Problem involving space elevators

[pepos04]

Homework Statement

The idea of the space lift dates back to the 19th century, and has been exploited by numerous authors of science fiction. Since the 1960s, it has also been considered scientifically.

It would consist of a cable of length $h$ and cross-section $\sigma$ anchored to the earth's surface and held tensioned by a counterweight of mass $m$ at the other end, in equatorial geosynchronous rotation, on which shuttles can be loaded to facilitate the most expensive initial part of space travel.

(a) Assuming the mass of the cable is negligible, calculate the length $h = L$ for which the voltage is minimum.

(b) Assuming now that the length of the cable slightly exceeds $L$, $h = L + l$ with$l \ll$L, calculate the cable tension $T(l)$.

(c) Recently, the hypothesis has been put forward of making the cable with carbon nanotube. It is estimated that a single perfect nanotube with a diameter of approximately $1 nm$ in diameter can withstand a tensile stress (tension force per unit of section) up to $\tau_{max} \approx 100 GPa$ before breaking (1 Pa= 1 Newton/m^2). Assuming $l = 100 km$ and again assuming cable mass negligible, estimate the maximum usable value for the mass $m$, for a cable with a diameter $d = 10 cm$, assuming that the cable has the same breaking strength of the single nanotube.

(d) Finally, set the problem when taking into account the mass of the cable. Useful data: Earth radius $R \approx 6 \times 10^6 m$; Earth mass $M \approx 6 \times 10^{24} kg$ ; angular frequency of rotation of the Earth $\omega \approx 7 \times 10^{-5} rad sec^{-1}$; gravitational constant $G \approx 7 \times 10^{-11}(N m^2/kg^2)$.

Relevant Equations

$\omega^2 L = \frac{GM}{L^2}$

(a) The length $h = L$ for which the tension is minimum is the length that corresponds to the geostationary orbit, where the angular velocity of the cable matches the angular velocity of the Earth. This is because at this point, the centrifugal force balances the gravitational force, and the cable is in equilibrium. So the value of $L$ can be found by solving the equation: $$\omega^2 L = \frac{GM}{L^2}$$ , where $\omega$ is the angular velocity of the Earth, $G$ is the gravitational constant, and $M$ is the mass of the Earth. Rearranging and taking the positive root, I get: $$L = \sqrt[3]{\frac{GM}{\omega^2}}$$, which is an already known value. (b) If we have an object moving in a circle, and the only forces acting on it are some force $T$ and gravity, then the sum (taking downward forces as positive) must be the force required for uniform circular motion: $$m\omega^2 h = T + GMm/h^2 \Rightarrow T(h) = m\omega^2 h- GMm/h^2$$, which is the most general solution. Taking $h= L+l$, we would have: $$T(l)=m\omega^2 (L+l)- GMm/(L+l)^2$$. But there is an issue. Eventhough (a) could be correct, we should simply work out the general solution for all lengths h> 0 and then find the specific L that has the minimum tension. We should even check that it is a minimum, and not the boundary between pulling and pushing. So, we would take the derivative of T(h) and equal it to zero. By making that, I obtain $h = L =-\sqrt[3]{\frac{2GM}{\omega^2}}$, which is a different solution for $R$. How is it possible? (c) If the cable is made of carbon nanotubes, then the maximum tensile stress that it can withstand is $\tau_{max} \approx 100 GPa$. The tensile stress is the ratio of the tension force to the cross-sectional area of the cable, so we have: $\tau_{max} =\sigma T$, where \sigma=\pi\frac{d^2}{4} is the cross-sectional area of the cable, with d = 10 cm being the diameter. Solving for $T$, we get: $$T=\tau_{max} \sigma= \tau_{max} \pi \frac{d^2}{4} $$. This tension must be more than or equal to the tension T(l) calculated in part (b), so we have: $$\tau_{max}\pi\frac{d^2}{4} \geq m\omega^2 (L+l)- GMm/(L+l)^2$$, from which we could find $m_{max}$. (d) If we take into account the mass of the cable, then the problem becomes more complicated, since the tension is not constant along the cable, but varies with the height. The differential equation that governs the tension $T(h)$ is: $$\frac{dT}{dh} =-\frac{GM\lambda}{h^2} + \omega^2 \lambda h$$, where $\lambda$ is the linear mass density of the cable, assumed to be constant. This equation can be solved by using an integrating factor… If I am wrong, where am I wrong? Can you give me advice? I came also across this paper (https://users.wpi.edu/~paravind/Publications/PKASpace Elevators.pdf), but I just can't understand all the variables involved, there are so many of them and I am slightly confused. Can you try and have a look at this? It seems helpful also.

 

[erobz]

Relevant Equations: $\omega^2 L = \frac{GM}{L^2}$

$L$ is the length of the cable going from earth's surface to the counterweight in orbit. Maybe you should check you are at the correct radius?

Writing a force balance would also be good for the mass $m$?

 

[haruspex]

$L$ is the length of the cable going from earth's surface to the counterweight in orbit.

True, but if we redefine L and h as from centre of the Earth then all the algebra looks ok to me.

We should even check that it is a minimum, and not the boundary between pulling and pushing.

The tension cannot be less than zero, so zero is a valid minimum. It does not have to satisfy the derivative being zero.

For part b, I would think you are expected to make an approximation for $l<<L$.

For part d, why do you need an integrating factor? How about just integrating straight off?

Note that the paper you link considers the mass of the cable as being far more significant than that of the distal mass. Indeed, it seems to ignore the latter, despite having defined a symbol for it. Also, it allows that the diameter varies. Judging from the statement that the solution involves an exponential, the assumption seems to be that the cross sectional area is proportional to the tension.

It also specifies zero tension at the ground.

 

[erobz]

True, but if we redefine L and h as from centre of the Earth then all the algebra looks ok to me.

From the text of the OP, it doesn't appear to be defined that way. Either that math or the definitions need corrected...Thats the point I was trying to address at the moment.

 

[haruspex]

From the text of the OP, it doesn't appear to be defined that way. Either that math or the definitions need corrected...Thats the point I was trying to address at the moment.

Understood, but it is a simple matter to map back to the given definitions at the end. It should not stand in the way of answering the question the OP asked.

 

[pepos04]

The tension cannot be less than zero, so zero is a valid minimum. It does have to satisfy the derivative being zero.

The problem is that by doing the derivative and setting it equal to 0 I do not get the same result.

For part b, you are expected to make an approximation for l<<L.

What does it mean? What should I do?

It would get rather more challenging if the diameter were varied to make all points on the verge of snapping.

This can be interesting. If, after you have helped me solve the problem, you want to provide advice and guidance on this, you are welcome.

For part d, why do you need an integrating factor? How about just integrating straight off? The paper you link is only relevant to part d since it considers the mass of the cable as being far more significant than that of the distal mass. Indeed, it seems to ignore the latter, despite having defined a symbol for it. It also specifies zero tension at the ground.

Could you go further into this?

From the text of the OP, it doesn't appear to be defined that way. Either that math or the definitions need corrected...Thats the point I was trying to address at the moment.

@haruspex @erobz I am attaching the figure attached to the problem to avoid any doubts. Now, you will be able to tell me whether I am doing it all wrong or not.

 

[Steve4Physics]

The problem is that by doing the derivative and setting it equal to 0 I do not get the same result. What does it mean? What should I do?

I'd like to add to what @erobz and @haruspex have said.

If $R$ is the earth’s radius the tension can be expressed as
$T(h) = m\omega^2(R+h)~ –~ \frac {GMm}{(R+h)^2}$
This is an equation you derived, except you incorrectly used $h$ instead of $R+h$.

You can see the dependence of tension on $R+h$ by considering the general shape of a curve such as $y = x – \frac 1{x^2}$ (where $y$ represents $T$ and $x$ represents $R+h$.

