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I'm working on a school project on calculating the Keplerian orbital motion of objects and then plotting/animating the trajectories. One feature I want to include is plotting a Hohmann transfer from one orbiting height to another one.

My question is do any of you know where I can find an equation of motion that would represent a Hohmann transfer "orbit". I've looked around online and the best I've found is the equation to calculate the change in velocity needed, which doesn't really help me plot the motion of this orbit. Thanks in advance for any help.

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2 Answers 2

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You can use:

$x=a\left(\cos\tau-e\right)$

$y=a\sqrt{1-e^2}\sin\tau$

to plot. $a$ is the semi-major axis and $e$ is the eccentricity. The central body being orbited (e.g. the Sun) is at $\left(0,0\right)$. For a Hohmann transfer, you are going from periapsis to apoapsis, or vice-versa, so run $\tau$ from $0$ to $\pi$, or $\pi$ to $2\pi$. You may need to rotate the $\left(x,y\right)$ coordinates to line up with your departure and arrival points.

$\tau$ is not time. It is the eccentric anomaly. If you want to put time ticks on the plot, you can use:

$t=\sqrt{a^3\over\mu}\left(\tau-e\sin\tau\right)$

to compute the time, where $\mu$ is the $GM$ of the central body.

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With the equation of motion I assume you mean the position as a function of time. However this doesn't exist for eccentric Keplerian orbits, at least not explicitly. Because you would have to solve Kepler's equation. This can be approximated numerically. But it would be possible to calculate it the other way around, so the time as a function of the position.

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  • $\begingroup$ You don't need position as a function of time to make a plot. You only need position as a function of something. $\endgroup$
    – Mark Adler
    May 1, 2014 at 0:58
  • $\begingroup$ @Mark Adler, that is why I said: "With the equation of motion I assume you mean the position as a function of time." And since time as a function of the position (true anomaly or radius) can be calculated analytically I thought I should mention this aswell. $\endgroup$
    – fibonatic
    May 1, 2014 at 9:05

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