How fast does the Hermes need to rotate to have gravity? | Nerd Math
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- čas přidán 18. 05. 2024
- I love a bit of nerd math in between main videos. This month I broke down the physics of space travel in The Martian, one of my favourite movies and one of my absolute favourite space ships in scifi!
I would go into space on the Hermes. But how fast does it need to rotate to give 1 g of gravity?
Check out my video about The Martian if you want more nerdy • Astronauts and Orbital...
About Me --
I'm Abi, a phd student at the University of Oxford working in plasma acceleration. I love to get super nerdy about pop culture and find ways to explain complicated topics in physics. It's a nerd revolution and I'm here for it! - Věda a technologie
What about other effects? Like the Coriolis effect, or the force decreasing as you get further from the floor? I wonder whether astronauts would find it very dizzying.
I agree
The force decrease along the height of a human is not enough to have any real issues. For the height of 1.57m the change in g over the height is 1 ms^-2 which won't cause any health issues, it may cause a slight change in weight senstaion but not a major one.
The coriolis effect can cause disorientation however it's just a case of adapting to the environment and learing how to navigate, the same way astronauts have to learn how to move around in zero-g and in spacesuits for EVAs.
I also believed they would experience dizziness but again, it's about adaptability. Studies have shown that the sensation is not extreme enough to render it and issue and that with proper training this set up is well withing the physical limitations of humans.
Interesting talk! But do they really need 1g? How about g/2 or g/3?
Probably it was just 1g because they didn't have to go over the numbers in the book/movie. It seems likely that structures using this sort of artificial gravity would operate at less than 1g because it would greatly reduce the engineering complexity needed to make it both structurally sound and minimize disorienting effects (like gravity varying at different heights and coriolis effect). However, if the goal is to prevent body degradation from low-g, I'm not sure what kind of gravity would be required.
There is no such thing as centrifugal force. What you are talking about is centripetal force experienced by objects trying to move in a straight line but are prevented from doing so by the wall of the rotating cylinder pushing on them and changing their motion into circular motion.
Centrifugal force is a 'pseudo' force but it is disntinct from centripetal force.
In a rotating reference frame, objects appear to experience an outward force due to their inertia, which tries to keep them moving in a straight line. This outward force is the centrifugal force. Meanwhile, the centripetal force acts inward, towards the center of rotation, and is responsible for keeping objects moving in circular motion.
Both forces are in place in a rotating system with the Centrifugal force being responsible for simulating the effect of gravity and creating the sensation of weight.
@@popculturescientist So, you are saying that force can be "relative" ?
Fyi: Sabine Hossenfelder says that: "Acceleration is absolute".
-So, doesn't that mean force is absolute❔
Surely, the real _inertial_ force is _tangential to the circle_ and the resultant _applied force_ is _diagonal_ ❔
(i.e. The resultant of the _friction_ that develops seems as if it were a normal reaction, but it's really diagonal but in line with a future circle tangent projection. Combine this friction with linear momentum and you get a vector parallel to a circle's radius, which is the reason why the orbiting mass orbits in a circle with "orbital angular momentum").
The rotation is way, way, way too fast to be comfortable for the astronauts. They would be getting nauseous and vomiting every time they turned their head. To make this work comfortably, the radius would need to be on the order of a hundred meters so that the rotation rate would be slow enough to be physically bearable.
This is a good video that details a lot of the experiements already conducted. czcams.com/video/nxeMoaxUpWk/video.htmlsi=HFYRaxTlT3QBLfz-The take away is that up to 10 rpm is adaptable. This is around 7.6rpm
Motion Sickness requires that you are relative to the Earth: that External Reference frame is gone: no inner ear dysphoria is caused, only visual from the stars spinning: your to far away from the Earth: for it to cause the experience of motion. As you leave the External Reference frame that includes the Earth, you enter an inertial frame: and you are no longer spinning against the mass of the Earth, but just the mass of the ship and your orbital component, and they do not present enough gravitation to be amplified by angular momentum and the centrifugal force, as it goes missing without an external and massive body to present the change against.
While motion sickness may indeed be influenced by reference frames, the rotational speed required for artificial gravity in a centrifuge is based on physics principles rather than Earth-centric experiences. The calculated rotation aims to simulate gravity for occupants within the rotating structure, irrespective of their distance from Earth.
@@popculturescientist So: 1G on Earth is 9.81 m/s^2. It is about the same on Venus, but on Mars: 3.71 m/s^2 is 1G, and on the moon 1G is 1.62. As 1G the force output it would require to levitate any mass, and continue to increase to provide the same in a constant acceleration. Without a Planet or mass body to reference: 1G is 0. Thus there can be no Centrifugal Force to imply. Ceres has 0.27 m/s^2 as 1G, Phobos provides 0.0057 m/s^2 as 1G.
