# Orbiting a tide locked planet



## Vertigo (Jul 24, 2015)

So here's another one I'm scratching my head over. I recently read a book with a spaceship in geostationary orbit around a tide locked planet. I'm thinking this is impossible? Geostationary orbit is determined by the planet's rate of rotation and if it's not rotating then you would also have to not rotate and would simply fall to the surface under gravity.


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## TØny Hine (Jul 24, 2015)

Tide locked would suggest a moon. The moons orbit would keep it in a geostationary position.


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## Ray McCarthy (Jul 24, 2015)

The moon is tidally locked. Always same face to Earth. This means it rotates to keep a face pointing at the Earth. Thus a geostationary orbit is possible.

A tidally locked planet to a Star is rotating once per orbit  (year). A Geostationary orbit would be a long way out! Such an orbit could only be possible for a while due to the Star.

All tidally locked objects (moons to planets or planets to suns or binary pair like Charon and Pluto) are actually rotating on their own axis.

More complex lock:
https://en.wikipedia.org/wiki/Mercury_(planet)


> Mercury is gravitationally locked and rotates in a way that is unique in the Solar System. As seen relative to the fixed stars, it rotates on its axis exactly three times for every two revolutions it makes around the Sun. As seen from the Sun, in a frame of reference that rotates with the orbital motion, it appears to rotate only once every two Mercurian years. An observer on Mercury would therefore see only one day every two years.



Mercury's orbital period is 7% wrong compared to Newton's laws due to how close it is to the Sun making relativity effects obvious.

Not the only thing odd:



> At certain points on Mercury's surface, an observer would be able to see the Sun rise about halfway, then reverse and set before rising again, all within the same Mercurian day. This is because approximately four Earth days before perihelion, Mercury's angular orbital velocity equals its angular rotational velocity so that the Sun's apparent motion ceases; closer to perihelion, Mercury's angular orbital velocity then exceeds the angular rotational velocity. Thus, to a hypothetical observer on Mercury, the Sun appears to move in a retrograde direction. Four Earth days after perihelion, the Sun's normal apparent motion resumes



So a tidally locked object is always rotating. But a Geostationary or Geosynchronous orbit (not quite same thing) might not be sensible for more than part of the orbital period and generally only makes sense for a moon or binary planet rather than a planet locked to a  star.


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## Vertigo (Jul 24, 2015)

It's the planet tidally locked to a sun case that I was thinking of. I initially thought it would have to be a long way out but then I decided it would be impossible. My reasoning is that you would have to orbit the planet, as you say, exactly once per orbit of the planet around the sun which means that you would have to be orbiting the sun as well. But that means that in order to get your orbital period right your radius of orbit about the sun would have to be the same as the planet and that's just not going to work. Maybe I'm getting it confused but the mechanics all seem to be wrong.

Take Mercury for example. I know it's not quite tidally locked but, supposing it was, then a geostationary orbit about Mercury would be at a radius greater than it's orbit around the sun which means the sun would totally mess up your orbit, as you would effectively be orbiting the sun not Mercury and then the only way to get your orbital period to match would be to be in exactly the same orbit as Mercury. That would be fine if you are chasing or leading Mercury but trying to orbit over any other point would simply not work. Or am I getting myself all confused...


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## Ray McCarthy (Jul 24, 2015)

Vertigo said:


> then a geostationary orbit about Mercury would be at a radius greater than it's orbit around the sun which means the sun would totally mess up your orbit, as you would effectively be orbiting the sun not Mercury and then the only way to get your orbital period to match would be to be in exactly the same orbit as Mercury.


Yes, but that is 88 days, I think. I suggested this problem.  Certainly a Geostationary orbit is possible for perhaps more than a week without major correction. But you would be far away.  For tidally locked or resonance objects like Mercury that are single planets rather than binary or a moon, I agree it's not useful due to the extreme distance. 

Geostationary is advantageous only for communications links on rotation much faster than orbital period. For survey / exploration a transpolar orbit is better. The special sun synchronous ones are best if the rotational period is fast enough (less than a week perhaps) as you can orbit just along the day/night terminator for best contrast and depth estimation. In a few local "days" you will have mapped entire surface at same resolution. A Geosynchronous or Geostationary orbit is useless for that. It's only good for Direct to Home broadcasts (as the satellite doesn't need tracked)!


