a 1 e 1 a 1 a 2 e 2 a 2 F 2 F 3 P When a transfer orbit is tangent to coplanar departure and destination orbits, velocity vectors point in the same direction. No delta V is needed for direction change, only speed change. A transfer orbit tangent to two coplanar, circular orbits is the well known Hohmann transfer orbit. But what if the destination orbit is an ellipse? For destination orbits, we’ll look at coplanar asteroid orbits with perihelion > 1 A.U. Like the Hohmann transfer orbit, we want our transfer orbits to be tangent to the destination as well as the departure orbit. This way no delta V is needed for direction change. F 1 F 2 f 3 P Ellipses 1 and 2 share a tangent line at point P. A ray emanating from a focus will be reflected to the second focus with angle of incidence equal to angle of reflection. This implies P, F 3 , and F 2 are collinear. Also angle F 1 PF 2 = angle F 1 PF 3 . Call this angle . Call the asteroid’s orbit Ellipse 1. Ellipse 1’s eccentricity: e 1 semi major axis: a 1 sun’s center: F 1 2nd focus: F 2 Call the red transfer orbit Ellipse 2. eccentricity: e 2 semi major axis: a 2 sun’s center: F 1 2nd focus: F 3 Call the point of tangency P F 1 Finding Low Delta V Paths to NEOs Page 1 Finding Low delta V paths to NEOs
Transcript
a1
e1a1
a2
e2a
2
F2
F3
P
When a transfer orbit is tangent to coplanardeparture and destination orbits, velocityvectors point in the same direction.
No delta V is needed for direction change,only speed change.
A transfer orbit tangent to two coplanar,circular orbits is the well known Hohmanntransfer orbit.
But what if the destination orbit is an ellipse? For destination orbits, we’lllook at coplanar asteroid orbits with perihelion > 1 A.U. Like the Hohmanntransfer orbit, we want our transfer orbits to be tangent to the destination aswell as the departure orbit. This way no delta V is needed for direction change.
F1F2
f3
PEllipses 1 and 2 share a tangent lineat point P. A ray emanating from afocus will be reflected to the secondfocus with angle of incidence equalto angle of reflection.
This implies P, F3, and F2 arecollinear.
Also angle F1PF2 = angle F1PF3.Call this angle .
Call the asteroid’sorbit Ellipse 1.Ellipse 1’seccentricity: e1
semi major axis: a1
sun’s center: F1
2nd focus: F2
Call the red transferorbit Ellipse 2.eccentricity: e2
semi major axis: a2
sun’s center: F1
2nd focus: F3
Call the pointof tangency P
F1
Finding Low Delta V Paths to NEOs Page 1
Finding Low delta V paths to NEOs
F1F2
The distance betweenF1 and P we call ka1.
1 - e1 ≤ k ≤ 1 + e1
The other sides of the triangle canbe deduced by properties of anellipse.
a1
e1a1 e1a1
A
ka12a 1 -
ka 1
So if know the asteroid orbit’s eccentricity we can determine angles alphaand beta for a given value of k.
a2
F1
F3
P
2a2-2
1a2
1
ka1
The transfer ellipse is tangent to earth’s orbit sowe know it’s perihelion is 1 A.U.
Therefore the distance between f1 and f3 is 2a2-2.
The third side can be inferred from a property ofthe ellipse.
Using the law of cosines we can deduce
a2 = (2 - k2a12(1+cos )) / (-2ka1(1+cos )+4)
= acos( ( ka1a2-2a2+1) / (ka1a2 -ka1) )
2a 2-ka 1
Now we look at the transfer orbit, ellipse 2.
2e1
k2 - k
Finding Low Delta V Paths to NEOs Page 2
We set up a spread sheet to doincrements of k from 1-e1 to 1+e1.
This gives us a spectrum of possibletangent transfer orbits.
For each transfer orbit we can finda, e, and ~ .
Since we’re assuming coplanar orbits,i and line of nodes are undefined.
How about , time of periapsis?
Recall we leave earth at the transferorbit’s perihelion. That the longitudeof periapsis coincides with earth’sposition indicates the date.
For example, a 0 degree longitude ofperiapsis implies departure duringthe autumnal equinox, aroundSeptember 21.
In the illustration to the left, departurefrom earth occurs at 307 degreeswhich would be around July 30.
F1F2
F3
P
b-
-
F1
P
Finding Low Delta V Paths to NEOs Page 3
We would like to know the date the transfer orbit arrives at it’s destination.
As the transfer vehicle moves between earth and the destination orbit, itsweeps out the grey wedge below. What fraction is this of Ellipse 2?
Vertically scaling Ellipse 2 by (1- e22)-1/2 gives a circle having radius a2.
Area triangle F1OQ is e2 a22 sin E / 2.
E
ka1
e2a2ka1cos(f1)+e2a2
F1O
Q
V
a2
a2
P
f2
S
1 + e2 cos f2
a2e2 + a2 (1 - e22) cos f2
1 + e2 cos f2
e2 + cos f2
Area wedge OQV is E a22 / 2.
Area FQV = Area Wedge OQV - Area triangle F1OQ
Area FQV = E a22 / 2 - e2 a2
2 sin E / 2
= (E - e2 sin E)a2
2
2
Finding Low Delta V Paths to NEOs Page 4
We now have arrival dates and longitudes for the tangent transfer orbits.
But how often will the target NEO be at the right place at the right time?
To determine this we graph the true longitude (L) of the tangent transferarrivals by date as well as the asteroid’s true longitude by date.
Where the paths on the graph intersect, we have a time and place for wherethe asteroid is amenable to receiving a tangent transfer orbit. In Excel, draggingthe cusor over a location on the graph gives information on the point:
Finding Low Delta V Paths to NEOs Page 5
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Scrolling through the various increments of k, I found the transfer ellipse thatarrives at the true longitude of 221 degrees at October 25. From this row Ican also get departure date and longitude as well as other information.
