There's something I don't understand:

When it comes to de-commissioned satellites, rocket bodies etc. I thought one way of re-entry was via a highly elliptical orbit: The perigee gets low enough (app. 100 km) to reach into the atmosphere and then re-entry happens.

Things yet don't seem to be that simple from what my observations of 1976-006D tell me: I started observing it a month ago, when its apogee was at 7767 km, perigee at 99 km. I expected re-entry to happen any day. Four weeks later, apogee has sunk to 3474 km, perigee is still at 100 km. How can this be?

2013-062C, on the other hand, re-entered at 3557/105 km. Perigee dropped from 110 to 105 km within four days and the object quickly re-entered.

Other recent re-entries happended at 244/228 and 213/167 km orbits.

How does this fit into one logic? What else do I need to understand or to factor in?

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    $\begingroup$ Have you played Kerbal Space Program? $\endgroup$ Commented Apr 24, 2017 at 0:22
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    $\begingroup$ When you slow down at periapsis it will lower your apoapsis. Think in terms of speed (kinetic energy), potential energy and flying so fast that you miss the ground. $\endgroup$
    – Michael
    Commented Apr 24, 2017 at 7:06
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    $\begingroup$ @eerie because "magic", but seriously, because it demonstrate re-entry quite realistically with deadly re-entry mod. $\endgroup$
    – Raze
    Commented Apr 24, 2017 at 16:46
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    $\begingroup$ @eerie ...and, more to the point, it also simulates orbital mechanics quite realistically. In particular, one of the first things you learn when playing with it is that deceleration at perigee lowers the apogee, and vice versa. $\endgroup$ Commented Apr 24, 2017 at 18:48
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    $\begingroup$ @eerie xkcd.com/1356 $\endgroup$
    – Steve
    Commented Apr 24, 2017 at 19:35

4 Answers 4


I'll give you an intuitive way to think about it, then a script to play around with if you like.

When the spacecraft is at periapsis, it is there, at periapsis. A bit of drag isn't going to change its position by much, but it will change its velocity.

The new velocity determines how high it can reach at apoapsis. If it has slowed down, it will climb to a lower apoapsis.

However, it's in an elliptical orbit, and there's no drag up there, so it's going to return to nearly the same periapsis whence it came.

caveat: That assumes there is no lift. The term "lift" stands for any aerodynamic force not in the direction of the relative velocity of the spacecraft with respect to the atmosphere. If the spacecraft is not spherical or sufficiently tumbling, or if it has wings and is space-shuttle-shaped and Clint Eastwood is in the "pilot's seat", then all bets are off.

Here's a python script for a simple approximate simulation that demonstrates this. It is not intended to be precise, but it's physical enough to show that this is what tends to happen.

Eccentric orbits "touching the atmosphere" at periapsis will tend to first circularize before they burn up.

And this has at least one important implication; it won't necessarily re-enter and burn up near the original periapsis. It can happen anywhere. Round and round it goes; where it stops, nobody knows.

enter image description here

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

    alt   = np.sqrt((x**2).sum()) - re
    rho   = rho0 * np.exp(-alt/hscale)

    Fdrag = -0.5 * rho * CD * area * v * np.sqrt((v**2).sum())

    return Fdrag/mass

def deriv(X, t):
    x, v  = X.reshape(2, -1)

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

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

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

halfpi, pi, twopi = [f*np.pi for f in [0.5, 1.0, 2.0]]

GMe  = 3.98600418E+14   # m^3 s^-2
mass = 5000.   # kg
area = 5.      # m^2
CD   = 1.0
hscale    = 7200.  # meters (ROUGHLY scale height fudged for 100km)
re   = 6378000. # meters (equatorial radius)
rho0 = 1.225    # kg/m^3

r_peri = re + 90000. # meters
v_peri = 10000.  # m/s  

X0 = np.hstack(((r_peri, 0, 0), (0, v_peri, 0)))

time = np.arange(0, 1000000, 100) # sec

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

theta = np.linspace(0, twopi, 360)
xe, ye = [re*f(theta) for f in [np.cos, np.sin]]

alti = np.sqrt((answer[:,:3]**2).sum(axis=1)) - re

if True:


    plt.subplot(2, 1, 1)

    x, y = answer.T[:2]
    plt.plot(x, y)
    plt.plot([0], [0], 'ok')
    plt.plot(xe, ye, '-k', linewidth=1.5)

    plt.subplot(2, 1, 2)

    plt.plot(time/(24.*3600), alti/1000.)
    plt.ylim(1, None)
    plt.xlim(0, 10)

