Differences between revisions 22 and 23
 ⇤ ← Revision 22 as of 2011-06-27 04:39:43 → Size: 18045 Editor: KeithLofstrom Comment: ← Revision 23 as of 2011-06-27 04:43:12 → ⇥ Size: 18068 Editor: KeithLofstrom Comment: Deletions are marked like this. Additions are marked like this. Line 170: Line 170: $v_p ~ = ~ { \Large \sqrt{ \mu \over r_p } } ~ \left( { { \sqrt{ { \omega_T {r_p}^3 {v_p}^3 } \over { \mu^2 } } - 1 } \over \sqrt{ 1 - { \Large \left( { v_{vc} \over { v_p - \mu / r_p v_p } } \right) } ^2 } } + 1 \right)$ $v_p ~ = ~ { \Large \sqrt{ \mu \over r_p } } ~ \left( { { \sqrt{ { \huge { \omega_T {r_p}^3 {v_p}^3 } \over { \mu^2 } } } - \Large 1 } \over \sqrt{ \Large 1 - { \huge \left( { v_{vc} \over { v_p - \mu / r_p v_p } } \right) } ^2 } } + 1 \right)$

# GEO Rail

## "Landing" Launch Loop payloads in a destination without rockets

Imagine a long and heavy orbiting tether, running vertically from 29900 km radius, through GEO (42164 km), to a counterweight above. The tether has a thin, passive conductive rail around it. It will be used to magnetically capture vehicles and move them to GEO, using the sliding rail to push payloads transversely while they magnetically slide upwards.

This is not the normal apogee capture system, where we launch into an orbit whose perigee transverse velocity matches a hanging tether at the right altitude. Instead, we launch the vehicle a little faster and a little later. That puts the vehicle in an orbit with an apogee trailing the tether. The vehicle will never get to that apogee, because we will slide in front of the tether below apogee, where we still have a high (1800 m/s) radial velocity. We will still match the tether's transverse velocity and altitude, though.

In the rotating tether frame of reference, the vehicle will approach from below at high speed, from the rear, and decelerate in the traverse ("horizontal") direction as it approaches the tether. The vehicle is rising radially, conserving angular momentum, and losing angular velocity as the radius increases. In the tether's rotating frame, this looks like Coriolis acceleration, equal to twice the velocity times the tether's angular velocity. The angular orbital velocity for a GEO tether is 7.2921E-5 radians per second, and the radial ("vertical") velocity will be around 1800 meters per second, so the Coriolis acceleration is 0.26 m/s2. If the vehicle misses the tether rendezvous, it will continue to accelerate retrograde in the rotating frame, with the backwards velocity turning into downwards Coriolis acceleration in the rotating frame, reaching apogee above the intended attach point but well below GEO.

The tether is made with not-excessively-tapered Kevlar. We will need many tons of it, especially at GEO, as an angular momentum bank. More mass is better for this system. THIS is where we put the hotels, the radiation shielding, and the heavy buffet tables for obese tourists.

We launch a vehicle off the launch loop at 10148.7 m/s, and with very good radar and some trim thrusters, we "land" on the bottom of the rail at 29979.7 km altitude. At that point, we have the same angular frequency and circular orbital velocity, and a large vertical velocity component, 1746.8 m/s .

As the vehicle slides up the tether, the weak gravity slows it down radially, as it accelerates forwards in orbit to remain on the "faster at higher altitude" tether.

As the vehicle slides upwards, it magnetically levitates/pushes against the east side of the rail with Coriolis acceleration. That is 26 centimeters per second squared at 1800 m/s vertically, and 7 centimeters per second squared at 400 m/s, our speed as we approach GEO altitude. The Coriolis acceleration and the eddy currents both reduce with relative velocity - most convenient! The vertical deceleration (gravity minus rotational centrifugal acceleration) decreases to zero as we approach GEO, still with significant velocity, though with less than the velocity of a frictionless ballistic because of the eddy currents.

As the vehicle approaches GEO, it is still moving fast, which is good, because it has 12000 kilometers to travel. The trip will take about 4 hours. About 10000 meters below GEO station, the rail surface changes to increase eddy current drag, slowing down the vehicle. The vehicle reaches the drag section at 400 meters per second, and deceleration increases from a few millimeters per second to 10 m/s2, a bit more than 1 gee (with passengers facing backwards, backs into chairs). The vehicle slows to 25 meters per second about 40 meters below the station, and with the aid of some linear motors, reduces deceleration and velocity to zero over the next 5 seconds. The payload or passenger compartment is plucked off the magnet-rail and wing, and pushed into the station through an airlock.

