Frame of reference question: Car traveling at the equator

AI Thread Summary
A car traveling west at 460 m/s at the equator experiences different speeds depending on the reference frame used. In an inertial frame centered on Earth, the car's speed is effectively zero as it moves along a curved path due to Earth's rotation. In a rotating frame, the car's weight is influenced by centrifugal and Coriolis forces, which alter the apparent gravitational force acting on it. The Coriolis force, which acts perpendicular to the velocity vector, has both vertical and horizontal components depending on the latitude, affecting the car's motion. Understanding these dynamics is crucial for accurately assessing forces and motion in different reference frames.
  • #51
jbriggs444 said:
Would it rotate like what? Yes, it rotates. Yes, it rotates faster than the Earth. Yes, the inside rotates faster than the outside.

What is it that you are trying to ask?
In inertial frame coriolis effect don't exist,so why then we agian see wind bend to the right which gives counterclok wise rotation at cyclon even in inertial frame?
If you look at missile form inertial frame ,missile is going in straight line ,so why wind will not do the same?
Why is my question so strange,isnt my question logic?
 
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  • #52
jbriggs444 said:
Earth rotates clockwise

I think you mean counterclockwise, at least if you are looking down on the Earth from above the North Pole.
 
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  • #53
John Mcrain said:
In inertial frame coriolis effect don't exist,so why then we agian see wind bend to the right which gives counterclok wise rotation at cyclon even in inertial frame?
If you look at missile form inertial frame ,missile is going in straight line ,so why wind will not do the same?
Why is my question so strange,isnt my question logic?
The individual parcels of wind are doing the same. They move in straight lines unless deflected by a real force.

The individual parcels of wind start at rest relative to a rotating earth. They are already rotating about a common center. If all things were equal (uniform temperature, etc), this would be an equalibrium situation. No wind would be blowing anywhere. The air would stay in a stable rotation, at rest above the rotating Earth. The Earth itself would (and does) relax into an ellipsoidal shape so that the required centripetal acceleration is provided for.

But if you apply a real physical force to draw a air parcels in toward the center of rotation, that inward movement means that their original tangential velocity at a large radius is now a greater tangential velocity at a smaller radius. The rotation rate of all of the parcels about their common center has increased. The air is no longer at rest relative to the rotating Earth. Winds blow.
 
  • #54
haruspex said:
On the contrary, it is crucial here.

Let the gravitational attraction of the Earth on the car be W.
In the frame of the rotating Earth you have to consider the centrifugal and Coriolis forces: https://en.m.wikipedia.org/wiki/Coriolis_force.
The centrifugal force, ##mR\Omega^2##, points up. Here, R is the radius of the Earth and ##\Omega## is its rotation rate.
The Coriolis force is given by ##-2m\vec\Omega\times\vec v##, where v is the velocity of the car in the rotating frame.
If the car were going West to East, this would also point up, giving a net upward force in the rotating frame of ##3mR\Omega^2##. To an observer in that frame, the car has an acceleration ##R\Omega^2## towards the Earth's centre. For all this to add up correctly, the normal force from the ground would have to be ##W-4mR\Omega^2##. Back in the inertial frame, the car's speed is ##2R\Omega##, so the centripetal force is ##4mR\Omega^2##, giving the same result.

But here the car is going East to West, so the Coriolis force points down, and its magnitude is ##2mR\Omega^2##. Gravitational force plus centrifugal plus Coriolis gives a downward total of ##W-mR\Omega^2+2mR\Omega^2=W+mR\Omega^2##. To the observer on the ground, the car's acceleration is the same as in the W to E case, so the normal force must be just ##W##. In the inertial frame the car is stationary, giving the same result.
What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?
 
  • #55
John Mcrain said:
What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?
Why does there have to be only one explanation for any given effect?
 
  • #56
John Mcrain said:
What is real reason why bullet hit traget high(Eotvos effect) when shoot at east;
1)is bullet hit higher at target,because coriolis has net upward force on bullet
or
2)the target has time to move down during the flight time below what the aiming point was, so the hit is at the top ?
It depends on your frame of reference. In the frame of reference rotating with the earth, the Eötvös effect (Coriolis) applies and you can consider that the reason. In an inertial frame there is no such effect, but the target is on a curved trajectory.
 