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    $\begingroup$ @eerie oh, the caveat is there mostly because if it wasn't people would leave comments and point this out. If a spacecraft has an unusual shape, like the space shuttle for instance, it will interact with the atmosphere in a more complicated way. Instead of just slowing down, it could get pushed up, or down, or sideways, depending on the way it is pointing relative to the direction of motion, and this makes everything much more complicated. This is why people like to make spacecraft tumble randomly as they start to reenter, because that actually makes things more predictable. $\endgroup$
    – uhoh
    Commented Apr 24, 2017 at 16:09
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    $\begingroup$ @eerie See this answer and/or search for the words "re-entry" and "tumble" and/or ask a new question - which someone more knowledgable about this subject than I will likely answer. For "Clint Eastwood" see this vintage clip: youtu.be/rKHW39mShF4 (purely for humor reasons) $\endgroup$
    – uhoh
    Commented Apr 24, 2017 at 16:10
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    $\begingroup$ Thanks! Geez, it really IS complicated. I guess that's why it's called rocket science ;-) $\endgroup$
    – eerie
    Commented Apr 24, 2017 at 16:14
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    $\begingroup$ @eerie - Seriously though, you should play some Kerbal Space Program and it'll all make sense. $\endgroup$
    – William
    Commented Apr 24, 2017 at 16:52
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    $\begingroup$ @uhoh: Whereas I've never seen the attraction of a language in which (just for one instance) editing a program with an editor that has a different definition of embedded tabs than the original author's editor can completely destroy its logic. $\endgroup$
    – jamesqf
    Commented Apr 26, 2017 at 3:39

Four weeks later, apogee has sunk to 3474 km, perigee is still at 100 km. How can this be?

The primary effect of drag on a vehicle in a highly elliptical orbit that just touches the atmosphere at perigee is, at least initially, to lower apogee. So long as the vehicle survives its brief encounter with the atmosphere near perigee, that encounter will have very little effect on perigee. As apogee decreases, the vehicle starts spending more than a tiny fraction of it's orbit in the atmosphere. This will eventually reduce perigee. With an apogee at 3474 km and perigee at 100 km, that means most of this vehicle's orbit is still outside the atmosphere.

How does this fit into one logic? What else do I need to understand or to factor in?

You need to factor in coefficient of drag, cross section area, and mass. Acceleration due to drag is proportional to coefficient of drag and cross section area but inversely proportional to mass. Another way to look at this is to combine these three factors into one, the ballistic coefficient: $BC = \frac M {C_d A}$ where $M$ is mass, $C_d$ is the coefficient of drag, and $A$ is the cross section to drag. Drag acceleration is inversely proportional to the ballistic coefficient.

1976-006D is the upper stage of a rocket. It has a nice aerodynamic shape, making for a smallish coefficient of drag a small cross section. It also has engines, beefy structure to support those engines, and empty fuel tanks, making for a high mass. Putting these together means a high ballistic coefficient. 2013-062C on the other hand is a marked as debris, either a tank or a piece of one. In either case (a tank or part of one), it will have a lousy coefficient of drag, a lousy cross section to drag, and a smallish mass. Putting these together means a much smaller ballistic coefficient that that of 1976-006D.

  • $\begingroup$ Thanks, this gave me a very good idea of the factors I did not know about. I've learned a lot by it. $\endgroup$
    – eerie
    Commented Apr 24, 2017 at 16:03

Effect of manoeuvring on perigeee/apogee radius

One of the issues is that during some instant of any orbital manoeuvre the orbiting object, if the manoeuvre were to end there, would complete one orbit and then pass through the same point on the next rev. Think of this as the idealised situation with no other orbital perturbations. The effect of the manoeuvre is to change the immediate velocity, not the immediate location.

Continuing with the idealised situation the effect of an impulsive manoeuvre at the apogee or perigee is to affect the radius at the opposite apps. Thus a retrograde perigee manoeuvre will lower the perigee velocity which in turn mainly reduces the apogee height. The perigee altitude is unaffected. This also happens in reverse for apogee manoeuvres.

In the case of a perigee passage through the upper atmosphere the braking effect of the atmosphere can be thought of as a whole series of retrograde manoeuvres spread out over a short arc. This has the affect of reducing the perigee velocity, which following the previous thought experiment, largely has the effect of dropping the apogee.

Relative drag effects on two objects

As to the second part of the question relating to why two objects have behaved differently - I don't know without looking more closely. It could quite plausibly be explained by a difference in Area/Mass for the two objects, making one behave much more ballistically than the other.

In this way it seems plausible that the object with the higher apogee and thus higher perigee velocity and thus more drag, might nevertheless experience less velocity change per perigee pass if it has a lower Area/Mass ratio. It would also have a longer orbital period and thus few perigee passes per day.

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    $\begingroup$ I like the idea of thinking of drag as a series of retrograde manoeuvres! $\endgroup$
    – eerie
    Commented Apr 24, 2017 at 16:05

Beside the dynamics of space object interaction with the atmosphere, one other important factor that contributes to re-entry from highly elliptical orbit is the luni-solar gravity attraction. The objects in HEO spend the larger part of their orbital passage near apogee where the luni-solar gravity forces are relatively dominant.

The perigee altitude oscillates, due to the luni-solar gravity, and results in either a rapid decrease in perigee altitude with catastrophic fragmentation (accelerated by atmospheric heating) at sharp re-entry angles or increase in the perigee altitude.


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