Since the tether rail is not magnetic or active (that would be far too heavy for a 12000 kilometer gossamer structure) the vehicle magnet-rail will need to wrap partly around it in some way. 26 cm/s2 Coriolis acceleration is far too weak to hold the vehicle against a rippling tether at high speed. For example, a ripple with a 20 kilometer wavelength and a 20 meter peak-to-peak amplitude will shake the vehicle back and forth at 0.3 gees ( 12x Coriolis ) and a period of 11 seconds - a mild roller coaster. Such a transverse ripple will not be a standing wave, but traveling up or down the tether at around 1 km per second. It will be a challenge to remove it, perhaps by periodic transverse shaking at GEO by the large momentum mass. It would be more difficult to remove by shaking the bottom counterweight vertically, inducing Coriolis accelerations, because a lot of power is required.

We can reverse the process. Using a magnetic accelerator to launch a vehicle down a rail on the west side of the tether, we falling off the end in a transfer orbit back to the upper atmosphere. This restores most of the GEO rail momentum.

The rocket thrust needed by GEO rail vehicles will be pure velocity correction, centimeters per second if we've done our radar and orbital mechanics calculations correctly. If we miss, we just reenter normally - a delay, but not a disaster. Besides a little correction exhaust, and whatever we need to add to for momentum restoration, the system is mass conservative and mostly energy-recycling.

More rockets bite the dust ...

## Restoring GEO Rail orbital momentum

If we de-orbit GEO station trash much faster than we receive payloads (aiming for empty and lifeless portions of the ocean) we can add more momentum by exploiting Coriolis force acceleration. We can also receive mass (and momentum) from the moon, or launch mass from the loop in slingshot orbits around the moon. However, the moon will not always be conveniently located for this, and vehicles may make many 20 day orbits before arriving with a suitable orbital position.

A slightly stronger tether can swing like a pendulum. That allows vehicles to arrive from the launch loop with higher angular velocity, and leave with lower angular velocity. This variation needs further study.

We will probably need to make up some energy and momentum losses with plasma rocket engines at GEO station, especially if we accumulate more upward vehicles than downward ones. But overall, the energy will be tiny compared to the the launch loop energies, and we can float a lot of solar cell around GEO as part of our "momentum anchor".

As part of an overall "space anti-litter" ethic, we should make sure the plasma rocket engines are offset so they are firing exhaust away from the rest of the geosynchronous ring, and that the rocket exhaust is traveling fast enough to escape the solar system ( >16 km/sec at noon when the rockets are pushing retrograde to the Earth's orbit around the sun, >76 km/sec (!!) at midnight, when the rockets are pushing retrograde ).

A Variable Specific Impulse Magnetoplasma Rocket (VASIMR) engine operates optimally at 50 km/sec, so operation at 80 km/sec is close to optimum. The reaction mass is argon, which can be frozen at -200C with a density of 1600 kg/m3 and a vapor pressure of 0.1 atmosphere for transit from earth. We will need about 1 kg of argon reaction mass shipped up for every 50 kg of uncompensated mass shipped up from Earth. Argon is 0.93% of the atmosphere, and costs about \$1/kg (liquid) in high quantities. It is a byproduct of the production of liquid oxygen.

We will expend about 40 kilowatt hours per "uncompensated vehicle kilogram" operating the plasma rockets, including coolers and radiator operation. Typical solar cell weights for existing satellites are 65 kg/kilowatt, so 1 kilogram of solar cells provides enough energy in a year to bring aboard 3.5 kilograms of mass - a power doubling time of 5 months. However, if we can use "server sky" style ultra-thin InP cells, closer to 1000 kg/kW, we can approach 50 kilograms of mass per kilogram of solar cell. In the very long term, solar cells made from lunar materials will bring both momentum and cheaper launch to the GEO rail, but solar cells are a high-tech undertaking, and it will be a long time before that level of manufacturing technology can operate practically in space.

In the shorter term, we may be forced to get by with lower ISP engines than 80 km/sec VASIMR, perhaps operating only around noon with an exhaust velocity of 20km/sec. That increases our argon needs 4X, but reduces our power needs 16X. We will only be operating our engines with about a 30 percent duty cycle, so our solar panel needs only drop 5.3X. Still, our power doubling time drops to a month or so. Probably worth it during the building phase. We may even be rascals, and run the VASIMR engines 24 hours for a while, dumping a few tons of midnight argon into orbits in the inner solar system.