  • #57
haruspex said:
It depends on your frame of reference. In the frame of reference rotating with the earth, the Eötvös effect (Coriolis) applies and you can consider that the reason. In an inertial frame there is no such effect, but the target is on a curved trajectory.
How can I set equation why bullet hit top of target using inertial frame?
I know that bullet absolute velocity is incresed (earth speed + bullet ground speed) so centripetal force is increased also,but centripetal force point down (not up), toward center

?

jbriggs444 said:
Why does there have to be only one explanation for any given effect?
Yes it can be more...
 
  • #58
John Mcrain said:
How can I set equation why bullet hit top of target using inertial frame?
The point from which the bullet is fired has some instantaneous velocity.
The bullet is aimed as though the target had the same velocity, but it doesn't. If firing East, the target's velocity is, from the shooter's perspective, angled slightly down, so by the time the bullet arrives the target will have descended relative to the trajectory.
Conversely, if firing West, the target's velocity is tipped up, causing it to rise above the trajectory.
 
  • #59
John Mcrain said:
I know that bullet absolute velocity is incresed (earth speed + bullet ground speed) so centripetal force is increased
Why do you say that centripetal force is increased? We need to examine that reasoning because it is faulty.
 
  • #60
John Mcrain said:
so centripetal force is increased
Centripetal force is not an extra applied force. It is that component of the resultant of all the forces that are applied that is normal to the velocity. In this case, it is just gravity, and that is not about to change.
What you mean is that in order to hit the target on the anticipated trajectory the centripetal force would need to increase - but it doesn't.
 
  • #61
haruspex said:
Centripetal force is not an extra applied force. It is that component of the resultant of all the forces that are applied that is normal to the velocity. In this case, it is just gravity, and that is not about to change.
What you mean is that in order to hit the target on the anticipated trajectory the centripetal force would need to increase - but it doesn't.

You proved in rotating frame how car/aircraft or bullet travel at east has reduced weight because of corioils net upward force.
In general I have trouble to prove reduction in weight of car/aircraft/bullet in inertial frame.

How can I set equations for this?
if car/aircraft/bullet travel et east with 463m/s (same velocity as Earth rotation )

Car: mg - N = mv2/r (where v is apsolute speed of car,v=car ground speed + Earth speed)
weight(mg) = mv2/r + N

Aircraft: mg - L = mv2/r (where v is apsolute speed of aicraft,v=aircraft ground speed + Earth speed)
weight(mg) = mv2/r + L

Bullet: mg = mv2/r (where v is apsolute speed of bullet,v=bullet ground speed + Earth speed)

in all three cases when v is increased,because of eastward travel(absolute speed increase) then weight(mg) is increased also,which is not consistent with results from rotating frame!

So what I am doing wrong?
 
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  • #62
John Mcrain said:
in all three cases when v is increased,because of eastward travel(absolute speed increase) then weight(mg) is increased
No, mg, the true weight, does not change.
How do you measure your weight?
 
  • #63
haruspex said:
No, mg, the true weight, does not change.
How do you measure your weight?
I stay on weight scale.

Than what is changed in inertial frame?
Do both frames must get same final results?
 
  • #64
John Mcrain said:
I stay on weight scale.
And what force does a scale actually measure?
 
  • #65
haruspex said:
And what force does a scale actually measure?
it show normal force = mg - mv2/r (centrifugal force)

But can I call it "mv2/r" centrifugal force?
 
  • #66
John Mcrain said:
it show normal force
Right, the apparent weight is the normal force, and what all of these questions concern is the change in apparent weight, not the actual weight.
Redo your work in post#61 on that understanding.
 
  • #67
haruspex said:
Right, the apparent weight is the normal force, and what all of these questions concern is the change in apparent weight, not the actual weight.
Redo your work in post#61 on that understanding.
But is my "mv2/r " centifugal force or centripetal force? that confused me

in one way, must be centripetal because I am working in inertial frame
in other way, doesn't make sense is centripetal because it act toward center of Earth so must be add it to mg!
I am fighting with terminology
 
  • #68
haruspex said:
Redo your work in post#61 on that understanding.
aircraft/bullet/car speed = 463m/s ,weight=1000kg
r=6371km
inertial frame;

