0
$\begingroup$

Supposing that another planetary system is in one plane like our solar system, if let's say that the transit of one planet is visible then is it safe to also assume that all planets in the system must transit?

Logically I feel like that this makes sense because since they are all in the same plane then they therefore must all transit as well. But then I feel like the transit probability equation could make this false? Could someone please clarify my doubts?

Also suppose that we are in another system where each planet doesn't necessarily orbit in the same plane. Is it easier to find transits if the planets are orbiting in the same or random planes?

I think that you have a better chance detecting a planet if they are in random orbits. but i am a bit lost in regards to detecting multiple transits.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

This is unfortunately not true.
If you already look at our solar system, even this is not co-planar.
The few degrees inclination that the planets have to each other would already be enough to make some of them undetectable, when others are detectable by transits.

The answer to your question therefore is not so much about the transit probability, but more about what the mutual inclinations are of the planets. Those can be stirred up by close-planet encounters or resonances after the gas disc dissipates in which the planets form.

$\endgroup$
2
  • $\begingroup$ What if hypothetically all of the planets in the system had the same inclination? Also do you have any suggestions for my second question? $\endgroup$
    – aa11
    Commented Nov 24, 2016 at 14:20
  • $\begingroup$ @aa11: a) If they were nicely co-planar (same inclination) then we would see them all transiting. In fact the multiple planet systems (en.wikipedia.org/wiki/List_of_multiplanetary_systems) mostly are discovered due to this effect, a few of them are radial velocity or TTV multiples. b) As you usually have finite mutual inclinations to each other, the transit probability doesn't help you too much. It just states 'if something is there, then...' but it can't tell you 'if we don't see anything, then with this probability there's something still'. $\endgroup$ Commented Nov 24, 2016 at 16:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .