RE: Cosmos 2221 and its rocket body

Ted Molczan (
Sat, 17 Jul 1999 09:57:51 -0400

Judy May asked:

> I observed for the first time tonight Cosmos 2221 (22236) and Cosmos
> 2221-r (22237).  They travel in tandem together, separated by
> about one
> degree.  I have never seen such a pair before, and was quite pleased
> with the view.  However, I was under the impression that after
> separation the rocket would quickly and permanently disassociate from
> the satellite.  This raised three questions for me:
>      1. What happened in this instance that kept these two
> together all these years?

Judy, the objects have not remained 1 deg apart all these years. They remain
in nearly the same orbital plane, but the payload currently is in a slightly
lower orbit than its rocket body, so it moves faster and has a shorter
orbital circumference, causing it to periodically catch-up, overtake, and
pull-away from the payload. It is analogous to what happens in racing - the
faster racer may lap the slower one.

Last night, you happened to observe near the time of an "overtaking." Before
long, the objects will be far apart, but you can easily estimate the
approximate date of the next overtaking, as I will describe.

Here are the objects' latest elements, obtained from NASA/OIG:

1 22236U 92080A   99197.18425502 +.00000997 +00000-0 +14489-3 0 03433
2 22236 082.5140 073.2352 0021208 325.1890 034.7931 14.74918934357237

1 22237U 92080B   99197.18392966 +.00000430 +00000-0 +61371-4 0 07755
2 22237 082.5147 073.3520 0021463 329.8423 030.1550 14.74340760357210

The payload is making 14.74918934 rev/d (revolutions per day), but the
rocket is making only 14.74340760 rev/d - a difference of 0.00578174 rev/d.
The inverse of this difference is the approximate time it will take the
payload to lap the rocket body, a period of nearly 173 days.

If the objects' orbits were not decaying appreciably, or decaying at the
same rate, the above calculation would be sufficiently accurate, but a look
at the elements shows that the payload's orbit is decaying more rapidly than
the rocket body's; therefore, it will lap the rocket in less than the 173
days calculated above.

A more precise estimate of the time-to-lap, can be made by solving this
simple equation:

a * t^2 + b * t + c = 0


a = one half the difference between the rate of decay of object 1 and object
2, in rev/d^2. This is simply the difference between their ndot/2, found in
the 2-line elements. Using the above elements, a = .00000997 - .00000430 =
.00000567 rev/d^2.

b = the difference between the mean motion of object 1 and object 2. Using
the above elements, b = 14.74918934 - 14.74340760 = 0.00578174 rev/d.

c = the difference between the number of revs of object 1 and object 2,
which equals -1 in this example.

t = equals time for object 1 to lap object 2, c times, with c defined as

The solution of this quadratic equation is:

t = (-b + (b^2 - 4 * a * c)^0.5) / 2 / a

Using the above values for a and b, I obtain t = 150.69 days.

So, assuming their rates of decay remain close to their present values, in
about 151 days, or about 14 Dec 99, the objects will repeat the
close-proximity event you observed last night.

> 3. Can anyone recommend some other neat pairs like this to observe?

I don't know of any off-hand, but the topic has come up before, and there
must be a large number of pieces from launches that have remained nearly
co-orbital. It would not be too difficult to write a computer program, based
on the principles I have outlined, to search a file of elements for nearly
co-orbital objects, and to estimate their future times of close proximity. I
don't plan to write one, but I would not be surprised if someone already

Ted Molczan