Car: N = mg - mv2/r .
trevel EAST 1000kgx9.81 - 1000kg x (926m/s)2 / 6371000m...N=9675 Newtons
trevel WEST 1000kgx9.81 - 1000kg x (0m/s)2 / 6371000m...N=mg=9810 Newtons
136N/9810N= 1.38% Car apparent weight is 1.38% less when trevel EAST

aircraft : same calulation as car case, just normal force(N) I will call L (lift)
(neglect increased of radius becasue of fligt altitude)

bullet: again looking from inertial frame bullet trevel et east with 926m/s and to west 0m/s
but here is see only one force,gravity mg,
so mg=mv2/r ??
so if I put v for east and west ,what now?
has bullet decreased apparent weight like car and plane when trevel east??
 
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  • #69
John Mcrain said:
But is my "mv2/r " centifugal force or centripetal force? that confused me

in one way, must be centripetal because I am working in inertial frame
in other way, doesn't make sense is centripetal because it act toward center of Earth so must be add it to mg!
I am fighting with terminology
Centripetal force is a component of the net force. Specifically, it is the component which is normal to the velocity. That definition works in all circumstances, including non inertial frames.
This leads the object to travel in an arc around some point, at least transiently. The radius of that arc is given by ##\frac{v_{tangential}^2}{a_{radial }}##.

Suppose there is a satellite circling at the equator at distance r from the Earth's centre, speed v, W to E. It's angular rate is v/r.
A ground based observer perceives the satellite's angular rate as v/r-Ω, so speed as |v-Ωr|, so computes the centripetal acceleration as (v-Ωr)2/r. The net force on it is calculated by taking mg and adjusting for centrifugal and Coriolis forces. If there is no fault in the math, this should equal m(v-Ωr)2/r.

In the inertial frame, the centripetal acceleration is calculated as v2/r, and no adjustment is made: the net force is just mg.

Now consider a geostationary satellite. v=Ωr. This makes the observed speed and acceleration zero in the non inertial frame, so no centripetal force and no Coriolis. But the centrifugal correction still applies, namely, mΩ2r. But that is the same as mv2/r.

In sum, centripetal force always equals mv2/r, but centrifugal force only equals that (with opposite sign) when the rotation of the non inertial frame makes the observed tangential velocity zero.
 
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  • #70
haruspex said:
In the inertial frame, the centripetal acceleration is calculated as v2/r, and no adjustment is made: the net force is just mg.

How then I write equation for bullet trevel at 463m/s(ground speed)for east and for west ,from inertial frame?
earth speed rotation is also 463m/s
 
  • #71
jbriggs444 said:
Why do you say that centripetal force is increased? We need to examine that reasoning because it is faulty.
When bullet fly to east his speed form inertial frame is 926m/s and when fly to west his speed is 0m/s,and if I put this numbers into centripetal formula mv2/r, I get higher centripetal force to east..

I know i am doing something wrong ,because my results are not same as in rotating frame
 
  • #72
John Mcrain said:
When bullet fly to east his speed form inertial frame is 926m/s and when fly to west his speed is 0m/s,and if I put this numbers into centripetal formula mv2/r, I get higher centripetal force to east..
The "r" that you are putting into the formula. You are using the radius of the earth, right?

The "r" that belongs in the centripetal force formula is the radius of curvature of the projectile's path, right?

What makes you think that the two are identical?
 
  • #73
jbriggs444 said:
The "r" that you are putting into the formula. You are using the radius of the earth, right?

The "r" that belongs in the centripetal force formula is the radius of curvature of the projectile's path, right?

What makes you think that the two are identical?
Because I assume that bullet follow Earth radius for make calculation easier..
But I know that they are no identical..

So how to set equation for east and west bullet shoot for inertial frame,can you write?
 
  • #74
John Mcrain said:
Because I assume that bullet follow Earth radius for make calculation easier..
Bullets do not follow Earth radius. Not even close. In order to follow Earth radius, a bullet would need to attain orbital velocity -- about 7.8 km/sec.
 
  • #75
jbriggs444 said:
Bullets do not follow Earth radius. Not even close. In order to follow Earth radius, a bullet would need to attain orbital velocity -- about 7.8 km/sec.
Yes I know that ,bullet radius is smaller then earth...

So how can I set equation for bullet?
 
  • #76
John Mcrain said:
Yes I know that ,bullet radius is smaller then earth...