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## Ray McCarthy (Jul 24, 2015)

Vertigo said:


> then a geostationary orbit about Mercury would be at a radius greater than it's orbit around the sun


Umm.. I'd have to do the sums based on the mass of Mercury. Mercury's orbital period is due to mass of Sun and distance from it (neglecting Mercury's mass). Your orbital period for a planet or moon (neglecting mass of spacecraft) is mass of planet and distance.

If a planet was tidally locked to a Star then indeed orbiting the Star at APPROXIMATELY the same distance, will APPROXIMATE to geostationary provided you are only "falling slowly" (or "drawing away" slowly) for a while. Hours, Days to weeks depending on distance from Planet without thrust.

Technically you add the masses and they orbit around a common centre of gravity. If that point is outside either object's surface you have a binary orbit.


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## Vertigo (Jul 24, 2015)

Okay so I did the sums and taking the example of Mercury, geostationary orbit (ie orbiting Mercury with a period of 88 days) gives an orbital radius (or semi-major axis) of 318,299 km so still tiny compared to Mercury's own orbit of the sun. So I guess maybe it might work. Interestingly this means that you could sit 'orbiting' Mercury whilst staying always in Mercury's shadow (and so not get cooked!!!). Though you'd be almost as far from Mercury than the moon is from Earth.


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## Ray McCarthy (Jul 24, 2015)

Vertigo said:


> Interestingly this means that you could sit 'orbiting' Mercury whilst staying always in Mercury's shadow


A sun-synchronous orbit across the poles will work for a planet uncomfortably close to a Star that has a much faster rotation than Orbital time.

I find this interesting stuff and I do try and check the physics and maths when writing SF.



Vertigo said:


> (ie orbiting Mercury with a period of 88 days)


That's the hypothetical tidal locked version rather than the real 3:2 resonance version?



> ... an orbital period of about 88 Earth days. Seen from Earth, it appears to move around its orbit in about 116 days ...
> Due to Mercury's 3:2 spin–orbit resonance, a solar day (the length between two meridian transits of the Sun) lasts about 176 Earth days. A sidereal day (the period of rotation) lasts about 58.7 Earth days.


Mercury is the sort of thing that would seem unlikely in Fiction.


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## Vertigo (Jul 24, 2015)

Hehe yes, it was for the hypothetical tidal locked Mercury that I did my calculations. And, as you've probably guessed, I find this sort of stuff interesting too. I actually wrote a program a year or two back that I made available here on the Chrons, for calculating the time dilation in relativistic travel. The interface isn't too pretty but I didn't have that much time to spend on it. I'll see if I can find the thread if you're interested.

I'm just an aged geek really


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## Mirannan (Jul 24, 2015)

I suppose one could say that an object at a Lagrange point for the sun-planet system is geostationary, if the planet is also tide-locked to the star. This isn't too unreasonable either, as a tide-locked planet would have to be fairly close to the star. If said planet was habitable, that would mean a red dwarf primary - which in turn means the Lagrange points L1 and L2 would be reasonably close.


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## Vertigo (Jul 24, 2015)

Well the L1 and L2 points would be similar to geostationary but they would always be directly above the same two points on the planet. If you wanted to be geostationary above a different point then you are back to true geostationary. Interestingly though, in the tidal locked Mercury example the L1 and 2 points are closer to the planet than the geostationary orbit.


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## Ray McCarthy (Jul 24, 2015)

Vertigo said:


> I'll see if I can find the thread if you're interested.


I remember that, and I downloaded it. You promised me the source to play with?

a) If you are visiting Mercury a short while, the rotation is slow anyway ...
b) Only TV or similar Satellites need/want Geostationary, ever, anyway. A visiting space ship will always find trans-polar orbits more interesting (surveys).


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## Vertigo (Jul 24, 2015)

Unless they need to remain over a particular location for some reason. But in general I agree with you.

Re the software I remember that now, did I never get it to you? PM me your email and I'll send it to you.


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