# Mathematics behind attitude maneuver

I found a paper that says: "Let's take the task of modifying the orientation of a 700kg mass spacecraft along one of it's principal axes ; a 5 deg rotation is to be performed. It is assumed that control torque is generated by two thrusters, operated at a specific impulse of 8000s, placed at 1m from the center of mass and opposite sidea of it, firing normal to the line between them " Then the results are My question is, knowing the mass and the specific impulse of the thruster and the angle to rotate, how can I get the necessary thrust for a certain time?

This paper is called "FEEP thrusters applications to constellations and small satellites"

• I don't think this can be answered without knowing the moment of interia of the spacecraft Jul 29 '20 at 14:47
• @SteveLinton I thought the same Jul 29 '20 at 15:00
• When linear acceleration of the spacecraft is considered, it may be modeled as a point mass of 700 kg. But for rotation of the spacecraft the point mass modell is useless.
– Uwe
Jul 29 '20 at 16:17
• @Uwe the title probably needs to be adjusted, this seems to be an attitude change since the thrusters are pointing tangent to the line between them.
– uhoh
Jul 29 '20 at 16:21
• update: the title has been adjusted!
– uhoh
Jul 29 '20 at 23:23

This paper is a sizing exercise, made before the thrusters in question had been flown on any spacecraft. The moment of inertia can be backed out from the numbers in the article. Given the bang-bang control scheme used in the article, the relation between angular acceleration $$\dot \omega$$, firing time $$t$$, and angular change $$\theta$$ is $$\frac12\dot\omega\left(\frac t2\right)^2 = \frac12\theta \quad\Rightarrow\quad\dot\omega = \frac{4\theta}{t^2}$$ The torque is related to angular acceleration via $$I\dot\omega = \tau \quad\Rightarrow\quad I = \frac{\tau}{\dot\omega} = \frac{\tau t^2}{4\theta}$$
With this, I get $$998\,\text{kg}\cdot\text{m}^2$$ for two 100 μN thrusters firing for 22 minutes, $$1031\,\text{kg}\cdot\text{m}^2$$ for two 500 μN thrusters firing for 10 minutes, and $$1010\,\text{kg}\cdot\text{m}^2$$ for two 1 mN thrusters firing for 7 minutes. Call it $$1000\,\text{kg}\cdot\text{m}^2$$ as a nice round number. (This is a sizing exercise, after all.)
What kind of spacecraft is this? It's not a one meter radius spherical cowcraft. Such a vehicle would have too low a moment of inertia, at most $$467\,\text{kg}\cdot\text{m}^2$$ if all of the mass was on the surface of the sphere. It's not a cylindrical spacecraft, either, with all the mass on the surface of the cylinder. (Once again, too low of a moment of inertia.)
It appears to be a cubical spacecraft with edges that are two meters long and with all of the mass on four faces of the cube; the top and bottom faces have negligible mass. The axis of rotation is about the axis passing through the middle of the top and bottom faces. The thrusters are at the center of opposite side faces. This gives a moment of inertia of $$933\,\text{kg}\cdot\text{m}^2$$ about the rotation axis of interest. Since this is a sizing exercise, it makes sense to round this up to $$1000\,\text{kg}\cdot\text{m}^2$$.