Sunday, 1 November 2015

forces - Will a machine or a technique ever be possible that allows artificial gravity in interplanetary space?


Will a machine or a technique ever be possible that allows artificial gravity in interplanetary space?



And I ask out of pure curiosity. I know there is the "Vomit Comet" that allows for a non-gravity 'simulation' (for lack of a better word) on Earth, and I was thinking about if there would be a way to reverse that and rather have gravity in space.


edit: Say I was building a space shuttle because I'm dying to go to the moon (true story). Unlimited budget, the smartest people you can fathom helping me build my space shuttle. I get motion sickness and the whole "anti-gravity" thing wouldn't really work for me. Would there be a way to have gravity on my space shuttle? Perhaps by using a type of machine or magnets or something of the like?



Answer



It sounds like you don't want the normal rotating spaceship like in "2001" because you get motion sickness. No one really gets "motion sickness" just from moving, though. That's impossible because moving with constant velocity is physically the same as being stationary. What you get is "acceleration sickness". You feel the bumpiness of a car ride. Even then, it might more-accurately be called "jerk sickness" or "tidal sickness". "Jerk sickness" would be when your acceleration is changing over time, as in going over a bump. "Tidal sickness" is when different parts of your body experience different accelerations at the same time.


Inside a rotating cylinder, what you feel would be identical to normal gravity to first order, so you wouldn't necessarily get motion sick like in a car. If you get motion sick because you see the stars rotate when you look out the window, don't build any windows and you'll never know.


A few effects break this illusion of perfect gravity, including tidal forces, Coriolis forces, and non-constant angular velocity of the spaceship. The bigger your ship, the smaller these effects will be.


Tidal forces arise when the gravitational field is not the same everywhere. To make tidal forces small, you need the ship to be big. The strength of the artificial gravity on your ship is $\omega^2 R$ with $\omega$ the angular frequency (how fast the ship is rotating) and $R$ the distance from the rotation axis. Tidal accelerations are then on the order of $\omega^2\Delta R$, with $\Delta R$ the size of the object we're considering (e.g. your body). Tidal forces compared to gravity are $\Delta R / R$, or the size of the object compared to the size of the ship. If you're 1000 times smaller than your ship, the tidal forces you feel will be 1000 times smaller than gravity - too small to notice. (On Earth they are much smaller still, of course, because you are so small compared to Earth).


Coriolis forces are forces that makes tossed projectiles appear to bend when you throw them inside the ship, even though nothing is pushing on them. The size of the Coriolis acceleration is on the order of $\omega v$, with $v$ the speed of the thing being thrown. In order to make them smaller, you need to decrease $\omega$. Remember that the strength of gravity is $g = \omega^2 R$, so $\omega = \sqrt{g/R}$. The make the Coriolis forces small, you need a big ship again. Suppose you want Coriolis forces to be under some fraction $\alpha$ of $g$ when you go at a certain maximum velocity $v$. Then the formula for the minimum size of the spaceship is $R > v/(\alpha^2 g)$. If $v = 10 {\rm m/s}$ and $\alpha = .01$, we need a spaceship 10 kilometers across. So we're going to have to tolerate fairly high Coriolis acceleration. But remember you don't feel this at all if you're stationary. If you're moving it's just a constant force pushing up/down/left/right depending on which way you're going. Only if different parts of you are moving different directions does it becomes a serious problem. Your free throws would all be off, too.


Finally, when stuff moves around in the spaceship, the rotation rate of the entire ship could change because the stuff pushes on the ship. This acceleration would be on the order of $am/M$, with $a$ the acceleration of the moving object, $m$ the mass of the object, and $M$ the mass of the ship. Most things on the ship would have a tiny $m/M$, so this wouldn't be a big deal.


If an object didn't accelerate, but moved towards the center at a constant speed $v$, the ship would accelerate like an ice skater accelerating as they pull their arms in. This acceleration would be on the order of $v^2m/RM$, and again would be small for small objects.



Another potential problem is that if the mass distribution of the ship is not symmetric, the ship would wobble. Suppose that the ship is balanced at some time, but then an object of mass $m$ is moved a distance $d$ along the side of the cylinder towards the end. Then the frequency of the wobble should be on the order of $\omega md/MR$ and the amplitude should be on the order of $md/M$. The jerk you feel would then be on the order of $\omega^3 m^2d^2/M^2 R$, which should be small as long as it's a fairly small mass moving a small distance.


How small can these effects be? That relies mostly on the size of the ship, as we saw earlier. If the ship is going to simulate one Earth gravity, it can't be too big. Eventually, the tension in the sides of the ship would be so great the ship would tear itself apart. If your ship were a loop, that critical radius would be $T/\lambda g$, with $T$ the tension and $\lambda$ the linear mass density of the wall. For carbon nanotubes this gets you a ship size of up to $10^7 {\rm m}$, a figure so large (bigger than Earth) that all effects could be made inconsequential. For steel it's about $5*10^3 {\rm m}$, meaning that the effects on to human-sized might be in general $.001$ or $.0001$ of gravity.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...