View Full Version : Space elevator questions

2006-Mar-09, 02:34 AM
Hi. I have a question regarding the concept of a space elevator. My maths just isnít up to solving it myself

If the elevator is designed so that the counter balance consists of a continuing length of ribbon that stretches out say 150,000km. A weight starting at the geostationary position and sliding freely out along the ribbon will be accelerated both tangentially and outwards from the earth. Iím curious as to what total velocity WRT earth it would achieve and the peak g-force it would experience.

Thanks in advance for any help.

2006-Mar-09, 02:46 AM
well, if the cable is 150000km long the end will trace a circle 2*pi*150000km in a day. That works out to 39268.75 kph.

2006-Mar-09, 02:48 AM
Actuall slightly faster, i was assuming 150000km from the center of earth not the surface.

2006-Mar-09, 02:57 AM
That's assuming the object is stopped at the end of the ribbon and allowed to come to rest (WRT the ribbon). I'm thinking if it was allowed to slide along the ribbon from the geostat position it will develop considerable additional velocity. I just don't know how to figure out how much. :(

2006-Mar-09, 03:11 AM
Hrmm not sure about that. The cable can only put a force on the climber perpendicular to itself, the climber moves 'outward' past geosync only due to its own inertia (unless it's a powered ascent). When it's released, wether it's stopped WRT the cable or not, it will always release tangental to the path swept by the cable at that height. So it should always release with the same velocity as the cable at that height.

2006-Mar-09, 04:09 AM
Oh dear. Iím even more confused now. I thought as the climber (slider) moved outwards from geostat the centripetal force is now greater than gravity so it is accelerated both tangentially and outwards. I've obviously got this badly screwed up in my head.

2006-Mar-09, 04:51 AM
Centripital force accelerates inward, not outward. It's the force keeping the end of the cable in a circular path when it's moving faster than orbital speed, the tension in the cable. There is no real 'outward' force, there is only a force perpendicular to the cable. The 'climbing' is just the climber trying to continue in a straight line instead of the circular path that the cable is sweeping, as soon as the climber is released it will continue in a straight line perpendicular to the cable at it's release point and moving at the same speed as the cable. The cable and earth will fall away below it only because they continue moving in a circular path while the climber is now moving in a straighter path. If it's not high enough to be at escape velocity, it will end up in an eliptical orbit with perigee at the altitude it was released.

Hrmm not sure if that cleared anything up. :) Hard to explain without pictures.