Here's a plot of $y = x – \frac 1{x^2}$ produced using Desmos:

Note that there is no minimum at the point where tension ($y$) is zero. (Also, negative values of $x$ have no physical meaning in the present context, but do give rise to a local maximum in the graph.)

That's easy to explain. Imagine a rigid rod rather than a cable. Negative values of ‘tension’ correspond to compression; so values of 'tension' which are less than zero are valid. Zero tension is not a minimum.

Minor edit to imrove readability of equation.

 

[pepos04]

I'd like to add to what @erobz and @haruspex have said. If $R$ is the earth’s radius the tension can be expressed as $T(h) = m\omega^2(R+h)\frac {GMm}{(R+h)^2}$ This is an equation you derived, except you incorrectly used $h$ instead of $R+h$. You can see the dependence of tension on $R+h$ by considering the general shape of a curve such as $y = x– \frac 1{x^2}$ (where $y$ represents $T$ and $x$ represents $R+h$. Here's a plot of $y = x– \frac 1{x^2}$ produced using Desmos: View attachment 336748 Note that there is no minimum at the point where tension ($y$) is zero. (Also, negative values of $x$ have no physical meaning in the present context, but do give rise to a local maximum in the graph.) That's easy to explain. Imagine a rigid rod rather than a cable. Negative values of ‘tension’ correspond to compression; so values of 'tension' which are less than zero are valid. Zero tension is not a minimum.

Good explanation. I obtain, therefore, for the first point: (a): $h = L =-(R+ \sqrt[3]{frac{GM}{\omega^2}})$. Does everyone confirm this? This result seems suspicious to me because it is negative... (b): For point (b), should I substitute, in the expression for $T(h)$, $L+l$ into $h$, with $L =-(R+ L = \sqrt[3]{{frac{GM}{{omega^2}})$? I.e., $T(l) = m\mega^2(R+L+l)- \frac {GMm}{(R+L+l)^2}$, with $L = \sqrt[3]{\frac{GM}{\mega^2}}$? I await further advice on points (c) and (d). I still can't come up with them.

 

[Steve4Physics]

I obtain, therefore, for the first point: (a): $h = L =-(R+ \sqrt[3]{frac{GM}{\omega^2}})$.

Disagree. Check your algebra and your LaTeX.

Note that the Preview icon is the one on the top-right of the toolbar when editing. Using the Preview icon will let you check that your LaTeX is correct before posting.

Also, note that in Post #8 you have lost a minus sign when quoting Post #7. The correct equation is
$T(h) = m\omega^2(R+h)~ –~ \frac {GMm}{(R+h)^2}$
not
$T(h) = m\omega^2(R+h) \frac {GMm}{(R+h)^2}$
I have edited Post #7 to put spaces on both sides of the minus sign for clarity.

 

[pepos04]

@Steve4Physics I had mistyped. My result is: $h = L =- (R + \sqrt[3]{\frac{2GM}{\omega^2}})$. What about points (b), (c), (d)?

 

[haruspex]

The problem is that by doing the derivative and setting it equal to 0 I do not get the same result.

Sorry, I left out the important word "not" in post #3. I did correct it four hours before your post #6, but somehow you got the old version.
I made numerous other edits to that post, mostly because I came to understand more about the linked paper, so please read the updated post.

What does it mean? What should I do?

You have a term in the equation of the form $(L+l)^{-2}$. Get that into the form $(1+l/L)^{-2}$ so that $l/L$ is very small, use the binomial theorem to expand that into a series, then discard all except the first two terms.

For part d, why do you need an integrating factor?

Could you go further into this?

You have an equation $dT/dr=f(r)$. Just integrate as $T=\int f(r)$.
Wrt varying the diameter, I think I made an error. I was fooled by the paper's use of T for stress instead of tension. More later.

you incorrectly used h instead of R+h.

True, but the algebra is a lot easier if we either redefine these measures as being from Earth's centre or, if preferred, make up some new variables for that, like "y". The map back to the original variables at the end is trivial.

