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 Click here to return to the Astronomy Archives

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