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How would you figure out the fuel cost or delta V for spinning up a S/C to match the orbital period of 27.32 days (moon orbit)? Given that the S/C is initially not rotating about any axes.

For simplicity say the S/C is in an equatorial orbit and only needs to rotate about one axis. If the max torque is 40 N-cm, max thrust is 3N, total mass is 2kgs and ISP is 200s. Just the general equations are sufficient.

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  • $\begingroup$ You need to know the moment of inertia of your spacecraft $\endgroup$ May 18 '20 at 16:22
  • $\begingroup$ if its 0.00045m^4? still not sure how to relate that back? Thanks $\endgroup$
    – Alberto
    May 18 '20 at 16:36
  • $\begingroup$ Those units don't work. Moment of inertia is in units of $kg\, m^2$, $\endgroup$ May 18 '20 at 18:00
  • $\begingroup$ Sorry you're right I got confused. It is 0.00045 kgm² $\endgroup$
    – Alberto
    May 18 '20 at 19:52
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OK. So given the torque and moment of inertia we can work out the angular acceleration.

$$\dot\omega = T/I = 0.4/0.00045 = 1000 s^{-2}$$

Now the angular velocity you want is $$2\pi/(27.32*86400) = 2.6\times 10^{-6} s^{-1}$$ So full thrust for about $2.6$ nanoseconds would be needed to spin it up. Now a thruster with an $I_{sp}$ of 200s and a thrust of $3N$ consumes $3/200g kg$ of fuel per second (where $g$ is the acceleration of Earth's gravity at the surface) or about $1.5gs^{-1}$. So your spacecraft would need roughly 4 nanograms of fuel to achieve the desired rotation.

In practice, controlling a thruster of this power sufficiently accurately to achieve such a precise spin rate is likely to be a harder problem than fuelling it.

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