 

[Steve4Physics]

True, but the algebra is a lot easier if we either redefine these measures as being from Earth's centre or, if preferred, make up some new variables for that, like "y". The map back to the original variables at the end is trivial.

The original question defines $h$ as the length of the cable. Parts a) and b) are posed in terms of $h$. So to avoid ambiguities I'd go for the $y=R+h$ approach.

 

[haruspex]

Wrt varying the diameter

If
$r$ is the distance of an element length $dr$ from Earth's centre,
$A=A(r)$ is its cross sectional area
$\rho$ is its density
$\sigma(r)$ is the stress at $r$ and $T(r)$ is the tension there
then we have
$T(r+dr)-T(r) +(A\rho dr)\omega^2r=\frac{GMA\rho dr}{r^2}$
$T=A\sigma$

Model I: $A$ constant
This is the model the paper uses in its section II.

$\frac{dT}{dr}=A\rho(-\omega^2r+\frac{GM}{r^2})$
We can integrate this immediately to obtain T as a function of r.
The paper specifies $T(R_e)$, i.e. at Earth's surface, =0.
This produces a tension which rises from zero to a maximum at geostationary orbit then declines. We can either terminate the cable with a mass m at $r_{max}$, so $T(r_{max})=mr_{max}\omega^2$, or dispense with that and define the cable length by $T(r_{max})=0$.

Model II: $\sigma$ constant
This is the "tapered tower" model the paper uses in its section III.
$dT=\sigma dA$.
$\sigma\frac{dA}{dr}=A\rho(-\omega^2r+\frac{GM}{r^2})$
This time we will get $\sigma\ln(A)=$ the integral of the function of r. This is problematic because it means the tension is nowhere zero. I am not sure how to interpret this. The paper simply plugs in an arbitrary nonzero value at $r=R_e$ and the same value at $r_{max}$.
Arguably, there would be a minimum practical value for A. The optimal solution would then be to use that area out to some radius a bit above ground level then transition to the tapered solution. Determining that radius would be a second stage optimisation problem.

 

[pepos04]

@haruspex @Steve4Physics I am very confused about your assertions on point (a). What is the correct choice? Write a general expression for $T(h)$ with $R+h$, take the derivative and set it equal to 0, and find the corresponding value of h accepting the possibility that there are negative values of $T$? Or simply write $T(h)$ with $h$, set it equal to 0 without differentiating it, and find the value of $h$ with respect to the center of the Earth?

The results to which these two methods lead (the first by @Steve4Physics, the second by @haruspex) are different, at least it seems so to me (see posts #1 and #10). Is there a third result that I'm missing? Is there some way these two methods find a point of convergence and I don't see it? I would like clarity on this point...

As for (b), the binomial expansion is a great idea that I hadn't thought of, but I can't do it if I can't figure out the correct form of $T(l)$, which depends on what we discussed. For points (c) and (d), do you have any advice? I don't understand if my development of (c) in post #1 is correct, while on point (d) I don't have a clear understanding of the text's requests, even if I outlined a small procedure in post #1. I await clarification.

 

[Steve4Physics]

@Steve4Physics I had mistyped. My result is: $h = L =- (R + \sqrt[3]{\frac{2GM}{\omega^2}})$.

Your algebra is incorrect. A negative answer is not physically meaningful. I shouldn't really do this but ...

$T(h) = m\omega^2(R+h)~ –~ \frac {GMm}{(R+h)^2}$ (do you agree?)

When tension is zero, $h=L$ and this gives:
$m\omega^2(R+L) =\frac {GMm}{(R+L)^2}$

Rearranging gives:
$L = \sqrt[3]{\frac{2GM}{\omega^2}} - R$
which you need to evaluate. Note there was no need to differentiate.

You haven't yet sorted out part a), so I won't comment on the other parts.

