# Could a manned maneuvering unit provide artificial gravity?

Could a Manned Maneuvering Unit-type device provide a few hours of light artificial gravity* for a small person on an object with minimal gravity of its own? What sort of fuel would it need?

* Comparable to moon gravity or more.

• What sort of path does the MMU need to take? It would actually be very efficient if you used a MMU to accelerate a person/counterweight pair in a circle. You would just need to accelerate up to the correct angular rate, then stop spending fuel. It would be less efficient if you wanted 1/6G of sustained linear acceleration, because you'd need to keep expending fuel. – Cort Ammon Dec 10 '14 at 3:50
• If anyone was thinking along the same lines as me: Unfortunately, the MMU is bigger than the Moon's Schwarzschild radius. – HDE 226868 Dec 10 '14 at 16:43

Short Answer: Yes it's possible. You need a large magnet and an existing magnetic field.

Gravity can be created, at close to 1 G, using magnetism. However, you can't do this outside of an existing magnetic field. So, on Earth, you can make something levitate using superconductors (Source). However, this would be more difficult to achieve in space, however, because in space you don't have a consistent magnetic field. In the solar system, possibly the Sun's magnetic field might be able to be used. Or a space ship could generate a magnetic field that could be used to create gravity.

Creating gravity using magnets requires superconductors or other extremely powerful magnets. Either several megawatts of electricity or cryogenics (very cold stuff) would be required for this to work. In am MMU environment, cryogenics would be more feasible. It is also not known if this method adversely affects humans.

Second option, spinning. You can always create gravity by spinning something around. This would require the MMU to have an axis different from where the astronaut was standing, so that the gravity would push the astronaut away. The astronaut would need a wall or platform to stand on when the gravity was in effect. This type of artificial gravity may not be very useful, because the astronaut will be hard pressed to interact with his environment.

Final option, forward acceleration. If something moves forward fast enough, it creates artificial gravity. Astronauts feel this all the time when an rocket leaves the Earth. However, to maintain this gravity, it requires large amounts of fuel. An MMU could probably not hold enough fuel to maintain gravity in this manner for more than a few seconds.

Summary: creating gravity in a small environment like an MMU is not easy. It can be done, but gravity either requires a magnetic field, large amounts of fuel, or dizzying spinning. All make it difficult.

• I think you misunderstood the page you linked to. Diamagnetism is not gravity (nor even a form of gravity), it's a magnetic field that's induced in matter, the strong magnets use diamagnetism to induce a -1G force in objects to counteract the +1G force of gravity which makes the object appear to be weightless. If done in a zero G environment, it would just propel the object away from the superconducting magnet. And this effect has nothing to do with the Earth's magnetic field, it's a quantum mechanical effect that occurs in all materials. – Johnny Dec 11 '14 at 16:20

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 $$F=m_pg=G\frac{m_bm_p}{r^2}$$ 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 $m_b$: $$m_b=\frac{gr^2}{G}$$ 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: $$F_c=\frac{mv_c^2}{r}=mg_{\text{Moon}}$$ 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: $$\frac{dL_{\text{system}}}{dt}=0$$ 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 $\frac{mv_c}{r}$ is a constant. We also know that $F$ is a constant. This means that $$\frac{d(\frac{mv_c^2}{r})}{dt}=0$$ So $\frac{mv_c^2}{r}$ 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 $dv \neq 0$, 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.

The straightforward approach seems to attach the object to the lasso and keep spinning it around as a cowboy does before throwing. The centrifugal force will simulate the gravity for the object.

The astronaut may rotate in the opposite direction but for the relatively light object this effect will be small. It can be compensated with the rocket thrusters, spinning another similar object in the opposite direction (tricky but may be possible) or holding to the massive spacecraft with another hand.

The usefulness of the idea depends on why does the object need the gravity. If, say, suspended particles must be brought down from the fluid, the approach would work.