orbital space elevator.

[cameron marical]

Hello, I came up with the idea of a space elevator being, instead of Konstantins original idea of a straight one,{lse},It Is curved outward and away from earth. In the shape of a sideways j.

Obviously the chord would be rooted to the ground, and with a little "push" to launch a load off and get going, it would get into space using earths revoloutions. It would work right?

 

getting back, you would just have to go faster than the chord Is spinning due to earths revoloutions right? Not that fast. Less energy overall? Is this good?

 

Thanks.

Edited May 26, 2009 by cameron marical

[CaptainPanic]

What you mean is to make a sling shot kinda elevator.

 

That only works if you achieve the escape velocity of the earth, which is 11 km/s. The surface of the earth, at the equator, rotates with:

 

2*pi*6.378*10^6 / (24*3600) = 463 km/s, which is not nearly enough (luckily for us).

[Sisyphus]

As CaptainPanic says, you can't just slingshot off from the surface. (If you could, stuff at the equator would be flying off into space). Of course, the higher you go, the less the escape velocity. ( The shape of your path doesn't matter.) At the top end of a space elevator, beyond geostationary orbit, it would indeed be a "slingshot" as the end is dragged along by the Earth. Indeed, that's the whole idea, and the reason why it can stand up in the first place. The counterweight, experiencing a centrifugal effect, pulls on the lower portions and supports their weight.

 

As for the shape of it, that's going to be determined by orbital mechanics. You can't specify a particular shape, because it wouldn't be rigid in the first place. It's just an extremely long, extremely strong rope. And it would be curved, not straight up and down, AFAIK.

[SH3RL0CK]

What you mean is to make a sling shot kinda elevator.

 

That only works if you achieve the escape velocity of the earth, which is 11 km/s. The surface of the earth, at the equator, rotates with:

 

2*pi*6.378*10^6 / (24*3600) = 463 km/s, which is not nearly enough (luckily for us).

 

I think you mean 463 m/s; or 0.463 km/s.

[CaptainPanic]

I think you mean 463 m/s; or 0.463 km/s.

 

Umm... yes... indeed Thanks!

 

I used Google as my calculator, and messed it up in copypasting, causing Google to give me the wrong answer.

It missed a dot, and I blindly and stupidly copied the answer, and then proceeded to claim that 463 is less than 11.

I guess I didn't score a lot of points there.