So if you have the Moon Phobos at the center of your ship: you can generate a rotation force of 0.0057 m/s^2 as the 1G of Force in angular momentum generated by rotation.
Now Jupiter has 24.79 m/s^2 as 1G, so you can reduce the rotation vector to simulate the same amount of Gravity as 1G would be on Earth. The Sun provides 274 m/s^2 of Gravity as 1G. So if you fly away from the Sun equivalent to 1G on Earth: and you can hold its gravity for longer.
Now a Black Hole has 19 Billion m/s^2 as 1G, so you can fly away from them for a very long time and still feel their presence: of course the speed you need to fly away is proportional to distance: so it will not be useful, as it requires the acceleration be greater than light in our solar system.
Once in Motion: 9.81m/s^2 sustained is 1G, as a constant acceleration to break away. Once away: you are not accelerating in reference to anything to notice the acceleration.
This force would only exist near a planet, and only when the rotational body is accelerating. When you are in orbit, you are still falling towards the planet, and still close enough to be engrossed of its gravitation. Without the external body of reference to spin against: no amount of acceleration will produce a gravitational effect.
Flying away from a Planet at 9.81 m/s^2, as in constant effect of distance: has a limited distance until the speed reached is irrational. You an add spinning to the acceleration away from a mass body to enhance the gravitational effect. No amount of spinning in true zero gravitation, when away from a planet will produce gravitation. Gravitation is an acceleration, and a spinning body does not accelerate against anything relative to be considered spinning at all.
So you need to accelerate the rotation to feel gravity, and you need to be close to a planet for the forces to transfer. Velocity of rotation at a constant is not an acceleration, that is just m/s and lacks the square.
Again, the object on earth is fighting earth's gravity to transfer the force. A heavy pressure shower with statically charged water accelerated to the ground: would allow for some gravitational simulation, but that is just simulation. The accelerating habitat: would also be a staircase and not a walkway, as it would work best, when accelerating near a planet: as an incline plane.
This comment is very confusing. You appear to be speaking about gravitational acceleration. The Centrifugal force does not mimic acceleration.
It’s an outward force that is caused by rotation and does not require the presence of a planet
@@popculturescientist
When I spin an object on the Earth: it is fighting gravity. The gravity of the Earth is transferred by momentum and moment of change acting against the Earth's Gravitational field. If you take away the planet: you are not spinning anymore: because your not spinning against anything, as the moment of change is not apparent without the point of reference the Earth provides. Dynamic equilibrium of two fluids in a container: would never return or settle into separated states if in Zero G.
We’re not talking about Earth or gravity. We are specifically speaking about Centrifugal Force
@@popculturescientist
Centripetal force, yes I know. Angular Momentum needs an external point of reference to function, an external point which you rotate against or in respect to. If you only have a center point and a fixed point: it is a linear connection, and no force is found between them without another body or coordinate system it is spinning against. A centrifuge in orbit is still spinning to escape the earth's pull. If you just have the center and the external rotating point: the angular momentum does not matter as it does not exist in the closed system: if their is nothing you are spinning against, and you are not spinning against the stars: the gravitation is not relative to be a condition of a force applied.
The Centripetal force is related to the mass of the object, but no mention of the mass of the center point is made, until it is used to explain planetary motion. If I am not fighting to spin against anything: if no force exists that would slow me down: I am not spinning at all in mathematical reference! The context of where the equation is used. You are losing your core as an axis and Delta Velocity: without the body of inertial reference. There is no longer an angular momentum if you do not have the external reference: as you are no longer in reference to be spinning against anything. Only when you are still falling into the Earth, as on it or above it: does your Delta Velocity have an external reference outside its own axis: to represent the spin rate.
When you pull 1 G in angular Momentum: your rate of change is counted against the force of gravity: without it that force: there is no change, and thus no force can be applied: you lack the external reference to be a spinning system pulling against anything: so your rotational body does not have a force anymore. Your no are never really accelerating against the center of the axis: your accelerating against the Earth: it is an essential part of the system! If you remove the Earth: you do not have enough mass to induce any notable force.
@@popculturescientist I will not be satisfied until a satellite is sent out into deep space with a Centrifuge on board to prove that the change in angular momentum, is a change in relationship to a major gravitational center. A ship like in the Martian: would not spin: the center would spin inverse of the external section just in the opposite direction: making a net zero system. Even if the entire thing was spinning: it is not spinning against anything it is relative too: to be considered spinning. It is a matter of relativity.
Off topic,,, sort of,,ish,,,,what happen to pregancy in space wth zero g,, yea i get it no ones given birth in space ,, but um,, mice.?