Upload
girithik14
View
231
Download
0
Embed Size (px)
Citation preview
8/12/2019 Lecture 2 Conic Sections
1/28
Con ic sect ions
r
M
m
V
8/12/2019 Lecture 2 Conic Sections
2/28
= Keplers constant = GM(m3s-2)
h = angular momentum = = constant
E = total energy = cons tant =
e = eccentr ic i ty =
2r
r
V 2
2
1
12
cos1
2
22
Eh
rh
8/12/2019 Lecture 2 Conic Sections
3/28
ra r
p
semi minor
axis b
Semi major axis a
8/12/2019 Lecture 2 Conic Sections
4/28
Solut ion
21 1 coser h 1
8/12/2019 Lecture 2 Conic Sections
5/28
Locat ion of per igee (closest app roach) can
be found from 1 by set t ing to zero:
2
1
p
hr
e
8/12/2019 Lecture 2 Conic Sections
6/28
Locat ion of apogee (furthest departure) can be
found f rom 1by sett ing to (no te that raonly
has real values fo r e < 1, circ les and el lipses)
2
1a
h
r e
8/12/2019 Lecture 2 Conic Sections
7/28
Combin ing perigee and apogee give a useful
means of f inding eccentr ic i ty :
pa
pa
ap
rrrre
ererh
112
8/12/2019 Lecture 2 Conic Sections
8/28
Semi major axis, a =
2
212
1
p ahr r
e
8/12/2019 Lecture 2 Conic Sections
9/28
Semi minor axis, b =
21p ar r a e
8/12/2019 Lecture 2 Conic Sections
10/28
fam i ly o f solut ions
The value of eccentricity determines the
category of orbit obtained
e 1gives captive orbits
e 1creates escape trajectories
e = 0
circular
e > 0 1 hyperbola
8/12/2019 Lecture 2 Conic Sections
11/28
8/12/2019 Lecture 2 Conic Sections
12/28
example 2: Connect w ith
el l ipt ical trans fer orb it
8/12/2019 Lecture 2 Conic Sections
13/28
72.046500
650040000
pa
pa
rr
rre
10
1 cos 1 1
3.88 14*6.5 6*1.72 6.59*10
p ah r e r e r e
E E
8/12/2019 Lecture 2 Conic Sections
14/28
smEr
hV
smEr
h
V
r
hV
p
p
a
a
/130,1065.6
10*59.6
/650,1640
10*59.6
10
10
8/12/2019 Lecture 2 Conic Sections
15/28
sm
sm
sm
v
v
v
/3804
/150816503158
/2296783410130
2
1
8/12/2019 Lecture 2 Conic Sections
16/28
Fuel requ irement for manouver
For LH2/LOX rocket
with c = 4500m/s
429.04500
3804exp
0
m
mf
8/12/2019 Lecture 2 Conic Sections
17/28
This was an example of a Hohmann transfer
It was bel ieved to have the min imum energyrequirement for moving between co -planar
c i rcu lar orb i ts , as no fuel is wasted onveloci ty vector rotation
Disproved by US student who pos tulated thegrav ity ass is t manoeuvre for an ass ignment
8/12/2019 Lecture 2 Conic Sections
18/28
Example 3: hyperbo l ic capture
into circu lar orb i t
Arr ival at mars w ith a relat ive app roach
veloc i ty (V) o f 3km /sec, w ith cap tu re
into a circular orb i t at an al t itude of
4000km
Findvrequired
8/12/2019 Lecture 2 Conic Sections
19/28
V
v
V2
Vcirc
8/12/2019 Lecture 2 Conic Sections
20/28
2
26
2
40008*10 /
2
VE
r
J kg
Adjust hyperbolic approach so that its per igee is at
the alti tude of the required circular orbit. The
requi red velocity change wil l then be along theline of f l ight and at maximum velocity, giving the
greatest change in E for a given v
8/12/2019 Lecture 2 Conic Sections
21/28
smE
EE
rEV
/611064
1327.4*268*2
2
22
8/12/2019 Lecture 2 Conic Sections
22/28
smE
E
rVcirc
/327064
1327.4
8/12/2019 Lecture 2 Conic Sections
23/28
Combinations of these manoeuvres enable all
bodies in the plane of the ecliptic to be visited.
Some moons rotate out of plane, and require
3D calculations of a more complex nature
8/12/2019 Lecture 2 Conic Sections
24/28
8/12/2019 Lecture 2 Conic Sections
25/28
LEO opt ions
7834 m/s
10130
8/12/2019 Lecture 2 Conic Sections
26/28
GEO options
1650
3158
Minimum v
8/12/2019 Lecture 2 Conic Sections
27/28
8/12/2019 Lecture 2 Conic Sections
28/28
Fuel requ irement for manouver
For LH2/LOX rocket
with c = 4500m/s
28.04500
5724exp
0
m
mf