Beanstalk for a tide locked world?

[Flyndaran]

What differences would there be for making one? Easier, harder, impossible?

[Captain-Captain]

slower = further out.

Use the Moon as an example. it has a Seleneosychrounous sattelite. Earth.

The center of mass of the beanstalk HAS to orbit the equator of the planet and be the distance of the primary. A tide locked Earth would need a beanstalk whose center of mass was located out 93 million miles or so. And obviously couldn't be built at the Noon position lest it touch the Sun.

Of course angular momentum and stress REALlY REALLY REALLY become issues for the beasntalk.

[Agemegos]

Originally Posted by Captain-Captain

The center of mass of the beanstalk HAS to orbit the equator of the planet and be the distance of the primary.

The centre of mass of the beanstalk HAS to be at a Lagrange Point, hasn't it?

[Phoenix_Dragon]

Originally Posted by Brett

The centre of mass of the beanstalk HAS to be at a Lagrange Point, hasn't it?

I'm pretty sure it doesn't have to be, else you could never have one on earth.

[Captain-Captain]

It has to be geosynchronous to be moving at relative speed 0 for purposes of high altitude and a little more aircraft docking on it and dropping off cargo etc. Anything else and it's not "stationary" and you'll need to use actual spacecraft to get to the bottom end.

So it's center of mass MUST be the synchronous orbital height (22,300 miles for Earth), orbiting the plane that is perpendicular to the axis and contains the planet's center of mass and moving the same direction as the rotation of the planet. ANYTHING else means it's moving relative to the planet's surface.

If it's symmetrical, an Earth orbiting beanstalk will be 44,600 miles long. The far end will be moving MUCH faster than orbital speed for that height, easily enough for anything dropped from it to be whipped into higher orbit or even escape velocity for the system.

[Agemegos]

Originally Posted by Phoenix_Dragon

I'm pretty sure it doesn't have to be, else you could never have one on earth.

Earth isn't tide-locked.

In the case of a tide-locked planet the planet's orbit is synchronous with its rotation. A beanstalk has to be in orbit in synchrony with the rotation, and in the case of a tide-locked planet that means in synchrony with its orbit. Now, the definition of Lagrange points is that they are the circular orbits in a two-body system that are synchronous with the rotation of the system. Therefore the beanstalk of a tide-locked planet has to be orbiting in a Lagrange Point.

This seems straightforward to me. Is it not a standard result?

[Captain-Captain]

Originally Posted by Brett

Earth isn't tide-locked.

In the case of a tide-locked planet the planet's orbit is synchronous with its rotation. A beanstalk has to be in orbit in synchrony with the rotation, and in the case of a tide-locked planet that means in synchrony with its orbit. Now, the definition of Lagrange points is that they are the circular orbits in a two-body system that are synchronous with the rotation of the system. Therefore the beanstalk of a tide-locked planet has to be orbiting in a Lagrange Point.

This seems straightforward to me. Is it not a standard result?

Earth MARKS the spot where the center of mass has to go for a lunar beanstock as it IS a synchronus sattelite for the moon. (Ok a beanstock placed above the part of the moon that is perpetually closest to the Earth.) Any beanstocks placed elsewhere over the Moon have to have their center of mass somewhere along the obrit of the Earth as it dances with Luna.

[Captain-Captain]

Brett:

How many geosyncronous sattelites are in a lagrange point?

Zero.

A sattelite's center of mass must a) be at a height where the orbital rotational speed is identical to the rotational speed of the planet. B) It has to be along the equator's plane and c) it must move the same direction that the planet is rotating.

The relative gravitational pulls of the two bodies determine the five Lagrange point positions. This has NOTHING whatsoever to do with the speed the planet rotates at.

[Agemegos]

Originally Posted by Captain-Captain

Earth MARKS the spot where the center of mass has to go for a lunar beanstock as it IS a synchronus sattelite for the moon. (Ok a beanstock placed above the part of the moon that is perpetually closest to the Earth.) Any beanstocks placed elsewhere over the Moon have to have their center of mass somewhere along the obrit of the Earth as it dances with Luna.

I think you are taking into account only the gravity of Luna. A lunar beanstalk is significantly affected by Earth's gravity, so you have to take into account the gravitation of both bodies. Which is what Lagrange did when he calculated the Lagrange points.

The rotation of the Moon is synchronous with the revolution of the Earth-Moon system. So a fixed position relative to the Moon's surface is a fixed position in the Earth-Moon system. The centre-of-mass of a lunar beanstalk must be in a fixed position with respect to the Moon's surface, and therefore with respect to the system. The only possible orbits in the Earth-Moon system that are fized with respect to the system are the Lagrange Points.

The same applies to a planet tidally locked to its star. The star is in a fixed position with respect to the surface of the planet, the beanstalk must also be in a fixed position with respect to the surface of the planet. That means the beanstalk has to be in a fixed position in teh (rotating) star-planet system, and the Lagrange (or Trojan, if you prefer) points are the only such orbits possible.

[Flyndaran]

Originally Posted by Brett

...
The same applies to a planet tidally locked to its star. The star is in a fixed position with respect to the surface of the planet, the beanstalk must also be in a fixed position with respect to the surface of the planet. That means the beanstalk has to be in a fixed position in teh (rotating) star-planet system, and the Lagrange (or Trojan, if you prefer) points are the only such orbits possible.

So while not technically impossible, pretty darn useless?