3
$\begingroup$

I am working on a satellite simulator for a project/game and I am stuck on this one bit of physics. So far I have a satellite that revolves around earth on a 2D plane following Keplerian motion using Kepler's equation. Everything is fine and the satellite will orbit many times without problem. I can also change the velocity of the satellite to manipulate apo/peri points as well as augment of periapsis.

However, the problem occurs if I change the velocity once the satellite passes the PI/2 mark, the entire orbit gets flipped (augment of periapsis is offset by PI) and the true anomaly resets to 0.

Here is how it is currently implemented: With the assumption that the satellite starts at the periapsis at t0, I find the eccentric and true anomaly given a time since periapsis $w$. Then I find the radius at that angle to get the final position.

$$t_0 = 0$$ $$E_0 = 0$$ $$M(0) = E_0 - e \sin E_0$$ $$n = \sqrt{ \frac{GM }{ a^3}}$$ $$E = {\rm kepler}(e, M)$$ $$v = 2 \tan^{-1}{(\sqrt{\frac{(1+e)}{(1-e)}}\tan\frac{E}{2})}$$ $$r = a\frac{1-e^2}{1 + e\cos{v}}$$

When I update the velocity of the satellite (flight path angle $\phi$ is recomputed after change in velocity), I recalculate a new true anomaly using $\phi$, I then find the difference between the old true anomaly and the new true anomaly, and add that to the augment of periapsis $w$. After that I recompute time since periapsis $t$ using the new true anomaly and recompute the orbital parameters as well.

$$vel = \sqrt{GM\frac{2}{r} - \frac{1}{a}}$$ $$\phi = \tan^{-1} ( \frac{e\sin{v}}{1 + e \cos{v}})$$ $$v_2 = \tan^{-1}{\frac{r v g m \cos{\phi} * \sin{\phi}}{r v g m \cos^2{\phi} - 1}}$$ $$w \mathrel{{+}{=}} v_2 - v$$ $$E = \cos^{-1}{ \frac{e + \cos{v}}{1 + e \cos{v}} }$$ $$M = E - e \sin{E}$$ $$n = \sqrt{ \frac{GM}{a^3} }$$ $$t = \frac{M}{n}$$ $${\rm calculateOrbit}()$$

I noticed if I were to $v = v + \pi$ if ever $v_2 < 0$ and $v > 0$, then the weird orbit/position flip won't occur until the $v = \pi$ Otherwise I do not know what the problem is. I would really appreciate it if someone can point me the way, as I have been scratching my head over this for a long time now.

$\endgroup$
4
$\begingroup$

A quick note, your equation for the radius $r$ as a function of the true anomaly is missing the semi-major axis $a$. I prefer to use the symbol $\theta$ for the true anomaly instead of $v$, since I will use that for the total velocity. So: $$ r(\theta)=\frac{a(1-e^2)}{1+e\cos\theta} $$ And the way I would calculate my new trajectory parameters after a velocity change is by evaluating the semi-major axis $a$ and eccentricity $e$: $$ a=\frac{\mu r}{2\mu-rv^2} $$ $$ e=\sqrt{1+\frac{r^3\omega^2}{\mu}\left(\frac{rv^2}{\mu}-2\right)} $$ where $\mu$ is the gravitational parameter (you use $GM$) and $\omega$ is the angular velocity, such that the tangential velocity is equal to $v_{\theta}=r\omega$.
From this you can calculate the true anomaly via: $$ \theta=\cos^{-1}\left(\frac{a(1-e^2)-r}{er}\right) $$ However the arc-cosine will return a value between $0$ and $\pi$ and to cover the rest of the arc (form $-\pi$ to $0$) you have to look at the radial velocity $\dot{r}$ or $v_{r}$. When this is negative it means that you have passed your apoapsis and that the true anomaly will be bigger than $\pi$ or negative (depending on which range you want to use for $\theta$), so the true anomaly would simply become: $$ \theta=-\cos^{-1}\left(\frac{a(1-e^2)-r}{er}\right) $$ PS: This might be a useful question since it contains an explicit expression for the the time as a function of your position.

$\endgroup$
  • $\begingroup$ I am using mean motion to calculate time. Otherwise the two changes I needed to make is to use the acos to find true anomaly instead of my atan, and account for the time when the true anomaly is negative. It now works as expected. Thanks a bunch! $\endgroup$ – omikun Jan 28 '14 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.