## Operation of an M288 rail base

A similar rail system can supply the M288 server sky orbits

MORE LATER

## M288 rail bases for orbital debris capture

Server sky thinsats are cheaper to launch if they are lighter, but below a minimum mass-to-area ratio, the orbits are unstable under light pressure. However, we can add mass ballast to a very light satellite, using gram-weight chunks of space debris. The processing is simple, just let the debris shatter as it is collision-captured inside a cornucopia-shaped funnel. Larger chunks can be cut apart at a simple processing plant at the M288 rail base, smaller chunks can be welded together.

Capture vehicles can drop below the tether, in intercept orbits where the debris crosses the equatorial plane. Depending on their orbit, they can return to the same rail base, but more likely a different rail base will be more convenient.

MORE LATER

## The Math

We will neglect oblateness, light pressure, and the gravity of other bodies for now. Simple Kepler two body orbits around the earth.

The vehicle leaves the launch loop at 80km altitude, the apogee of an elliptical orbit with a perigee of r_p = 6458.137 km and velocity v_a . The earth's gravitational parameter \mu = 398600.4418 km3/s2 = 3.986004418e14 m3/s2. Given those three values, we define the orbit:

h ~ = ~ r_p v_p                       angular momentum

a ~ = ~ { \Large { r_p \over { 2 ~ - ~ \LARGE { { r_p {v_p}^2 } \over { \huge \mu } } } } }                 semi-major axis

e ~ = ~ 1 - { \Large { { r_p } \over a } } ~ = ~ { \Large { { r_p {v_p}^2 } \over \mu } } - 1        eccentricity

v_0 ~ = ~ { \Large { { v_p } \over { 1 + e } } } ~ = ~ { \Large { { \mu } \over { r_p v_p } } }             characteristic velocity

r_0 ~ = ~ a ( 1 - e^2 ) ~ = ~ \Large { { {r_p}^2 {v_p}^2 } \over \mu }       characteristic radius

r( \theta ) ~ = ~ \Large { r_0 \over { 1 + e \cos( \theta ) } }              orbit radius as a function of \theta , the angle from perigee or true anomaly

{v_T}( \theta ) ~ = ~ v_0 ( 1 + e \cos( \theta ) )         orbit transverse velocity

{v_R}( \theta ) ~ = ~ v_0 e \sin( \theta )               orbit radial velocity

{\omega } ( \theta ) ~ = ~ { \Large { h \over{ r^2 } } } ~ = ~{ \Large { { \mu^2 ( 1 + e \cos( \theta ) )^2 } \over { {r_p}^3 {v_p}^3 } } }   orbit angular velocity

r_a ~ = ~ ( 1 + e ) a ~ = ~ { \Large 1 \over { \LARGE { { \huge 2 \mu } \over { r_p {v_p}^2 } } ~ - ~1 } }    apogee

The angular velocity of the tether in geostationary orbit (defined by the stellar day ) is \omega_T = 2 \pi / 86164.098903691 = 7.29211515E-5 radians per second.

At the capture point, the orbit has the same angular velocity as the tether, \omega = \omega_T .

### Classical Tether Capture

For the classical ( zero vertical velocity ) tether capture, v_R = 0 and \theta = \pi .

So:

{\omega_T} ~ = ~ { \Large { { \mu^2 ( 1 - e )^2 } \over { {r_p}^3 {v_p}^3 } } }

~ ~ ~ ~ ~ ~ = ~ { \Large { { \mu^2 ( 2 - r_p {v_p}^2 / \mu )^2 } \over { {r_p}^3 {v_p}^3 } } }

~ ~ ~ ~ ~ ~ = ~ { \Large { { ( 2 \mu - r_p {v_p}^2 )^2 } \over { {r_p}^3 {v_p}^3 } } }

The perigee velocity v_p can be computed by iterating on this equation:

v_p ~ = ~ \sqrt{ 2 \mu / r_p - \sqrt{ \omega_T r_p {v_p}^3 } }

Given v_p , we can compute the capture radius, which is r_a as above.

But we aren't done! We need to do two more things - climb up the tether, and restore momentum to the tether. Assume the tether is massive compared to a vehicle, so it does not deflect much. The will slow down by a tiny amount, and we must restore the momentum, perhaps with rocket engines. We must also supply energy to lift the vehicle up the tether against the small gravity (minus centrifugal force) at that high altitude.

