Why Does Tension Peak at Geostationary Height in a Space Elevator?

Katie1990
Messages
6
Reaction score
0

Homework Statement



I am tasked to summarise this paper http://chaos.swarthmore.edu/courses/PDG07/AJP/AJP000125.pdf

I am struggeling to understand the statements of tension at the bottom of the first side namely

For an element at geostationary height that is, at a distance from the Earth’s center equal to the radius of geostationary orbit the weight and centrifugal forces are equal W=FC, and therefore the tension forces at the two ends must also be equal FU=FD for equilibrium. For an element below geostationary height, the weight force W exceeds the centrifugal force FC and one must have FUFD for equilibrium. These two preceding statements imply that the tension in the tower increases with height from ground level to geostationary height. In contrast, for an element above the geostationary height, the centrifugal force FC exceeds the weight W and hence FUFD for equilibrium, implying that the tension in the tower decreases as a function of height past the geostationary height. A free standing tower is one for which
the tension drops to zero at both ends, requiring no restraint at either end to keep the tower in place. The overall picture of a free standing tower is thus of a structure in which the tension rises from zero at ground level to a maximum value at geostationary height, and then decreases to zero again at the upper end.

Homework Equations





The Attempt at a Solution



I can't understand why tension is maximum at geostationary orbit surely the two tensions at this point sum to zero? And shouldn't the tensions at each end be maximum and not zero as stated?
 
Physics news on Phys.org
Katie1990 said:
I can't understand why tension is maximum at geostationary orbit surely the two tensions at this point sum to zero? And shouldn't the tensions at each end be maximum and not zero as stated?

Consider a chain hanging from the ceiling with the first link fixed to the ceiling and the final one dangling just above the floor. There are n links, each with a weight of 1 N. What's the tension on the first link? How about the final (dangling) link?
 
I managed to solve this I had forgotten that tension has no direction, I had been cancelling tensions rather then summing them!
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...

Similar threads

Back
Top