# Elliptical orbits transformed to “equivalent” circular orbit

I'm trying to transform Mars' elliptical orbit into its circular "equivalent". I understand that if the semi-major axis is used as the radius, and the mean angular rate and period are kept the same, there is the bones of the answer.

My question is around propagation of the position. If Mars is at position $x_{1}$ at time $t_{1}$ on the elliptical orbit, and moves to $x_{2}$ at time $t_{2}$ through an angle theta measured at the centre of the ellipse, is its "equivalent" circular position just the same position extended outward to the circle? Apologies if that's not clear, I hope I made myself somewhat make sense.

Edit Credit @uhoh for their constant advice! Here is some important info.

So, I'm trying to plot an Earth - Mars transfer, and I have a bunch of transfers that place me in the same position as Mars, and then I need to match Mars' flight path angle at that point. As a verification, I need to plot an equivalent transfer as if Mars' orbit was circular, and see which transfer has the lowest final flight path angle (as it will be zero for a circular orbit). It is the 2D case!

• So, I'm trying to plot an Earth - Mars transfer, and I have a bunch of transfers that place me in the same position as Mars, and then I need to match Mars' flight path angle at that point. As a verification, I need to plot an equivalent transfer as if Mar's orbit was circular, and see which transfer has the lowest final flight path angle (as it will be zero for a circular orbit). It is the 2D case! – Harvey Rael Jul 18 '18 at 12:10
• Are you looking for bi-tangential transfer orbits? – HopDavid Jul 18 '18 at 14:40
• @HopDavid not explicitly as this is a low thrust transfer problem. Though I have heard that transfers involving a transfer angle of 180n degrees provide the lowest velocity at infinity's meaning the Delta-V's required for corrections are low – Harvey Rael Jul 18 '18 at 22:32

The period $T$ of some satellite in orbit is determined by its semi-major axis $a$, so for a given $a$ (and central body), the orbital period will always be the same:

$$T = 2\pi \sqrt{\frac{a^{3}}{GM}}$$

However, the angular rate of the orbit will change over time depending on the eccentricity of the orbit, but the mean motion $n$ will remain the same, since it is simply given by $n = \frac{2\pi}{T}$

I believe you are interested in the orbital anomaly which is the 'angle' around the orbit. There are three (main) types of anomaly:

Eccentric Anomaly

This is the angle to a point $P'$ around an imaginary circular orbit with radius equal to the semi-major axis of the orbit in question - its auxiliary circle. The point is found by projecting the satellite's position onto the imaginary circle from the semi-major axis line. This is labelled $E$ in the diagram below.

True Anomaly

This is the angle as measured from the semi-major axis line around the central body. This is labelled $f$ in the diagram.

Mean Anomaly

This is the angle by which an imaginary satellites would have travelled around a circular orbit with the same period as the one in question. The change in this over time is the mean motion.

In your case, I believe you would want to use the mean anomaly which will give you a circular orbit with approximately the correct timings. However, since all solar system planets have relatively low eccentricities, you can get a rough approximation by taking any of the anomalies and using them as a constant angle around a circle. It's also worth noting that all three anomalies are equal at $t=0$ and $t=T/2$, with angles of $0$ and $\pi/2$.

The Wikipedia pages for each anomaly has the equations for converting between them. This post has a fairly detailed (and brightly coloured) explanation of calculating orbital anomalies. Image by y CheCheDaWaff - This file was derived from: Eccentric and true anomaly.PNG:, CC BY-SA 4.0, https://commons.wikimedia.org/w/index.php?curid=48384905

• So, is P' the "equivalent" circular position? I am measuring my angle from the center of the ellipse to Mars' position. Don't know the name for that angle (is it Mean Anomaly?) That angle must be the same when measured on to the circular orbit - that leads me to think that I have to just extend my vector to the circle. However I don't know if that would be where Mars would be if it was on a circular orbit, does that make sense? – Harvey Rael Jul 18 '18 at 12:55
• @HarveyRael I think I understand what you're saying. Are you measuring the angle PCF? In the diagram, point F is the focus of the ellipse where the parent body (Sun) is located so we usually measure around that - the true anomaly f. – Jack Jul 18 '18 at 13:02
• @HarveyRael Saying where Mars would be is kinda difficult to define since it's not there, but we usually use the mean anomaly as a measure. If you google 'orbital anomaly' you'll find lots of diagrams that explain it much better than I can in words. – Jack Jul 18 '18 at 13:04