I am currently learning about the physics and the mathmatics behind gravity assist. I have a question regarding the coordinate transformation between the planet frame and the sun frame. I want to simulate the gravity assist maneuver of a probe mass A in the gravity field of Planet B. So i took a look the planet System, solved the differential equation: $$\vec{r}'' = -\gamma M_p \frac{\vec{r}}{\|\vec{r} \|^3}$$ and get my position vector $\vec{r} = \vec{x}_P(t)$ (w.r.t the planet frame) depending on the time $t \ge 0$. Now i want to take a look at the trajectory in the sun frame $\vec{x}_S(t)$. I tried the following for a two-dimensional sun system: Let $\vec{x}_P(t) = ( x(t), y(t))$ and assume the planet is moving with constant velocity $v_P$ parallel to the x-Axis (I assume constant velocity because the time my gravity assist happens is very small compared to the period of circulation). Therefore, considering my very little understanding of coordinate transformation, the location in the sun frame should be: $$\vec{x}_S(t) = \vec{x}_P(t) + (v_P \cdot t, 0) $$

My Questions:

  1. Is this the correct approach?

  2. How to do this in 3D?

  3. Are the assumptions of constant velocity $\vec{v}_p$ justified?

  4. I read that one could get two times the velocity of the planet and i am missing this here. If i know my probe mass is moving with velocity $\vec{v}_A$ in the sun frame. How do i transform the velocity in the planet frame? Is the following approach the right way?

    1. Transform movement of probe mass from sun frame to planet frame
    2. solve ODE
    3. transform movement of probe mass from planet frame to sun frame

I havent found a good source, explaining coordinate system transformations. If you have one, please share it with me.

  • $\begingroup$ A bit confused with the notation: $r$ is what? the coordinates of the probe w.r.t. the planet? $\endgroup$
    – caverac
    Commented Nov 6, 2016 at 1:17
  • $\begingroup$ Yes, I am going to edit it to make things clear. $\endgroup$
    – user123962
    Commented Nov 6, 2016 at 7:55
  • 1
    $\begingroup$ Remember, angular momentum is conserved and so is the kinetic energy of the probe in the planet rest frame (i.e., the probe's kinetic energy only changes in certain frames). $\endgroup$ Commented Nov 7, 2016 at 15:38

1 Answer 1


You do not need to make any assumption about the orbit. Let me use this notation:

$$ \vec{r}_{A/B} = \mbox{position of system } A \mbox{ w.r.t. to system } B $$

It is easy to see that

$$ \vec{r}_{U/S} = \vec{r}_{U/P} + \vec{r}_{P/S} $$

where $U$ is the probe, $P$ is the planet and $S$ is the sun. At a first approch you can do something like this

  1. Solve the problem $\vec{r}_{U/P}$:

$$ \vec{r}_{U/P}'' = -\gamma m_P\frac{\vec{r}_{U/P}}{r_{U/P}^3} $$

  1. Solve the problem $\vec{r_{P/S}}$:

$$ \vec{r}_{P/S}'' = -\gamma m_S\frac{\vec{r}_{P/S}}{r_{P/S}^3} $$

which is basically the same problem with another initial conditions. Once you solve both, you use the first equation and you are done. It naturally extends to 3D


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