Rotating (Revolving) skyhook

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neilsox

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How about an orbiting tether about 628 kilometers long? It rotates (revolves) once per hour, so the end passes within about 50 kilometers of the ground while the other end is at an altitude of 678 kilometers. A hypersonic airplane attaches the payload at an altitude of about 51 kilometers, 30 minutes later the pay load is released at an altitude of about 678 kilometers. The circumference of the tip travel is 2000 kilometers, so the speed of the tip is 2000 kilometers per hour, add the speed of rotation (revolving) of Earth's equator = about 1600 kilometers per hour subtract the orbital speed of about 28,000 kilometers per hour = -24,400 kilometers per hour = the required speed of the hypersonic airplane. The minus sign has little or no meaning. The release speed is the sum = 31,600 kilometers per hour, so the pay load can go anywhere in the inner solar system, except Mercury and the Sun with a minor mid course correction. So a sling shot maneuver around Jupiter is possible, allowing anywhere in the solar system.
The tether strength requirements are moderate, unless we increase the rotational speed, which decreases the hypersonic airplane speed. Retiring ICBMs = intercontinental ballistic missiles can be modified to attach small pay loads, so we can test the concept without building the hypersonic airplane. There are some more details: 1 We need to reduce the jerk which occurs shortly after attachment, to avoid breaking the tether 2 The tether will be a lazy S instead of straight due to air resistance at 50 kilometers, but the payload and the air resistance will stretch the tether. These tend to cancel. 3 The lift and air resistance are not free, so we will need to restore the 2000 kilometers per hour and circularize the orbit after a few payload lifts. 4 Up to 48 payloads per day can be lifted. 5 The orbit can be semi-polar allowing easy access by most of the nations of Earth and a wide range of throw directions for the payload. 6 The tether will have stretch transients which can aggravate or help the attachment and the throw accuracy. 7 Making the tether longer reduces air resistance and the speed of the hypersonic airplane, but initial cost rises rapidly with length. 8 This concept was published in an old Analog perhaps 20 years ago. 9 An Edwards type climber should travel the tether to repair damage by space junk, atomic oxygen and micro meteorites. Neil editited to add (revolves)
 
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Boris_Badenov

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Re: Rotating skyhook

What kind of "G" loads will the payload experience?
 
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MeteorWayne

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Re: Rotating skyhook

Just FYI, a 678 km orbit is about 98.3 minutes, not an hour :)

A 50 km orbit is 85.5 minutes.

So to begin with, the 50 km end will be moving faster than the 678 km end.

And don't forget, at 50 km, the lower end is still in the atmosphere, so will be losing energy to friction all the time.

How are you going to supply the energy to keep this contraption in orbit?

A 60 minute orbit is about 1300 km below the earth's surface, where the frictional loads will be considerable greater.

An orbit right at the earth's surface is 84.5 minutes, but you'll have to make sure everyone ducks, and with a speed of 7908 meteors/sec (7.9 km/sec), the sonic boom and heating will be additional problems as well as the friction.
 
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neilsox

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Re: Rotating skyhook

I don't know how to calculate the G load, but I think it is about 2 g max if we minimise the jerk. G load will decrease slowly as the pay load rises ending about one g at release. A longer tether decreases the g load. Less than one hour per rotation means, more g, and faster release speed, but a stronger tether is needed. 200 kilometers per hour attachment speed at low altitude may be practical, when, and if, CNT = carbon nanotubes with great specs are used to make the tether. As much as 10g is possible unless the tether is about 35,000 kilometers long = maximum to clear the GEO satellites. Neil
 
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neilsox

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Hi Wayne: I agree, about 97 minutes. I possibly should have said revolves once per hour. If the tether revolves (rotates) with a 194 minute period it will come closest to the same spot on the equator repeatedly, or 4 spots on the equator will repeat if the tether revolves (rotates) with a 48.5 minute period. At 60 minutes (especially semi polar) the close approach location is almost random, but predictable.
The period of rotation is determined approximately by the altitude of the center of mass = 364 kilometers, so perhaps 97 minutes, instead of 98.3. The tether will experience tide type forces, which are close to negligible.
Fuel is an important concern. Perhaps simplest is a tip rocket at each tip, which fine tunes the attachment and the release, reduces the jerk, restores the 60 minute period and re-circulizes the orbit. It can even assist lifting the payload if the tether is at risk of breaking. Occasional pay loads are tip rocket fuel. Neil
 
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MeteorWayne

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Well the orbit for a center of mass at 364 km is 91.8 minutes, but that doesn't solve the problem of the faster orbit for the bottom (though it will be slowed by the atmosphere, hence dragging the whole thread down) and the slower orbital speed of the top. I suspect the bottom will be slowed enough by friction with the air so that while dragging the whole thing down, the whole will have a reversed C shape.

With so much drag on the bottom, it's orbital lifetime will be measured in weeks or months.

As a reference point, the perigee of the shuttle after deorbit burn (which causes it to reenter in less than half an orbit) is 31 km.
 
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neilsox

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I agree 50 kilometers may be too low at about 25,000 kilometers per hour wind speed, even though the tips are only below 60 kilometers about 10% of the time. Fuel consumption might be lower if the tip motors keep the tips at about 60 kilometers except when there is a pay load to attach.
The tip motors can also make the C lazier or stretch the tether some when payload attachment is expected. The lazy C will straighten some after each tip leaves the atmosphere, but may retain significant S shape until just before reentering the atmosphere an hour later.
Since the tips are connected, by the equivelent of a very long speedometer cable, the speed can only be different briefly. The average speed must be the same unless the tether breaks. Neil
 
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