# Transit time calculation

1. Oct 3, 2007

### natski

Hi,

I've been looking for some of the early work done on transitting theory for the detection of extrasolar planets. I am especially interested in the calculation of the transitting time. The only equation I have so far come across is...

T/P=(1/pi)*arcsin(R*/a)

in the paper by Rivera et al 2005 on the detection of GJ 876d.

It strikes me that this equation (although not formally stated in the paper) is only an approximate (and was unreferenced so I can't trace where it came from). If anyone knows of a more exact equation and some work which has been done on this subject, please post a reference. Thanks.

2. Oct 3, 2007

### tony873004

Are you referring to computing when the transit will happen, or how long it will last?

If it's how long, I don't know why it wouldn't be as simple as v=d/t, where v is the orbital velocity of the planet, and d is the (diameter of the star + diameter of planet).

For v, if you wanted to assume a circular orbit, you could use sqrt(GM/r) where r is the semi-major axis of the planet and M is the (mass of star + mass of planet), or if the orbit were elliptical, you could compute the velocity at periastron or apiastron for an upper and lower range of your velocities. (Equations 3.16 and 3.17 on Bob B's webpage: http://www.braeunig.us/space/orbmech.htm )

If you want to compute when it will happen, it should simply be multiples of its orbital period from the last observed transit.

I suspect I am misunderstanding the question, because there's no trig functions in either of my two suggestions.

3. Oct 4, 2007

### natski

The duration of the transit. I don't think t=v/2(R+r) since this assumes that not only is the viewer is at infinite distance (often valid) but the planet orbits as infinte distance from the parent star (not valid in my opinion).

4. Oct 4, 2007

### tony873004

I wouldn't say "often valid" when referring to exosolar planets since they're all light years away. In the case of Venus transiting the Sun as viewed from Earth, I can see where this would come into play.

I don't really see why the planet's distance to the star would make a difference, unless the planet were so far from the star that it's distance was a significant percentage of the Earth-star distance. I tried drawing it on paper, and with Earth at a non-infinite distance, the lines of sight to the star and planets at various distances to the star are non-parallel.

But placing Earth at an infinte distance, all lines of sight to the star and planets orbiting it at various distances, are parallel to each other, essentially turning this into a 1-D problem. So I don't really see a problem with accurately approximating it as v/(2(R+r)).

You do have me curious, since Rivera uses a trig-based formula in his paper. You might want to try e-mailing him. I've tried this in the past with various authors and have found that most of them are delighted that someone is expressing an interest in their work.

Last edited: Oct 4, 2007
5. Oct 5, 2007

### tony873004

Just an interesting update...
I volunteer to run the observatory on public nights at San Francisco State University on Thursday nights, and tonight I met my new partner. He told me he was doing research under Chris McCarthy at SFSU, writing code to compute transit times. What a coincidence!

So obviously I asked him this question. I asked him if it was simply t=v/d. He said for a circular orbit that would be true, or for an elliptical orbit with a large semi-major axis, that would be true. But for a planet in a close-in elliptical orbit, which are the types of planets they study, you can not assume that the velocity at ingress will be the same as at egress. i.e. the planet can accelerate across the face of the star. He didn't know the formula, as this is not his part of the research he does, but didn't doubt that it contained a trig function. Maybe I can pick some brains over there and get the precise formula explained to me.

6. Oct 5, 2007

### natski

Hmm interesting, thanks for the update on that....

I derived a similar trig function by considering the Earth-star and star-exoplanet distances as finite, hence giving two triangles. I then assumed a circular orbit and hence constant velocity and then used T/P=L/2*pi*a

where a =orbital radius of the planet, T=transit time and P=orbital period

L is the length of the arc along the planet's orbit which is inside the triangle formed by the the two edges of the star in question and the observer (the Earth).

I then corrected L to L+delta where delta is the length of the arc which corrects for the fact the planet is not a single point but has a finite radius (i.e. this length is approx 2r).

7. Oct 5, 2007

### FTL_Diesel

Natski,

The equation you reference from Rivera is correct (though only valid for central transits that cross the center of the star). To derive it, consider the portion of the planet's orbit that it spends in front of the star. If you draw a diagram for a circular orbit, the angle of this arc is 2*arcsin(R_* / a), since we have two triangles of height 'a' and base 'R_*'. As a fraction of the total orbit, this is (2*arcsin(R_*/a)) / 2*pi = (1/pi)*arcsin(R_*/a).

For the time spent in transit, just multiply by the period: t = (p/pi)*arcsin(R_*/a), which is Rivera's equation rearranged.

This is incomplete however, because it doesn't take into account the fact that transits are rarely across the exact center of the star. If you know the impact parameter 'b', you can see that the transit time should go as t * sqrt(1-b^2).

For first approximations though, this is often not a large correction.

On the issue of circular vs. elliptical orbits, it is generally safe to consider short period (<20 day) planets as always having circular orbits. Within this distance, we would expect tidal forces to circularize the orbit fairly rapidly. In detail this is not exactly true, but for practical purposes of calculating transit durations, you should probably just assume that it is, especially considering that pretty much all known transiting planets have periods of less than twenty days.

Interestingly, the transiting planet announced yesterday, HD 17156 b, is the longest period transiting planet known (p~21 days), and does not have a circular orbit (e=0.7, or something). Which is interesting from a dynamical standpoint, since it is not immediately clear why such a large eccentricity has not already been damped out tidally.

8. Oct 5, 2007

### natski

Yes, I understood where Rivera's eqn came from but it neglects many factros and I was hoping for a much more detailed analysis in a paper somewhere. I read about the new 21 day planet, very interesting.

9. Oct 5, 2007

### tony873004

lol. Why did I assume that all transits were central transits? Of course there has to be a trig function in there as it's crossing a disk. Thanks, FTL.

10. Oct 5, 2007

### FTL_Diesel

No problem Tony! I forget that too sometimes.

Well actually, for the case of circular orbits that equation is the exact solution. I can't think of anywhere that it's derived in the literature; usually people just take it as a given. Are you worried about elliptical orbits?

If you're looking for something about the details of transit timings (like for timing variations?), I'd go for Holman & Murray 2005 ("The Use of Transit Timing to Detect Terrestrial-Mass Extrasolar Planets") and its related papers.