mfb
Mentor
- 37,398
- 14,233
This.Simon Bridge said:I'm reading mfb's proposal as "having paid the enormous cost of getting one in the first place - we can use the space elevator to draw off energy from the Earth rotation ..." which would satisfy OPs question - sort of.
Building a space elevator purely as power plant is highly impractical. But if that structure is ever built for cheap space launches, it would be possible to use it as power plant, too (engineering issues are a different thing). The first cable would extend to some point a bit above GEO to keep the whole thing in place, afterwards you can extend it outwards with material lifted on the space elevator itself.
I think you have to extend the equation a bit in the general case, as the center of mass and rotation shifts, too. Assuming I of Earth already contains the space elevator (or neglecting its mass):K^2 said:For a sufficiently massive object with sufficiently low angular velocity, the second equilibrium will actually be at higher potential than surface potential, despite your formula still saying you can gather energy from it.
The initial center of mass relative to the geometric center of Earth is Rm/(M+m), assuming that the mass to lift just lies around on the surface. It becomes important once you lift your mass over large distances as done in your calculation:
The center of rotation will be at x=r m/(M+m), this gives a total moment of inertia of I + Mx^2 + m(r-x)^2 (neglecting momentum of inertia of m).
Using the reduced mass \mu=\frac{Mm}{M+m}, the r-dependent angular momentum becomes
\left(I+M\frac{r^2m^2}{(M+m)^2}+m\frac{r^2M^2}{(M+m)^2}\right)\omega(r)<br /> = (I+\mu r^2)\omega(r)Short:
(I+\mu R^2)\omega_0 = (I+\mu r^2)\omega(r)
Solving this for ω(r) gives
\omega(r)=\omega_0\frac{I+\mu R^2}{I+\mu r^2}
The total potential+kinetic energy at distance r now is given by
U<br /> =\frac{-GMm}{r} + \frac{1}{2}(I+\mu r^2)\omega_0^2 \left(\frac{I+\mu R^2}{I+\mu r^2}\right)^2<br /> =\frac{-GMm}{r} + \frac{\omega_0^2(I+\mu R^2)^2}{2}\frac{1}{I+\mu r^2}<br />
Fixing m to some value and introducing positive constants a,b and c, the general shape is
U=\frac{-a}{r} + \frac{b}{c+r^2}
Derivative:
U'=\frac{a}{r^2} + \frac{-2br}{(c+r^2)^2}
Solving for U'=0:
a(c+r^2)^2 = 2br^3
Small angular momentum or high mass of Earth gives small b or large a, while c is constant (given by the geometry of earth), therefore the equation will not have any solution.
Ok, I can confirm your result.
Bending is significant only if you want to lift a lot of mass at the same time.And if we are talking about practical limitations, there is no material out of which you can build the elevator like this. Nothing will provide sufficient resistance to bending over such distance.