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Circular and Elliptical Orbit Common Points
Name: Kathy
Status: other
Grade: other
Location: NV
Country: N/A
Date: N/A
Question:
Is there a common point at which the velocity of a
satellite in elliptical orbit equals the velocity in a circular
orbit? If so, where?
Replies:
This is going to require (a lot of) analysis. There are several sets of
conditions:
1. The circular trajectory lies completely inside the elliptical trajectory.
2. The circular trajectory lies completely outside the elliptical
trajectory.
3. The circular and elliptical trajectories cross, but not at the same time
and point in space --
no collision. This may or may not occur depending upon initial
conditions of the two objects.
4. The trajectories of the two satellites cross, but not on a collision
path. This also may change with
time, and depends upon initial conditions.
5. As stated in the question, a "common velocity" does not necessarily imply
a common location in
in space. So there could be one (or more) common locations and/or common
velocities without a
common velocity but not a common location at any time.
Other assumptions: only gravitational forces and with no "drag" from the
atmosphere.
This is a very good question, but a correct and full answer, would require a
very complicated answer.
This is the level of sophistication encountered by astrophysicists. Not to
be worked out on the back of an envelope.
Vince Calder
Hi Kathy,
The answer to your question is yes. For an elliptical orbit with
eccentricity "e" and semi-major axis "a", the speed in the orbit
will be equal to the circular orbit speed whenever r = a (since by
definition, r = a for all points on a circular orbit). Using the
orbit equation, r = a*(1-e^2) / (1 + e*cos(nu) ), where nu is the
true anomaly, or angle from periapsis, we substitute r for a and
solve for the value(s) of nu to determine where on the orbit these
points lie. When you do, what you find is that nu = acos(-e), which
has two solutions - in the 2nd and 3rd quadrants of the orbit. For
an eccentricity of 0.2, nu = 101.5 degrees or 258.5 degrees. For an
eccentricity of 0.5, nu = 120 degrees or 240 degrees.
Aaron Brown
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Update: June 2012
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