What's the path of something dropped from a space elevator

Question

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?

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:

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.