So how can I set equation for bullet?
It depends. What are you trying to calculate?
 
  • #77
jbriggs444 said:
It depends. What are you trying to calculate?
I want prove with equation why bullet hit east target high and west traget low using inertial frame.
in same maner like I do for aircraft and car case,in post #68
 
  • #78
John Mcrain said:
I want prove with equation why bullet hit east target high and west traget low using inertial frame.
in same maner like I do for aircraft and car case,in post #68
I repeat: What is it that you want to calculate?

I am guessing that you want to know how far left and how much further up (or down) you will need to aim to hit your target.

So we have a target a distance d away on a flat and level plane. It is at the same height as the barrel of the rifle. Your muzzle velocity is v. The local acceleration of gravity is g. We start with a baseline situation of a stationary Earth. You may neglect air resistance. You may pretend that the Earth is flat.

How much above the horizontal do you need to aim?

When we get to the next step, by your own stipulation, you will be required to do all calculations in the inertial frame -- the gun will be moving, the target will be moving and the Earth will be spherical.
 
  • #79
jbriggs444 said:
I repeat: What is it that you want to calculate?

In short, is bullet apperent weight smaller when shoot east compare to west?(just like is case with plane or car which travel east,which I proved in post#68 )
I need equation that prove this in INERTIAL FRAME
 
  • #80
John Mcrain said:
In short, is bullet apperent weight smaller when shoot east compare to west?(just like is case with plane or car which travel east,which I proved in post#68 )
I need equation that prove this in INERTIAL FRAME
You have forbidden us to use the rotating frame. That means that the concept of "apparent weight" is off the table. Not allowed by your own rules!

In the inertial frame, both shooter and target are moving. The moving target means that one must lead the target, aiming counter-clockwise (left of target). The moving shooter means that one must compensate for the shooter's movement as well. This is also a leftward correction.

If the target is eastward, there is also a downward correction required because the shooter is moving up and the target is moving down relative to a point at rest in the middle. If the target is westward, there is an upward correction for the same reaason.

If you want an equation, you have to put in the work. Let's see the equation for a stationary shot.
 
  • #81
jbriggs444 said:
You have forbidden us to use the rotating frame. That means that the concept of "apparent weight" is off the table. Not allowed by your own rules!

@ Haruspex call apparent weight the normal force .
normal force at plane is lift,for car is ground that push you up...
So what is normal force at bullet? don't exist or what?
I hate inertial frame grrrrrrrrrrrrrrrrrrrrrrrrrrrrr grrrrrrrrrrrrrr, I'm starting to go crazy
 
  • #82
John Mcrain said:
@ Haruspex call apparent weight the normal force .
normal force at plane is lift,for car is ground that push you up...
So what is normal force at bullet? don't exist or what?
Correct. A bullet is ballistic. It is in a free fall trajectory. There is no normal force.
 
  • #83
jbriggs444 said:
Correct. A bullet is ballistic. It is in a free fall trajectory. There is no normal force.
So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?
 
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  • #84
last 30 posts I ask for that simple equation...
there is no any " mv2/r" in bullet case

Fg = ma

 
  • #85
John Mcrain said:
So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?
For a projectile, we can observe the trajectory and compare it with the known value for its gravitational acceleration. Doesn't matter whether it's parabolic (g effectively constant), or a satellite in orbit.

In an inertial frame there is no discrepancy, so no difference between true weight, mg or ##\frac{MmG}{r^2}##, and the apparent weight deduced from its trajectory.

In a rotating or accelerating frame, if we don't consider centrifugal and coriolis forces then we will deduce a different weight (i.e. gravitational force). This is the apparent weight in that reference frame. But if we do take those forces (and the Euler force if the rotation is not uniform) into account then once again we will deduce mg as the weight.
 
  • #86
haruspex said:
For a projectile, we can observe the trajectory and compare it with the known value for its gravitational acceleration. Doesn't matter whether it's parabolic (g effectively constant), or a satellite in orbit.

In an inertial frame there is no discrepancy, so no difference between true weight, mg or ##\frac{MmG}{r^2}##, and the apparent weight deduced from its trajectory.

In a rotating or accelerating frame, if we don't consider centrifugal and coriolis forces then we will deduce a different weight (i.e. gravitational force). This is the apparent weight in that reference frame. But if we do take those forces (and the Euler force if the rotation is not uniform) into account then once again we will deduce mg as the weight.
So final conlusion is:

Bullet don't has any reduction in weight in any frame, just like car or plane have...
 
