sup /sci/ quick question
Say the space elavator was built
The rope leading all the way down to the ground was cut and only 1 man is holding it
Since the station in orbit is in a friction less, 0-gravity Newtonian environment would it not be feasible to begin pulling on the rope eventually causing the station to be dragged back down into earth, de-orbiting it and causing it to burn up in atmosphere?
>>8120344
The cord is under an ungodly amount of tension. Cutting it at the bottom makes it fly away with the force of a thousand suns
>space elevator
>rope
>0 gravity Newtonian (kek) environment
>>8120344
>Since the station in orbit is in a friction less, 0-gravity Newtonian environment
>>8120344
Opie, if there was very little tension on the "rope", why would it exist at all?
t. codemonkey
>>8120344
>can 1 man pull an entire space station
Thanks for that OP, that made my day
>>8120360
Not op, and correct me if I'm wrong, but wouldn't the space elevator be at geostationary orbit? Other geostationary satellites don't have cables attaching them to earth and they don't fly away.
>>8120344
>>8120344
The station is weightless, but the rope within the atmosphere is not. Imagine how much that big ass rope weighs.
There is so much wrong with this thought, but even if a man could hold onto the rope without the rope flinging the man far into the atmosphere, it is still not in a space of zero g's. As previously stated, the station would have to be in geostationary orbit and because of that it would have a ton of angular velocity. It would have so much angular velocity that it would make it impossible for someone alone to alter its course. It'd be like suggesting it'd be possible to pull the moon towards Earth with a rope.
>>8120518
Then the cable will pull down the anchor.
>>8120666
The station can be at geostationary orbit but it would require a counter weight past geostationary orbit.
The station can actually be at any point along the path but it would have gravity and anything departing the station would be placed into an elliptical orbit.
>>8120360
Actually, it's not under any tension aside from its own weight, which, assuming linear mass density of
[math]
\sigma
[/math]
is
[math]
F_g = -\int_0^H_{station} \frac{GM_e \sigma}{(y+R_e)^2} dy
[/math]
Aside from that, the force of gravity acting on the station is already such that it does not require additional force to keep it in orbit. If the rope were not detached, the tension would increase as the station progresses, since it is now undergoing circular motion not with radius
[math]
H_{station} + R_e
[/math]
but with radius
[math]
H_station
[/math]
and with the same velocity (at apoapsis in the new trajectory). The velocity would be a function of y, and the tension a function of velocity, height, and the angle between apoapsis and the station's current position.
Explicitly;
[math]
\frac{\sqrt{frac{GM_e}{R_e + H_{station}}}^2}{2} - \frac{GM_e}{R_e} = \frac{v(y)^2}{2} - \frac{GM_e}{y}
[/math]
and
[math]
F_t = -\frac{mv^2}{H_{station}} + m\frac{GM_e m}{(y)^2} cos(\theta)
[/math]
Pretty simple really.
>>8120997
Ugh. Fucked up my TeX.
[math]
F_g = - \int_0^H_{station} \frac{G M_e \sigma}{(y+R_e)^2} dy
[/math]
[math]
\frac{\sqrt{\frac{GM_e}{R_e + H_{station}}}^2}{2} - \frac{GM_e}{R_e} = \frac{v(y)^2}{2} - \frac{GM_e}{y}
[/math]
>>8120997
And one last thing, the last line of TeX is the force the guy feels, and has an extra term, which is the same as the negative integral expression for F_g
>>8121008
And there's an extra m in the second term of F_t
