# Firing an AK-47 while anchored to the ISS [closed]

The Setup

I magnetically anchor myself to the floor of the ISS. (yes I know there's no "floor", any wall will do)

Then, I fire a 5-round burst out the airlock.

The Question

Can the recoil from this action significantly alter the rotation of the ISS in any measure, or is this too weak to have a noticeable effect?

## closed as off-topic by L.Dutch♦, jdunlop, Trish, MichaelK, AifyJun 27 '18 at 16:39

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question does not appear to be about worldbuilding, within the scope defined in the help center." – L.Dutch, jdunlop, MichaelK, Aify
If this question can be reworded to fit the rules in the help center, please edit the question.

• This is pure physics question. As such, I see it better suited for Physics.SE – L.Dutch Jun 27 '18 at 5:10
• not worldbuilding, stroy based. – Trish Jun 27 '18 at 8:33
• "I magnetically anchor myself to the floor of the ISS". You know it's aluminium, right? Also, read this: what-if.xkcd.com/21 – Madlozoz Jun 27 '18 at 8:52
• Orbital rotation or own rotation? – Alexander Jun 27 '18 at 18:08

Let me analyse this for you: You anchor yourself to the station. So your weight is now in the ballpark of $m_1=420\times 10^3\text{kg}$. The shot bullets are about $m_2=8\times 10^{-3}\text{kg}$ each and travel at $v_2=710 \frac {\text m} {\text s}$.

Insert these into these:

1. $F_1=5\times F_2=m_1\times a_1 = m_1\times\frac {dv_1}{dt_1}$
2. $F_2=m_2\frac {dv_2}{dt_2}$

Assume $dt_2=0.1\text s$, $dt_1=5\times dt_2$ and do some simple math that I won't give here and you get...

1. $dv_1=\frac{5\frac{8\times10^{-3}}{420\times10^3}\frac{710 \frac {\text m} {\text s}}{0.1\text s}}{5 \times 0.1\text s}=0.00135\frac {m}{s}$

Up to here, only the numeric value of the change in speed was relevant, NOT the direction (vector) at all. To see how this affects the rotation, you'll have to insert some data: your own position from that rotational axis and the vectors of both the shot and the station.

1. $\vec {v_1}=\vec\omega_1\times\vec{r_2}$
2. $\vec{d\omega_1}=\frac{\vec {dv_1}} {|\vec{r_2}|}$

This change in orbit (which is dependant on the part of $\vec{v_1}$ along its orbit only) and rotation ($\vec {d\omega}$) is tiny and the next automated correction burn will fix more than this.

• 0.00135 m/s = 1 millimeter per second = 3.6 meters per hour. This was for entire weight of the station, and rotation computation should probably use lower weight (maybe half?), leading to even more speed. If station is 100 meters wide, this is enough rotation to be noticeable. But I assume station has GYROSCOPES to compensate for people pushing off the walls as they move around the station, and they will probably cancel out most of the rotation from AK-47 rounds. – Bald Bear Jun 27 '18 at 14:47
• @BaldBear that's not how physics work. All or nothing. The further out you shoot, the less impact it has. – Trish Jun 27 '18 at 15:42