 Classical Tether Capture \mu gravitational parameter 3.986004418e14 m3/s2 \omega_T GEO angular frequency 7.29211515e-5 radians/s r_{GEO} GEO radius 42164172.4 m v_{GEO} GEO velocity 3074.66 m/s E_{GEO} GEO energy / kg -14180301.17 J/kg H_{GEO} GEO angular momentum/kg 1.296e11 m2/s r_p perigee radius 6458137.0 m v_p perigee velocity 10074.5754 m/s e eccentricity 0.64446 a semimajor axis 18164241.20 m r_a apogee radius 29870345.40 m v_a apogee velocity 2178.1800 m acc_a apogee radial accel. 0.2879 m/s^2 E_a apogee energy / kg -15716587.30 J/kg H_a apogee ang momentum/kg 6.506e10 m2/s \Delta r radius increase 12293826.96 m \Delta v velocity increase 896.4800 m/s \Delta E radial energy increase/kg 1536286.13 J/kg \Delta H radial ang. mom. incr./kg 6.458e10 m2/s

If we haul the payload up with a solar cell power source weighing as much as the payload, then our "hauling power" will be about (16/2) = 8 watts per kilogram using current satellite technology. Against a 0.2879 m/s2 acceleration field, we can move at (8/0.2879) = 28 m/s near the bottom, faster near the top. At 8 watts per kilogram, and 1536286 joules per kilogram, climbing the tether will require at least 53 hours. However, at some point we are velocity limited - perhaps 200m/s for our motor. That occurs when the acceleration force is ( 8/200 ) = 0.04 m/s2, about 39800 km high. The velocity limit increases climber time to 55 hours .

If we can use a pulley system without tangling cables, running at 400m/s, climber time is 8.5 hours. We can go a lot faster if we are not moving the motor and the solar cells.

MORE LATER

### Rail Tether Capture, with vertical velocity

This is a little trickier. The system moves faster if we have a lousy "lift to drag" ratio, because we can travel the tether faster. It also makes the math a little simpler if we pick a fixed starting speed, as long as it is more than enough to make the climb. Lets start with the energy increase for the classic case, 1.54 MJ/kg , double it, and turn the result into a velocity: about 2500 meters per second. We will arbitrarily choose that as v_{vc} , the vertical speed the vehicle will be moving up the tether when it is captured. Since we are launching from the loop with about 52 MJ/kg anyway, an extra 3% energy loss will not be a show stopper, and we will get to the GEO rail station faster. Time is money!

The capture radius r_c is below orbit apogee by some as-yet undefined amount. The angular velocity will be \omega_T as before, so the transverse velocity will be \omega_T r_c . The total velocity squared will be the sum of the squares: {v_{vc}}^2 + ( \omega_T r_c )^2 . That is a function of a and r_c :

{ \Large { { {v_{vc}}^2 + ( \omega_T r_c )^2 } \over \mu } } ~ = ~ { \Large { { 2 \over r_c } - { 1 \over a } } }

We also know that:

\omega_T r_c ~ = ~ v_0 ( 1 + e \cos( \theta ) )

v_{vc} ~ = ~ v_0 e \sin( \theta )

r_c ~ = ~ r_0 / ( 1 + e \cos( \theta ) )

So:

\sin( \theta ) ~ = ~ { \Large { v_{vc} \over { v_0 e } } }

~ ~ ~ ~ ~ ~ ~ ~ = ~ { \Large { v_{vc} \over { v_p - \mu / r_p v_p } } }

\cos( \theta ) ~ = ~ \sqrt{ 1 - \sin( \theta )^2 }

~ ~ ~ ~ ~ ~ ~ ~ = ~ \sqrt{ 1 - { \Large \left( { v_{vc} \over { v_p - \mu / r_p v_p } } \right) } ^2 }

\omega_T ~ = ~ ( v_0 / r_0 ) ( 1 + e \cos( \theta ) )^2

~ ~ ~ ~ ~ ~ = ~ { \Large { \mu^2 \over { {r_p}^3 {v_p}^3 } } } ~ \left( 1 + \left( { \Large { { r_p {v_p}^2 } \over \mu } } - 1 \right) \sqrt{ 1 - { \Large \left( { v_{vc} \over { v_p - \mu / r_p v_p } } \right) } ^2 } \right)^2

Let's solve this mess for v_p . We know all the other terms. It will be iterative, as before:

v_p ~ = ~ { \Large \sqrt{ \mu \over r_p } } ~ \left( { { \sqrt{ { \huge { \omega_T {r_p}^3 {v_p}^3 } \over { \mu^2 } } } - \Large 1 } \over \sqrt{ \Large 1 - { \huge \left( { v_{vc} \over { v_p - \mu / r_p v_p } } \right) } ^2 } } + 1 \right)

MORE LATER

CaptureRail (last edited 2021-06-20 03:27:51 by KeithLofstrom)