Conic sections - PowerPoint PPT Presentation

1 / 28
About This Presentation
Title:

Conic sections

Description:

semi minor axis b. Semi major axis a. Solution. 1 ... e =1 parabola. e 1 hyperbola. Example 1: circular Earth orbit. e = 0, r = 6500 km and 40,000 km ... – PowerPoint PPT presentation

Number of Views:85
Avg rating:3.0/5.0
Slides: 29
Provided by: Morg181
Category:

less

Transcript and Presenter's Notes

Title: Conic sections


1
Conic sections
V
m
r
?
M
2
  • µ Keplers constant GM (m3 s-2)
  • h angular momentum constant
  • E total energy constant
  • e eccentricity

3
Semi major axis a
semi minor axis b
ra
rp
4
Solution
1
5
  • Location of perigee (closest approach) can be
    found from 1 by setting ? to zero

6
  • Location of apogee (furthest departure) can be
    found from 1by setting ? to p (note that ra only
    has real values for e lt 1, circles and ellipses)

7
  • Combining perigee and apogee give a useful means
    of finding eccentricity

8
Semi major axis, a
9
Semi minor axis, b
10
family of solutions
  • The value of eccentricity determines the category
    of orbit obtained
  • e 1gives captive orbits
  • e 1creates escape trajectories

e gt 0 lt1 elliptical
e 0 circular
e gt 1 hyperbola
e 1 parabola
11
Example 1 circular Earth orbit
  • e 0, r 6500 km and 40,000 km

12
example 2 Connect with elliptical transfer orbit
13
(No Transcript)
14
(No Transcript)
15
(No Transcript)
16
Fuel requirement for manouver
  • For LH2/LOX rocket with c 4500m/s

17
  • This was an example of a Hohmann transfer
  • It was believed to have the minimum energy
    requirement for moving between co-planar circular
    orbits, as no fuel is wasted on velocity vector
    rotation
  • Disproved by US student who postulated the
    gravity assist manoeuvre for an assignment

18
Example 3 hyperbolic capture into circular orbit
  • Arrival at mars with a relative approach velocity
    (V8) of 3km/sec, with capture into a circular
    orbit at an altitude of 4000km
  • Find ?v required

19
V8
?v
V2
Vcirc
20
Adjust hyperbolic approach so that its perigee is
at the altitude of the required circular orbit.
The required velocity change will then be along
the line of flight and at maximum velocity,
giving the greatest change in E for a given ?v
21
(No Transcript)
22
(No Transcript)
23
  • 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

24
Example 4 transfer of a satellite left in LEO
inclined at 30 degrees to the equator to GEO
  • This requires the same operations as in example
    2, plus an orbital plane change of 30
  • the plane change is best done at a point where
    the orbit crosses the equator, where a simple
    velocity rotation will suffice
  • It may be done at LEO, before or after perigee
    burn
  • or GEO, before or after apogee kick burn
  • or it may be combined with either of the 2 burns
  • Choose whichever gives the smallest ?v

25
LEO options
10130
7834 m/s
26
GEO options
3158
Minimum ?v
1650
27
(No Transcript)
28
Fuel requirement for manouver
  • For LH2/LOX rocket with c 4500m/s
Write a Comment
User Comments (0)
About PowerShow.com