(1— a) so
(1+ 5)
r = [1——
1+ ecosﬂ
From equation 6.5.18b. with a = 1 for a parabolic orbit:
2p
r 2
1+c050
72' . . 2p
At 6 = 3 the distance to the comet 15 r =0: = — = 2p
1+ c031
2
. ml2 I2
From equatlon 6.5.19. 0: = , where k = GMm, so p =
ZGM
As shown in Example 6.5.3. GM = ave2
7
For a circular orbit. ve = "7m
1 yr I 3 1 l
I: (2G1pr)3 =(2c1p)3 v9 = (20)? pzzry)’l ‘0 I‘ZdE) 9 4p2 3 —' g
T: = ———, 2a 2 271' '10 ‘
L) I [a (1+c080)' ( ) p L y, where 60 = eos'l (1 +2—p from Problem 6.27 (I 16 ﬁpg [9 d6 2
7m; ’1 (l +0030) T = yr (Lt 1 x 1 3x
=tan—+—tan — (1+cosx)2 2 2 6 2 0° 1 3 6:
tan +  tan yr
2 3 From a table of integrals. I win» it a 2
x l—cosx 3
an —=
2 1+cosx
2p 5 I T is a maximum when (2}) + a)(a  p)5 is a maximum. §;[(2p+a)2 (a—p)]= 2(2p+a)(2)(a—p)+(2p+a)2 (—1) =(2p+a)(3a—6p) . . (I
T IS a maxnnum when p = E . I
T = ﬂail]: = 1 yr = 77.5 day 37: 2 37!
When p = 0.60
T : g(2.2)\/.04 = 0.2088yr = 76.2 day
71
6.29 V(r) = £—k—f i‘ I' 17 mar
m) r2+6£ ‘5
=7! 3—2 =7;
W l: [I'Z+36]: From equation 6.14.3, y/ = 7r:3 + a r2 + 38 r2 38 2
For £=7RAR‘ R=4000mi, AR=13mi
D
7 1
g=§(4000)(13)=2.08x10‘mi
For I‘zR, r2=l.6x107mi2
y/ = 1.00397! =180.7°
2
6.30 Vn,,=£— I ,[E+—]£
I 2mc I
. dV k 2 k
r :———=——1+ 7 E+— —
/( ) dr 1" 2mc [ 7 ,
/'(r)=:f[1+ 12[E+£]]——’i—[ ‘
1 mac r I" ma
f’(r)=i3[2+ 1 2[ZE+EI:):
I mc r
1 3k
., 2+ 2E+—
f (a) _ _1 mac2 [ a
f(a) a 1+ 12[E+£]
mic a
, ) [3+ 32(E+£)—2—
3+“ j (a : mc a m,
f(a) 1 18 1 + ,
mac" ._.—_.._—— 1+ I,[E+£)]
mac” 0
k
(I
) , / 3+a fa mvc‘ + E+— : l
a) _[ mac: + E
) _'
2 ml 1
k 2 "l
(
y/=n‘[3+afi(u ulzzr l+—2——
mac +E . 1— 2 . . .
6.31 From equation 6.5.1851 r = LL) (Here 0 IS the polar angle 01 come
1+ scosO
section trajectories as illustrated by the coordinates in Figure 6.5.1) and the data on ellipses in Figure 6.5.] lb = (1(1—6) so
1+ 5
’mm = ’2 ———'
1+ acosﬂ
. _ a a a
From equation 6.3.18b r =— and at 0 = 0 1;, = —
1+ ecosﬂ 1+ 6‘
. _ mlz ml2
And from equation 6.3.19 6: =— so r0 =
k k(1+ g)
m _ m l I _ Gil/[m = GM
From Example 6.5.3, GM = aev: and 12 :19 1:2 sin2 ¢ CU," Com
1'2 v2 Sin2¢(1+£) r com com
com — aev:(1+£)(l +gc036)
l 1: Rstinz¢——
l+£c039 1 2 . 3
(2030—;(RV sm (15—1) 1
v _ 2 l_ _L 2 ~ 2 _ 2 E
sm0—(l—cos 0)2 —[l 82(RV 5m (15 1)] l9 I sing =1[ 2 —(RV2 sin2 ¢)2 +2(RV2 sin2 ¢)—1]2
5 Again from Example 6.5.3 l
.5 =[l +(V2 —%](RVSlIl¢)2]2
£2 =l+(RV2 singzﬁ)2 —2Rstin2¢ sinO=l[(RV2 singzﬁ)2 —(RV2 sinZ qty]? 8 sinO = lRV2 sin¢ cos¢
6 c056 _ Rstinz¢—l
sin0 Rstin¢cos¢ 2
cot0=tan¢———,—_—
RV‘sm2¢
2
6: ot'l tzi — ,csc2
c [ u (15 RVL For V=().5.R=4.¢=30": 6=cot"[tan30"— 2 csc60”] = cot'I Vii—22:; = cot" 2
6 = —306 6.32 The picture at left shows the orbital transfer and the position of
the two satellites at the moment the transfer is initiated.
