Sun-exposure on a binary planet system

[Longbear]

Hello, I'm writing a story set on a planet that is in a tidally locked very close orbit with a sister-world of equal mass. (I know this may not be scientifically possible, but it's a fantasy story with dichotomy as a central theme, so binary worlds.) I'm having trouble visualizing the day/night cycle. I know at the equator you would essentially get a morning, 1-2 hours of eclipse as the sun passed behind the sister-world, and then a normal afternoon. How would this look in the northern/southern hemispheres? Additionally, the sister-world would only be visible on one side of the planet, since they're tidally locked.

Is there a free simulator I could use to help visualize this?

What kind of star would be best for this scenario?

 

[JunkMonkey]

As eclipses only happen in quite narrow bands on Earth, where the Moon is the same apparent size as the Sun, I would guess that, for most of both of the the worlds, day / night would be just as on Earth. It's possible that above (and below) a certain latitude the sister planets would not be visible from one another as they would be below the horizon.

A bigger problem I would have thought would be that - if they are tidally locked at their equators - wouldn't this have the tendency to (eventually) flatten out precession and reduce the difference between winter and summer to merely (!?) the distance from the sun. Instead of the northern and southern hemispheres there would be a global - or double global - colder season when the planets are furthest from the sun and a warmer when closest. So unless you have a very eccentric orbit you are going to have a very stable environment.

I'm presuming from the statement, "at the equator you would essentially get a morning, 1-2 hours of eclipse as the sun passed behind the sister-world, and then a normal afternoon" you envisage the worlds having their equators on the plane of the ecliptic. If they weren't and the binary's axis was not at 90 to the ecliptic, you would not always have the pattern of daylight you suggest. That would only happen (at the equator) once a year at the equinox. The path of the eclipse would then go south AND north (following the winter on one planet, and the summer on the other) till the solstice(s) and then back again... I think... Complex.

 

[JunkMonkey]

thinking about this further - it's keeping me awake! - days would be much longer than on Earth as the planets would have to make a complete orbit of the common centre of gravity to complete a day. Our tide locked Moon takes 29.5 or so days to do this. The day on Earth is shorter because we're revolving in relation to the moon. If both planets are tidelocked the days on them are going to be horrendiously long.

 

[JunkMonkey]

...

 

[Mon0Zer0]

gsim might be able to help you.

Gravity Simulator, by TestTubeGames

Fling planets and change physics in this gravity sandbox. Want to add 1000 planets? Go ahead. Curious about Lagrange Points? Make em. Hate the inverse square law? Change it.

www.testtubegames.com

 

[Venusian Broon]

You'll have to come up with a touch more precision before you can get a bit more precise answers.

For example what is the plane of the dual planet system relative to the ecliptic plane (i.e the plane of 'main' planet around the star)? I think you are assuming that the dual planet plane is the same as the ecliptic - there may be real physical reasons for the star's tidal forces squeezing the system into the ecliptic - but as you say this is fantasy. You could have a 'wobbling barbell' system so that neither planet causes eclipses on each other (or only very rarely as everything lines up 'once in a blue moon').

As JunkMonkey points out - how long is the period of 'day'?....

This will have a knock-on effect on the distance between both bodies. The closer they are to each other, the faster the day cycle will be. But then the closer they are the bigger the effect on the day/night cycle each will have.

I did a very dirty back of the envelope calculation and for your system - say two Earths - to have a 24 hour day then the planets need to be ~53,000 km apart. (The Moon is currently about 384,000km apart.) So, a little trig shows that standing on one planet would see something with an angular diameter of about 12 degrees (Moon is ~0.5 degrees). That's about 6.6% of the entire sky.

(I also calculated the Roche limit of such a system, and 53,000km is well above the limit, so said system should be possible and they shouldn't pull themselves apart.)

Anyway, that angular diameter is BIG.

And that will have an large effect on nights. So, for example - taking the above - if we assume that the tidally locked plane of the planets is not on the ecliptic, it means that generally speaking, when it is night on the tidally locked part of Earth 1, then the sky will be lit up by sunlight bouncing off Earth 2 - at a maximum when the star is directly 'behind' you on Earth 1, but even crescents should shine sufficient light to disrupt nocturnal life. So some parts of the planet will have 'true' night, whereas the other side could exist in a quite bright 'reflected twilight'. (I am also assuming that the Earth 2 ain't a completely black ball that absorbs all light as well!)

This then gets complicated if you want the planets orbit to just be the same as the ecliptic as you will then have to figure out how the day cycle then interacts with the year period of the orbit of the dual system around the star to see how they would block each others view of the star.

I wonder also what effect such as system might have on geology, geography, oceans etc...A very slow spin of the system could give the Earths a "central mountain ridge" along the plane of the 'spin', as water would tend to collect at the poles, and perhaps the tidally locked world might bulge out at each other, so that additional highlands are forced up facing each other? So perhaps mountains on one side, oceans on the poles and on the opposite sides? (I'm guessing a bit here, I may be wrong )