# If there was a non-rotating skyhook in Earth orbit, what would re-entry be like after dropping from its foot?

I have been looking at a space station that orbits Earth at an altitude of 3500 km above the equator, and has tethers that extend from it downwards to 250 km above the surface, and outwards to 6500 km altitude - a non-rotating skyhook. Right now I'm considering what would happen if a Dragon or Soyuz type re-entry capsule was dropped from the foot of the lower tether and returned on a ballistic trajectory.

I used the Tether Tool (written by our own Hohmannfan) to find the velocity of the foot. It was written first for the Moon and the rest of the solar system has recently been added to it. Don't forget to change the selection to Earth at the top.

The station at 3500 km altitude rotates with a period of 2.711 hours. At the foot (altitude = 250 km), this corresponds to a speed of 4.26 km/sec, or about half of the re-entry speed of a satellite from LEO. Since it's moving in the same general direction that the atmosphere is rotating, the speed relative to it is about 0.5 km/s less than that. Assume this is all in the plane of the equator.

A Dragon 2 weighs 6400 kg empty. If it was returning one person perhaps it's fair to increase that to 6500 kg. On such a steep descent, would it need to have a larger area facing into the wind, (perhaps an inflatable decelerator)? What sort of descent time would be involved before the parachutes deploy, if it was set up to slow properly? What would be the peak deceleration?

Together, that is maybe a lot to ask. It seems like it all belongs together. All input appreciated.

Related question: Roughly what is the rate of energy loss of survivable capsule re-entries?. (New)

• I just rediscovered my copy of Coming Home - Reentry and Recovery from Space. Doing a little reading.... – kim holder Aug 25 '17 at 15:50
• Wow, Hohmann fan has crafted a number of neat online tools. They are somewhat similar to some tools I had helped make but his are much easier to use. I plan to compare the results of his tools to mine and if we agree I will link to his tools from various pages I have on the internet. – HopDavid Aug 26 '17 at 4:30
• +uhoh, there is also so called centrifugal force, $\omega^2r$. In Kim's scenario $\omega^2r$ is about $2.7 m/s^2$ at the tether foot. $GM/r^2$ is about $9.1 m/s^2$. Net acceleration is about $6.4 m/s^2$ – HopDavid Aug 26 '17 at 6:55
• @kimholder as a side note, it is possible that 250 km is not the ideal altitude to let go. You might want to release somewhat closer to 100 km so you can loose some velocity while still moving forward and not down. You may stay cooler that way. – uhoh Aug 26 '17 at 14:53
• @HopDavid btw, those scripts are under a GNU v.3 public license. You are free to use them for any purpose, as long as you allow others to do the same, state changes, and give credit. Details are here – kim holder Aug 26 '17 at 16:54

I recreated your scenario using Wolfe's Spreadsheet. Here's a screen capture:

The numbers you cited match mine fairly closely. A very nice tool Hohmann fan has made available. I intend to link to his page from several of my tether blog posts.

The bottom of the spreadsheet answers some of the your questions. I called 70 kilometers the altitude where the craft starts serious aerobraking. Which is just a guess, to be honest.

The ship would be moving 4.6 km/s at that altitude. It would 7.3e-1 hours to reach this altitude which comes out to 43 minutes.

Scratch that - it is a mistake. Thank you for calling that to my attention, uhoh. The bottom of the spreadsheet was a transfer orbit from a 250 km apogee to a 70 km perigee. But in this scenario perigee is well below earth's surface and the ship re-enters well before perigee.

Here is a pic of Kim's scenario:

The eccentricity of the suborbital ellipse is .698. Semi major axis of that ellipse is about 3902 kilometers. Given that semi major axis, the period would be 2425 seconds.

But only a fraction of that period is swept out from release at apogee to atmospheric entry at 70 km. That fraction of the ellipse is shaded blue.

To get area of the fraction of an ellipse swept out, Kepler scaled the ellipse to make it a circle. Which is what I did. I also scaled radius of our circle to 1 to eliminate some arithmetic.

To get area swept out, Kepler would add area of a triangle and a wedge. Which is what I did above. In this case a little more than 1/10 of the ellipse.

.104*2425 seconds is about 252 seconds. Or about 4 minutes and 12 seconds.

After release from a 250 km altitude, the spacecraft would start serious aerobraking in about 4 minutes.

At 70 km altitude I get a flight path angle of 19.2º. Horizontal velocity would be be about 4.38 km/s. Vertical velocity would be 1.53 km/s.

The short path from a 70 kilometer altitude to earth's surface doesn't give much time to shed the 4.64 km/s velocity.

As the capsule descends gravity will increase slightly, centrifugal acceleration decreases rapidly while deceleration from dynamic pressure increases rapidly.

I whomped up a first order Runge Kutta model of this drop from the tether foot at a 250 km altitude to earth's surface. A few different scenarios:

Dropping in a Vacuum

Time to Impact: 279 seconds
Impact Velocity: 4.63 km/s.

This is within about 4% of what I got using Kepler's method to get time of flight and the Vis Viva equation to get speed at 6378 km (earth's radius).

Dropping into earth's atmosphere

To get atmosphere density at an altitude I use $exp(-altitude/scale height)*1.225 kg/m^3$. I call 8500 meters the scale height. Dynamic pressure is $.5*\rho*v^2$). Force newtons is dynamic pressure x drag coefficient x cross sectional area capsule.

