# Why doesn't the space station use centrifugal force for artificial gravity?

by Many A-Sun 2 months ago in astronomy
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Because it's difficult

Space Station

We don't have a large ring-shaped space station that uses centrifugal force (or more accurately, centripetal acceleration) to generate artificial gravity because it would be difficult.

We know from the laws of physics that it can be done - we just have a few issues to deal with.

1 - Cost

A centrifugal effect space station would have to be very large. According to the study, the station should rotate at 2 rpm (revolutions per minute) or slower so that those inside do not feel the adverse effects of tidal forces. The tidal force is the difference in gravity between a high point and a low point. For example, suppose you are on that space station. If it is too small, then your head (near the center) and your feet (on the floor/edge) will rotate at different speeds. This will feel different intensities of gravity in different parts of your body. To avoid this, the difference must be minimized by making the station very large.

For a 2 rpm space station, Earth's gravity (9.81 m/s²) can be replicated when the station has a radius of 223.64 meters (about 733.73 feet). This is much larger than the ISS.

The ISS is 108 meters long and costs about $150 billion. Consider how big a centrifugal effect space station would be. If its radius is 223.64 meters, then according to the perimeter formula, its length is 1,405 meters (4,610 feet), which is 13 times longer! This means that we can expect it to be at least 13 times more expensive, but probably more (>$2 trillion). All this money is used to produce the materials to build the structure, the onboard systems from life systems to electrical systems, the onboard equipment for space station experiments, and all the items the astronauts will need like food and fitness equipment.

2 - Stability

The space station is spinning so fast that we need to make sure its axis of rotation is stable. If it wobbles, it could make the people on board sick and disoriented, and it could destroy the technology inside. We needed to come up with technology that would keep it stable. Gyroscopic stabilizers come in many shapes and sizes. Accelerators had to be designed for such a large structure. There are many more.

In addition, we needed it to be able to avoid space debris. There are thrusts on the main body of the International Space Station. If it detects a piece of space junk or a meteoroid and calculates that it has more than a 1:10,000 chance of colliding with the station, it will change its orbit to move away. The ISS is easy because it does not rotate, so it is a simple "transnational" approach (a transnational orbital maneuver starts the engines while the spacecraft is not rotating).

For a rotating space station, panning is much more complicated. How can the station keep shooting in one direction to avoid debris? If the engine is on the outer edge of rotation, the engine will also rotate. This makes the co planar rotation operation easy, but it does not help at all because it changes the rotation speed of the station and changes the internal gravity. All transnational thrusts must be located on the axis of the space station. How would you design such a system? Would we put a literal "pole" on the rotating pole and put engines on it? How would we synchronize them with the rotation?

3 - Molecularity

The International Space Station is modular. This means that it is not a complete structure. It is a structure of many small pieces that are stacked on top of each other like giant Lego blocks. It is a large composite made up of many smaller units. For example, there are laboratories, airlocks, storage rooms, etc.

Just like Lego buildings, you can add more parts by simply adding them to the applicable joints. This is easy for the ISS because it doesn't rotate; you just attach a piece to the docking port and you're done.

Things will be more difficult with the Centrifugal Effects Space Station. It's spinning, and it's going very fast. It needs to be balanced. The center of mass and the center of rotation need to be as aligned as possible, or the rotation can become unstable and unbearable, or worse, the whole station is torn apart!

How can you add pieces without causing the station to become unbalanced? How do you even add debris when it rotates? The process of docking two spacecraft together takes a lot of time. Everything had to be perfectly aligned and the movements had to be very slow so they could be precise. This is almost impossible when the edge of the space station will be rotating at 47 meters per second. One possible solution (considering its huge size) was to install counterweights in the empty room. When the empty room is converted into a laboratory (or any room), the counterweights can be popped out and replaced with equipment.

These are the problems that would need to be solved to build such a space station. It is possible, but it would be difficult. We need to come up with new technologies on Earth that will make the process cheaper, the structure safer, and the experience more practical.

Extra Content.

You can calculate your own centrifugal effect space station! Use any one of these formulas.

Note: Humans need 2 rpm or less to not feel any tidal forces, although experiments have shown that it is possible to adapt to 23 rpm. I would still go well below that upper limit and leave the maximum at 10 rpm. people prone to motion sickness may need to spin at less than 2 rpm.

Equation 1: You know the rpm, you want to know the radius

r = 900 πsir = 900 a πs

RR is the radius in meters. One by one is the internal acceleration (aka gravity). Earth's gravity uses 9.81 m/s². $s$ is the speed of rotation in revolutions per minute.

Multiply the answer by 2 to get the diameter (span) or by 2π to get the circumference (the distance around; the full length inside the edge)

Equation 2: You know the radius, you want to know the speed of rotation

$s=\frack{30}{\pi} \sort{\frack{a}{r}}$

$s$ is the speed in revolutions per minute. $a$ is the internal acceleration (aka gravity). Earth's gravity uses 9.81 m/s². The $r$ is the radius in meters.

Formula 3: known radius and speed, want to find the gravity inside

$a=\frack{rs^2{\pi}^2}{900}$

$a$ is the internal acceleration (aka gravity). Divide your answer by 9.81 m/s² to see how it relates to the Earth's gravity. The $r$ is the radius in meters. $s$ is the rotational speed in revolutions per minute.

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