Take the 2-minute tour ×
Space Exploration Stack Exchange is a question and answer site for spacecraft operators, scientists, engineers, and enthusiasts. It's 100% free, no registration required.

What sort of paths would payloads follow after being released from a space elevator such as the one described in Clarke's Fountains of Paradise?

share|improve this question
1  
Just a note: Great job answering your own question (not sarcastic). This kind of thing works on Physics SE, and it looks like it will work here, too. –  HDE 226868 Sep 1 at 12:34
    
@HDE226868 Thanks. I was hoping answering my own question is kosher. Posting and answering questions seems a good way to convey info. –  HopDavid Sep 1 at 17:48
    
Yeah, I've been waiting a long time for someone to do something like this. I hope it will help the site. From the look of the votes, others agree. –  HDE 226868 Sep 1 at 17:50

1 Answer 1

After being released from an elevator, the payload and earth are a 2-body system. The paths in 2 body systems are conic sections: ellipses, parabola, and hyperbolas. What sort of conic section depends on where it is released from the elevator.

The elevator makes a revolution each sidereal day. The angular velocity (aka ω) is 2 π radians/sidereal day.

The speed of an object in uniform circular motion is ωr. In this case r is the point’s distance from earth’s center. So called centrifugal acceleration is $ω^2r$. This is a push away from the earth.

Besides $ω^2r$, there is another acceleration in the opposite direction: gravity. This acceleration is $GM/r^2$. When r=42,164 kilometers, $ω^2r$ and $GM/r^2$ exactly cancel. This is the orbit radius of geosynchronous satellites.

For convenience let’s choose sidereal days as our time unit and geosynch radius as our unit of length. Then orbital period is $r^{3/2}$ sidereal days. For example if the semi-major axis of an orbit is 4 geosynch lengths, then it’s period is 8 sidereal days. If a = 9, the period is 27 sidereal days.

Using these units our speed ωr becomes 2πr/sidereal day.

An orbit’s specific angular momentum is r X v where r is position vector and v is velocity vector. The magnitude of the specific angular momentum is rv sinφ where φ is the angle between position and velocity vectors. For objects traveling a circular path, the angle between position and velocity vectors is 90º. Sin 90º is 1, so the magnitude of angular momentum is simply rv.

Recall the speed is ωr. So $rv= ωr*r = ωr^2$ which is $2πr^2$/sidereal day

Angular momentum is also equal to twice the elliptical orbit’s area divided by it’s period. An ellipse’s area is πab where a is semi major axis and b is semi minor axis. $b=(1-e^2)^{1/2}$ where e is eccentricity. So twice the ellipse’s area would be $2πa^2(1-e^2)^{1/2}$ With the units we’re using the orbital period would be $a^{3/2}$ sidereal days.

So the magnitude of the specific angular momentum is $$\frac{2πa^2(1-e^2)^{1/2}}{a^{3/2}}$$ square geosynch lengths per sidereal day.

$$\frac{2πa^2(1-e^2)^{1/2}}{a^{3/2}}$$ reduces to $$2π(a(1-e^2))^{1/2}$$

So now we have $2πr^2= 2π(a(1-e^2))^{1/2}$

which reduces to $r^2= (a(1-e^2))^{1/2}$

Since the velocity vector is perpendicular to the position vector, r is either the perigee or apogee of the orbit. If released below geosynch, it’s apogee. If released above geosynch r is the perigee.

If r is the perigee, r = (1-e)a. If r is the apogee, r=(1+e)a.

These equations can be rewritten a = r/(1-e) or a = r/(1+e)

Substituting these values of a into $r^2= (a(1-e^2))^{1/2}$ we get

$r^2= (r(1-e^2)/(1-e))^{1/2}$ or $r^2= (r(1-e^2)/(1+e))^{1/2}$

These reduce to

$e=1-r^3$ or $e=r^3-1$ depending on whether r is more or less than 1 geosynch length.

Knowing the eccentricities that correspond to a specific r, we can draw orbits of payloads released from various locations on the elevator:

enter image description here

When r=1, the orbit is circular. I colored this orbit blue.

You may notice I cut the illustration off at $r=2^{1/3}$ geosynch lengths. As r approaches $2^{1/3}$, e approaches 2-1. When e reaches 1, the orbit is a parabola. When e exceeds 1, the orbit is a hyperbola. My argument rests on regarding specific angular momentum as twice an elliptical orbit’s area over the orbital period. For parabolas and hyperbolas those quantities aren’t defined.

Does $e = r^3-1$ when $r >2^{1/3}$? I suspect it does. But this is only conjecture on my part.

share|improve this answer
2  
Yes, in general $e=\left|r^3-1\right|$ for your dimensionless $r$. Also for both cases, $a=\left|{r\over r^3-2}\right|$. –  Mark Adler Sep 1 at 16:28
    
@MarkAdler Thanks. Could you point me to arguments that work for parabolas and hyperbolas? –  HopDavid Sep 1 at 17:40
    
Does this mean that a payload on an elevator (and the elevator itself) are experiencing continuous acceleration (force) to keep them in the siderial day orbit? It would be interesting to hear how that works! –  GreenAsJade Sep 1 at 23:55
    
Yes. The only point on the elevator that experiences no (or very nearly no) forces is right where the orbital period equals a sidereal day. Below that, there is a force down on the cable and the elevator. Above that, there is a force up. These forces, the required mass of the cable, and the tensile strength of known materials are what make a space elevator at Earth nearly impossible to build. –  Mark Adler Sep 2 at 1:39
2  
Much like the elliptical case, we have the energy equation for the hyperbola with just a sign change: ${\mu\over 2a}={v^2\over 2}-{\mu\over r}$. As you noted, for the hyperbola at closest approach, which is where you let go, we have $r=a(e-1)$. So $e=1+{r\over a}$. From the energy equation, and letting $v=\omega r$, $a={\mu\over \omega^2 r^2-{2\mu\over r}}$. Combining, $e=1+{\omega^2 r^3\over\mu}-2$ or $e={\omega^2 r^3\over\mu}-1$. In your dimensionless units, this reduces to $e=r^3-1$. –  Mark Adler Sep 2 at 2:02

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.