Payload on a Rotating Tether: Tension and Release
A 1200 kg payload is attached to the end of a 2.0 km tether whose other end is anchored at the rim of a space station rotating with period 90 s. The payload orbits with the station at radius 60 m from the spin axis during the first test, then the tether is paid out to 2.0 km while the system is spun up so that the payload's tangential speed about the axis becomes 85 m/s. Engineers observe that the measured tether tension at the 2.0 km configuration is 3.1 kN, which is lower than their simple centripetal prediction. They must decide whether to release the payload at the 2.0 km configuration or keep it attached for a further spin-up, and they worry the tension reading indicates a fault.
1. Compute the centripetal acceleration and required inward force for the payload at the 2.0 km configuration, and compare it with the measured 3.1 kN tension.
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At radius r = 2000 m and tangential speed v = 85 m/s, the centripetal acceleration is a = v²/r = 7225/2000 ≈ 3.61 m/s². The required inward force on a 1200 kg payload is F = m a ≈ 1200 × 3.61 ≈ 4.34 kN. The measured tension is 3.1 kN, which is about 1.2 kN short of the full centripetal requirement. That gap is a real signal: either the tension sensor is miscalibrated, or some other inward force (gravity gradient, aerodynamic drag, or a partially slack tether segment) is contributing. A slack tether cannot pull, so the most likely physical explanation is that the tether is not fully radial or the sensor is under-reading. The decision hinges on resolving this before release.
2. If the tether is released at 85 m/s tangential speed while the station continues rotating, what happens to the payload's path, and what release speed would give a circular orbit at that radius?
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Release at 85 m/s tangential speed means the payload leaves with that speed at r = 2000 m. For a circular orbit at that radius, the required speed is v_c = sqrt(GM/r). Around a station-scale central mass this is not the relevant comparison; the payload is in the station's rotating frame, not a free orbit about a planet. In the station frame, releasing at 85 m/s sends the payload outward along a straight tangent in the inertial frame, so it recedes from the station and does not stay at constant radius. To remain at constant radius after release, the payload would need to be at rest in the rotating frame, i.e., zero relative speed, which is impossible if it is being flung outward. The practical release speed should be chosen so the resulting inertial velocity matches the desired transfer trajectory, not simply the largest achievable speed.
3. The engineers propose doubling the spin rate to raise tension. Evaluate whether this is safe given the tension scales as v².
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Tension scales as m v²/r, so doubling v quadruples the centripetal force: from about 4.34 kN to about 17.4 kN. The tether and anchor must be rated for that. If the current 3.1 kN reading already suggests a partial fault or misalignment, quadrupling the load is risky. A safer path is to diagnose the 1.2 kN discrepancy first: check sensor calibration, verify the tether is radial, and confirm no slack segment. Only after the tension model matches measurement should spin-up proceed, and then in increments with tension monitoring. Doubling spin rate without resolving the anomaly is not justified.