Scenario 1: 6500 kg, 1.85 meter radius, drag coefficient .5
Time to impact: 354 seconds
Impact velocity: 150 meters/sec which is about 333 mph.
Max Q: 217 kilo pascals.
Max aerobraking acceleration: 18 g's

I've been told 90 kilo pascals has been a Max Q for descent. I am guessing 217 kilo pascals is unacceptable.

Scenario 2: 6500 kg, 2.9 meter radius, drag coefficient .5
Time to impact: 416 seconds
Impact velocity: 92 meters/sec which is about 205 mph.
Max Q: 87 kilo pascals.
Max aerobraking acceleration: 18 g's

Here Max Q is less than 90 kilo pascals. This may be acceptable.

I don't know what how to figure heating.

If made of Zylon with a safety factor of 3, the lower tether would have a taper ratio of about 1090. Tether mass to payload ratio would about about 6,670.

I think with that great length it would have a large enough cross sectional ratio that impacts would be inevitable in the LEO volume which has lots of debris and satellites. As much as I love vertical tethers, I don't believe they could survive close to earth.

A rotovator might be a way to catch suborbital flights as well as release re entering craft into the atmosphere at suborbital speeds. John Carmack has already tweeted to Musk if he can land a booster on an ocean platform, he should be able to rendezvous with a rotovator. Musk agreed, sort of.

tl;dr Everyone will die within five minutes. Your plan is a non-starter. You must consider letting go from a much lower altitude, or higher forward velocity, or convert your capsule into a spaceplane the likes of which do not yet exist.

Here is some physics. To get an approximate idea what the re-entry would be like, I've used a simple scale-height model of atmospheric density (7, 7.5 and 8 km), constant drag coefficient (0.2, 0.5, and 0.8) and a constant lift-to-drag ratio (0, 0.15, and 0.3). With all 27 combinations you get a peak heat production between 3 and 4 gigawatts, and a peak deceleration of between 15 and 20 gees.

These should be compared to values from survivable reentries.

In all cases the capsule hits the ground in roughly five minutes. If the Earth were flat that would be a bit less; the curvature buys you some extra seconds. You could even ballpark the fall time with:

$$x = \frac{1}{2} a t^2$$

The problem is that it's dropping for nearly 200 km before it hits enough air to really start losing energy. It would be better to let it go in the ballpark of 100 to 80 km so it can get rid of a lot of energy before it starts sinking too fast and the air gets too dense.

def deriv(X, t):

x, v = X.reshape(2, -1)

r, speed = [np.sqrt((thing**2).sum()) for thing in x, v]

acc_g = -x * GMe *((x**2).sum())**-1.5

alt = r - re

rho    = rho0 * np.exp(-alt/hscale)
Fdrag  = -0.5 * v * speed * CD * Area * rho
n_lift = np.hstack((-v[1], v[0]))/speed   # definition of lift
Flift  = LDR * 0.5 * n_lift * speed**2 * CD * Area * rho

acc_d = (Fdrag + Flift)/m0

return np.hstack((v, acc_g + acc_d))

import numpy as np
import matplotlib.pyplot as plt
from scipy.integrate import odeint as ODEint

pi  = np.pi
GMe = 3.986E+14

alt = 250000.   # meters
re  = 6378000.  # meters
v0  = 4260.     # m/s
hscales = [7000., 7500., 8000.]   # meters
CDs     = [0.2, 0.5, 0.8]
LDRs    = [0, 0.15, 0.3]
Area    = pi * 1.1**2             # m^2
m0      = 6800. # kg
rho0    = 1.25  # kg/m^3

X0   = np.array([0, re+alt, v0, 0])
dt   = 1.0      # seconds per reported value by the solver (internally variable timesteps)
time = np.arange(0, 301, dt)

for CD in CDs:
for hscale in hscales:
for LDR in LDRs:

answer, info = ODEint(deriv, X0, time, full_output = True)

km = 1E-03

gee = 9.8  # m/s^2

plt.figure()

x, y, vx, vy = answer.T
r = np.sqrt( x**2 +  y**2 )
v = np.sqrt(vx**2 + vy**2)
KE = 0.5 *m0 * v**2

plt.subplot(5, 1, 1)
plt.plot(time, km*vx)
plt.plot(time, km*vy)
plt.plot(time, km*v )
plt.title('vx, vy, vtot (km/s) versus time (seconds)', fontsize=16)
plt.subplot(5, 1, 2)
plt.plot(time, km*(r-re))
plt.title('altitude (km) versus time (seconds)', fontsize=16)
plt.subplot(5, 1, 3)
plt.plot(time[:-1], KE[:-1] - KE[1:])
plt.title('Watts dissipated', fontsize=16)
plt.subplot(5, 1, 4)
plt.plot(time[:-1], ((v[:-1] - v[1:])/dt)/gee)
plt.title('gees', fontsize=16)
plt.subplot(5, 1, 5)
plt.plot(km*x, km*(y-re))
plt.plot(km*x, km*(np.sqrt(re**2 - x**2)-re), '-k', linewidth=2)
plt.title('y versus x (km)', fontsize=16)
plt.show()

• Let me rephrase that whole mess, what exactly does the hscale control in this, and what are the LDRs in reference to? I can find this information but it doesn't explicitly explain what it means. I'm trying to figure out the lift for a parachute during supersonic reentry and it seems as if at first it would start negative, then slowly become positive due to the "angle of attack" for lack of a better phrase. – Magic Octopus Urn Aug 31 '18 at 17:32
• At the beginning of the answer, just after the tl;dr at the top, those parameters are described. – uhoh Sep 1 '18 at 0:22