Satellite B is “ahead’~ of satellite A by the angle 00 r +r . . . . . . .
a = ‘ 2 2 IS the semimajor ax1s of the elliptical transfer orbit. From Kepler‘s 3rd law (Equation 6.6.5) applied to objects in
orbit about Earth .,
2 47f" ‘; 2' = , a'
GM E“
The time to intercept is r l 72' g . GM F
7: =—=7I a : a smce 2’ =g
2 GME IN; RE 20 ﬁnasm 3 H mm +9 Ea aw n km +3 2:20 F 25 3 98 En Elms; cm Ea N maﬂzgom uses” 90 @655. Emwzmsm Emma 38 :6 E995 m.me u
r +w m K b + F
N.“ Ill \N N + _ N H ‘N N + — I
kmkm km 3mm km
183 @3an Pow” 5. u 8.: F? x: n Noe a: H mma Ed 25 F. H J  km N ANLOO E: 
wobmc 5:. 353m 3 :8 35608 5 n “rue rm lulu w
m E is? 9 “:39 TM “Sm: Pu.» :8 308533 m3 :5 Eganécca 380 3% . \m
6 3V 2 Ml
A V NRN
roam—Hm 2 H 2. . <8 :36 Amncmao: 0.0.3
_ N a: N u , _
MEN % += +~Az Vum
«3
am" 202. maamwmasm $2.: x H 8 A: H 3 cc 8 x H was A: H :2; .. s u .ng
o NSAW I “Q I Ewmzu _ .
wE Mnlién NUS? mo N
:2!
we H ... :55. Q:
Q # N H N U .v 0
NE mi; IM»: IS @322.
til
Q N ._. 595 2.:
o
o H u. _ u w H. N N
NE M55 IM»: IE v 5 : Before evaluating this integral, we need to ﬁnd um(=r min “), in other words. the distance of closest approach to the scattering center. E:T(rm)+V(rmin)=émvz+érk2 =gmv02 min But. the angular momentum per unit mass [is I = va = rmv and substituting for v into the above gives 7 7
ml' k 2 l mvo' 7
1 + , :mvo so , = , =um'
‘ ' ‘ r ' ml ' +k min r min ’min Solving for u mux ' ' ‘ l
um = ———k
I)2 + 1
\i mvo' Now we evaluate the integral for ()0 ll 90 = ~[———b—du =—b—sin'I u =;%
0 l[bz'i' k,]llz [b2+ l‘ 7] "max 0 [b2+ k7]
,nvn‘ Invob DIVO"
Solving for b k 26
b 0 )= ,—0
( 0 mvo‘ ’71? 4903 But (90 =%(7r—t95). Thus, we have k If — (9 {as (was) We can now compute the differential cross section [’(05): b db kzrz 36.
o—(gs)=  : 7 7 ( 32) .
Sings 616’s mv0’65‘(2/7—t95) Sings
Since (19 = 27min 6; day we get , 3 _
0(03)dQ=77rbdb=ki (” 63,) a (165
E (mam; ...
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This note was uploaded on 11/21/2009 for the course CEE 432 taught by Professor Ahmed during the Spring '09 term at Rowan.
 Spring '09
 Ahmed

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