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In the movie Gravity, two characters are dangling from the international space station by a long tether. I've previously wondered exactly how you could calculate the tidal forces that act on an object connected to another object in orbit.

I think I've got most of a solution... the orbital speed of an very small object orbiting a very large one is defined as $$ v=\sqrt{\frac{\mu}{a}} $$ where $\mu$ is standard gravitational parameter for the large object and $a$ is the semi-major axis. $\mu$ for the Earth is 398,600.4418, and the semi-major axis for the ISS is 6775 km. This gives us an orbital velocity of 7.670333 km/s. We can substitute in the new semi-major axis by subtracting in the length of the tether (be very generous and say 500 meters) and get 7.670616 km/s, which would be the orbital velocity of something that is floating free at that distance from the ISS... but that doesn't really help me.

What I need to be able to do is calculate, given an apogee of 6774.5 km, and an orbital speed of 7.670333, what is the perigee height. If I have that I have both a distance (apogee - perigee) and a time (one half the orbital period of about 90 minutes). This will tell me how many meters per second the tethered object wants to move, but I'm not sure how to turn that into an effective acceleration, or how to calculate the orbital heights.

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  • $\begingroup$ IIRC, 2 tethered objects that would be in stable orbits if untethered, have a rotational period equal to their orbital period. I cannot remember where I learned that though, has something to do with perturbations I think... $\endgroup$ – RBarryYoung Dec 7 '13 at 16:57
  • $\begingroup$ @RBarryYoung yes this stands to reason as the tidal forces will keep the tether pointed at the center of mass of the object around which they orbit. What I'm interesting knowing is how to calculate the exact force the tether exerts on each object, which should be a function of the difference between the orbital speed they would have untethered and the orbital speed of the combined system. $\endgroup$ – Jherico Dec 7 '13 at 19:32
  • $\begingroup$ actually, I believe that it rotates backwards. Remember that the lower end must always move faster than the upper end. Therefore the high point is always moving relatively backwards against the orbit to be in balance. $\endgroup$ – RBarryYoung Dec 8 '13 at 5:15
  • $\begingroup$ No, it doesn't rotate backwards. The axis through the two objects will always point to the center of mass of the object around which they are orbiting. It's the same effect that causes the moon to always show the same face to the earth, and is called tidal locking: en.wikipedia.org/wiki/Tidal_locking $\endgroup$ – Jherico Dec 8 '13 at 23:00
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You can calculate this using the conservation of specific orbital energy and angular momentum. But I not think that this will help you find the resulting tidal force.

The center of gravity of the system of the two assumed rigidly connected bodies will remain on the same trajectory. To find the tidal force, you just need to look at the resulting torque around this center of gravity. But without any dissipation of energy this system will not become tidally locked, since it will keep oscillating/rotating.

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