Average distance among pieces of space debris in low earth orbit?

[Kevin Le]

Hello all,

I have a question about space debris.

The number of space debris in the low Earth orbit is huge.There are more than 170 million pieces of space debris which is greater than 1mm. So in the orbit of the Earth, what is the average distance between 2 pieces of space debris?

 

[ShayanJ]

Hello
A very naive way of thinking about it is to consider they're uniformly distributed in the low Earth orbit. Considering the altitude to be between 160 km and 2000 km (from wikipedia), the volume of a hollow sphere with radii equal to those numbers is $3.35 \times 10^{10} km^3$. So the density is $n=\frac {17 \times 10^7}{3.35 \times 10^{10}} \approx 5 \times 10^{-3} km^{-3}$. Which means there is 1 debris in every $200 km^3$ of the llow Earth orbit volume. Assuming those are almost in the middle of a cube with that volume, the distance between them becomes $\sqrt 2 \times 5.84 km=8.27 km$. Assuming other shapes and places will give different numbers but at least it tells us that distance is of the order of several kilometers.
This is very crude but is a good starting point for further work.

 

[jim mcnamara]

Latex is driving me insane - sorry.
Volume recalculation --
with r(Earth)=6371km
r1 = r(Earth) + 160 == 6531
r2 = r(Earth) + 2000 == 8371

V = ( 8371^3 - 6531^3) * 4.188 ~~ 1289960242829 ( 1.3 x 10^12) km^3

 

[NTW]

Latex is driving me insane - sorry.
Volume recalculation --
with r(Earth)=6371km
r1 = r(Earth) + 160 == 6531
r2 = r(Earth) + 2000 == 8371

V = ( 8371^3 - 6531^3) * 4.188 ~~ 1289960242829 ( 1.3 x 10^12) km^3

For 170 million pieces of debris, that would mean one piece for every 7647 km^3. To get a linear distance between pieces, we may take the cubic root, and that results in 19,7 km of distance between any two pieces. Quite close...