Could a manned maneuvering unit provide artificial gravity?

Alert: Mathematics ahead.

Just being humorous. I know that not everyone likes math, so I figured I'd add that in. Just so I can say 'I told you!' if you complain about the math. In which case just read DonyorM's answer. Anyway . . .

The dimensions of the MMU are as follows:

• $0.846\text{ meters}$
• $0.711\text{ meters}$
• $1.27\text{ meters}$

The issue is, I don't know which of the first two is width and which is depth! Judging from picture, though, the MMU is wider than it is deep, so the depth appears to be $0.711\text{ meters}$.

The other bit of information we need to know is the surface gravity of the Moon. The reason we don't want the mass of the Moon is because the astronaut is much closer to the center of mass of the MMU when s/he is strapped in than an astronaut on the Moon would be to the Moon's center of mass. To replicate the effects, we need to replicate the surface gravity.

• $g_{\text{Moon}}=1.6249{\text{ m/s}}^2$

The formula for the force experienced on an person with a mass $m_p$ from a second body $m_b$ is $$F=m_{p}g=G\frac{m_b{m}_p}{r^2}$$ Cancelling out the $m_p$s leads us to $$g=G\frac{m_b}{r^2}$$ We have $g$, $G$, and $r$. Let's find $m_b$: $$m_b=\frac{g{r}^2}{G}$$ Plugging in our values, we get $$m_b=\frac{1.6249\times (0.711{)}^2}{6.673\times {10}^{-11}}=1.23096219\times {10}^{10}\text{ kilograms}$$ The volume ($V$) of the MMU is $$V=0.846\times 0.711\times 1.27=0.76391262{\text{ meters}}^3$$ Density ($\rho$) is $\frac{M}{V}$, so we have $$\rho =\frac{M}{V}=\frac{1.23096219\times {10}^{10}}{0.76391262}=1.611391353\times {10}^{10}{\text{ kilograms/meter}}^3$$ That's roughly 10 times as dense as a white dwarf. Not doable by humans.

As DonyorM discussed, the best choice for your question (because you asked about a human-made source) would be to spin the MMU. And what do you know - more math!

The centripetal force on an astronaut must be the same as the surface gravity of the moon: $$F_c=\frac{m{v}_c^2}{r}=m{g}_{\text{Moon}}$$ Once again, the $m$s cancel out, and we re-arrange to get $$v_c=\sqrt{g_{\text{Moon}}r}$$ Here, $r=0.711\text{ meters}$, so $$v_c=1.074850641\text{ meters/second}$$ The angular velocity $\omega$ is $\frac{v_c}{r}$: $$\omega =\frac{v_c}{r}=1.511744924\text{ radians/second}$$ Treating the entire apparatus as a block with the dimensions of the MMU (okay, I'll add half a meter to the height to account for the astronaut's legs sticking out, and the MMU has a mass of $148\text{ kilograms}$, and the astronaut has a mass of, say $75\text{ kilograms}$), we have the moment of inertia $I$ as $$I=\frac{1}{12}m(w^2+d^2)=\frac{1}{12}(75+148)({.846}^2+{.711}^2)=22.69465425$$ Angular momentum $L$ is defined as $$L=I\omega =22.69465425\times 1.511744924=34.30852836$$ We know that we have to obey the law of conservation of momentum: $$\frac{d{L}_{\text{system}}}{dt}=0$$ which becomes $$\frac{d(\frac{1}{12}m(w^2+d^2)\omega )}{dt}=0$$ Everything here but $m$ should, optimally, we a constant, because $m$ changes as fuel is released. So something else has to change, too. Let's go back to the definition of $\omega$: $\omega =\frac{v_c}{r}$. So now we have $$\frac{d(\frac{1}{12}m(w^2+d^2)\frac{v_c}{r})}{dt}=0$$ Divding out a whole bunch of constants, we have $$\frac{d(m\frac{v_c}{r})}{dt}=0$$ This means that $\frac{m{v}_c}{r}$ is a constant. We also know that $F$ is a constant. This means that $$\frac{d(\frac{m{v}_c^2}{r})}{dt}=0$$ So $\frac{m{v}_c^2}{r}$ is also a constant. Yet $m$ changes. Could $v_c$ change? No, because in the first case, $v_c$ is raised to the first power, while in the second case, it is raised to the second power. In both cases, it would have to match $dm$ . . . and if $dv\ne 0$, then this would be impossible in both cases. So $r$ must change, and so $$\frac{dm}{dt}=\frac{dr}{dt}$$ Whether or not this is possible for humans to engineer depends on how easily the radius is changed (impossible on the MMU!), and how quickly the mass changes (e.g. how much propellant is expelled in a given time.

posted over 10 years ago

HDE 226868‭

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