 

[haruspex]

take the derivative and set it equal to 0

Setting the derivative to zero is not going to work here.
Consider this question: A ball is dropped from a height of 1m; find its maximum speed.
Is setting a derivative to zero going to answer that?
A tension cannot be less than zero. If you can find somewhere that it is zero you have found a minimum.

the correct form of T(l),

Not sure what your issue is with this.
You obtained a correct equation for that in which L is measured from the centre of the Earth. Using the binomial expansion and keeping just the first two terms will give constant + term linear in $l$. Substitute the value for L found in (a) (the value which gives zero tension) and you should find the constant term vanishes.
If you still have any L remaining you will need to replace it with (L+R) to match the definition given in the question.

For c, your work is correct except you have a typo here

we have: τmax=σT,

The sigma is on the wrong side of the equation.

For (d), your ODE has a sign error. Start from the force balance equation. If the tension at h is T(h), what is it at h+dh?

 

[pepos04]

Setting the derivative to zero is not going to work here. Consider this question: A ball is dropped from a height of 1m; find its maximum speed. Is setting a derivative to zero going to answer that? A tension cannot be less than zero. If you can find somewhere that it is zero you have found a minimum.

@haruspex I understand. However, if you assert this, how can this be reconciled with what @Steve4Physics wrote in post #7? There it is explicitly stated that the voltage can take on negative values and an intuitive explanation of this is given at the end. So, does this mean that what @Steve4Physics said is wrong? That is: the two statements are irreconcilable, but I cannot refute either of them, given your valid explanations. Any clarifications? Moreover, I’ll try the next points following your advice.

Your algebra is incorrect. A negative answer is not physically meaningful. I shouldn't really do this but ... $T(h) = m\omega^2(R+h)~–~ \frac {GMm}{(R+h)^2}$ (do you agree?) When tension is zero, $h=L$ and this gives: $m\omega^2(R+L) =\frac {GMm}{(R+L)^2}$ Rearranging gives: $L = \sqrt[3]{\frac{2GM}{\omega^2}}- R$ which you need to evaluate. Note there was no need to differentiate. You haven't yet sorted out part a), so I won't comment on the other parts.

Two objections: 1) My final result is $L = \sqrt[3]{\frac{GM}{\omega^2}}- R$, not $L = \sqrt[3]{\frac{2GM}{\omega^2}}- R$. In fact, I don't see where factor 2 could come from. 2) You say that “tension is zero”, but in post #7 you explicitly said that “there is no minimum at the point where tension is zero” and that “values of 'tension' which are less than zero are valid. Zero tension is not a minimum”. My confusion arises from that.

 

[Steve4Physics]

@haruspex I understand. However, if you assert this, how can this be reconciled with what @Steve4Physics wrote in post #7?

As far as I can see, @haruspex has said nothing that conflicts with what I have said and vice versa.

There it is explicitly stated that the voltage can take on negative values

That is simply untrue. Why do you say 'voltage'? Note that I did not say tension in a cable can be negative; 'tension' in a rod can be negative - it is then compression.

and an intuitive explanation of this is given at the end. So, does this mean that what @Steve4Physics said is wrong? That is: the two statements are irreconcilable, but I cannot refute either of them, given your valid explanations. Any clarifications?

There is nothing irreconcilable; it appears that you have misunderstood what has been said.

Moreover, I’ll try the next points following your advice. Two objections: 1) My final result is $L = \sqrt[3]{\frac{GM}{\omega^2}}- R$, not $L = \sqrt[3]{\frac{2GM}{\omega^2}}- R$. In fact, I don't see where factor 2 could come from.

It is you who incorrectly used the factor 2. The correct formula (without the 2) is mine. You have swapped them!!!

What you are posting is (IMO) so incorrect and muddled that it makes a sensible reply impossible. I won't be posting further on this thread. Good luck.

 

[berkeman]

Thread closed for Moderation. @pepos04 -- check your PMs and respond please.

 

[berkeman]

Okay, thread will remain closed. Turns out pepos04 was a sockpuppet of a previously banned member. Thanks to all who tried to help them.