3 ms·
Colin, you are right. Of course if you are sliding along it you have some option to grip it and thus provide additional force needed to keep you in the vicinity
by lmilcin 8y ago
Colin, you are right. Of course if you are sliding along it you have some option to grip it and thus provide additional force needed to keep you in the vicinity of it.
Of course, if you decide to actually fall to Earth from L1 that starts becoming problem. I'm too lazy to calculate exact terminal velocity you would reach when falling L1 to Earth (it won't be the 11km/s you would reach if you were falling from Moon orbit because you also have Moon tugging you). I estimate it not lower than 9km/s (Moon has 2.4km/s escape velocity).
Now, if you moved along the pole at 9km/s you would feel much more tug (noticable 2.5cm/s2) or about 1/40th of Earth surface gravity. If you were 80kg person that would mean about 2kg apparent force.
Of course, by that time you wouldn't be thinking much about it as you would have other problems in mind...
- HopDavid 8y agoAltitude of EML1 is about 320,000 kilometers. An ordinary 320,000 x 300 km earth orbit would be moving 10.81 km/s at perigee. As an object falls towards earth it will surge ahead of the moon and the moon will be pulling it backwards. This makes effective apogee more in the range of 300,000 km. A 300,000 x 300 km orbit has velocity of about 10.8 km/s at perigee. In other words, dropping from the neighborhood of L1 will still give you speed very close to escape velocity by the time it reaches earth. In the case of orbital tether, an elevator car would have gripping wheels in contact with the tether. If its's descending the wheels can power a generator to charge the elevator car's battery.