Deriving the space elevator energy saving formula for geostationary orbit

[bksree]

Hi
I read in a paper that the energy saving in taking a payload to geostaionary orbit with a space elevator is (R/Rg)*(2-R/Rg) where R- radius of earth, Rg - radious of geostaionary orbit.

How is this obtained ?

TIA

 

[ZapperZ]

Hi
I read in a paper

Please make a complete citation of the paper!

Zz.

 

[Bill_K]

It's easy enough to calculate, and the formula you give is wrong, but more importantly the premise is wrong. Reaching orbit requires an increase in potential energy and an increase in kinetic energy. These are the same regardless of how you get there, whether you use a rocket ship or a space elevator.

 

[bksree]

Thanks for the replies. The paper is : P.K Aravind, 'The Physics of the space elevator', Am. J. Phys., 75(2), Feb 2007.
The eqn actually gives the percentage saving of energy w.r.t that required if rocket prpoulsion is used. I think it is related to the energy required to accelerate to escape velocity with rocket propulsion wheras the space elevator uses the centrifugal force ofthe rotating Earth to accelerate the satellite.

TIA

 

[nasu]

You don't need escape velocity to put a satellite in geostationary orbit. If it has this much velocity it won't be neither geo- nor stationary.

The paper may be about the energy "saved" from the point of view of not using fuel but rather some other way to get the energy required.
Similar to the energy "saved" by connecting an appliance to a solar cell or wind generator rather than the wall plug.