  • #87
John Mcrain said:
So bullet shoot to the east at equator DO NOT REDUCED WEIGHT (even in both frames?),just like aircraft weigh less when fly to the east at equator?
WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?
 
  • #88
jbriggs444 said:
WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?
example:
Car: N = mg - mv2/r .
trevel EAST 1000kgx9.81 - 1000kg x (926m/s)2 / 6371000m...N=9675 Newtons
trevel WEST 1000kgx9.81 - 1000kg x (0m/s)2 / 6371000m...N=mg=9810 Newtons
136N/9810N= 1.38% Car apparent weight is 1.38% less when trevel EAST

so car normal force "weight" is reduced by 136N when trevel east,which is not case in bullet

proffesor Matt say centripetal force is not real force,he say it is sum of forces in radial direction at 6:37
 
  • #89
John Mcrain said:
example:
Car: N = mg - mv2/r .
trevel EAST 1000kgx9.81 - 1000kg x (926m/s)2 / 6371000m...N=9675 Newtons
trevel WEST 1000kgx9.81 - 1000kg x (0m/s)2 / 6371000m...N=mg=9810 Newtons
136N/9810N= 1.38% Car apparent weight is 1.38% less when trevel EAST

so car normal force "weight" is reduced by 136N when trevel east,which is not case in bullet

proffesor Matt say centripetal force is not real force,he say it is sum of forces in radial direction at 6:37

I repeat: WHAT DOES WEIGHT MEAN WHEN YOU USE THE TERM?

I am not asking for a formula. I am not asking for someone else's video. I am asking for a definition. What do you mean when you use the term? You have used the term. Surely that means that you know what it means when you use it.
 
  • #90
jbriggs444 said:
I am asking for a definition

I am confused with all this terminology and you ask me what is definition?
before this topic i didnt know that i am talking about normal force,because I didnt know what is normal force...I just understand that if airplane fly to east ,lift at wings is reduced compere to fly to west.That mean plane can fly with lower AoA,this reduce induced drag,lower fuel consumption etc etc

so "my aircraft logic" was;weight is reduced,because lift must be equal to weight

So my definition of weight was,it has opposite direction from lift and has equal magnitude as lift,to keep plane in level flight.
 
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  • #91
John Mcrain said:
if airplane fly to east ,lift at wings is reduced compere to fly to west

Why do you think this is true?
 
  • #92
John Mcrain said:
I am confused with all this terminology and you ask me what is definition?
before this topic i didnt know that i am talking about normal force,because I didnt know what is normal force...I just understand that if airplane fly to east ,lift at wings is reduced compere to fly to west.That mean plane can fly with lower AoA,this reduce induced drag,lower fuel consumption etc etc

so "my aircraft logic" was;weight is reduced,because lift must be equal to weight

So my definition of weight was,it has opposite direction from lift and has equal magnitude as lift,to keep plane in level flight.
So for you, "weight" is the real, physical support force that would be required to keep the vertical component of an object's velocity constant.

Is that correct?
 
  • #93
John Mcrain said:
So final conlusion is:

Bullet don't has any reduction in weight in any frame, just like car or plane have...
Your difficulty seems to be that you keep switching what you mean by weight.

On the one hand, there is mg (or more generally ##\frac{GMm}{r^2}##), the force of gravity. On the other, there is what an observer may measure as its value, perhaps not taking into account all the complications.
If you stand on the bathroom scales, what you directly measure is the normal force the scales exert. This is your apparent weight, and it may be less than mg because of Earth's rotation, i.e. centripetal force.
If you observe a ball or bullet in flight, you can deduce a value of g from its acceleration in your reference frame, but the value will be inaccurate if you do not adjust for Coriolis as well as centripetal force.

Bottom line, the true weight is always mg; the apparent weight differs only because the observer in a non inertial frame fails to make all the appropriate corrections. It makes no difference whether the object is a car, bullet, plane...
 
  • #94
jbriggs444 said:
So for you, "weight" is the real, physical support force that would be required to keep the vertical component of an object's velocity constant.
I don't understand well this combinations of words.I am not physicist.

I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...whatever that mean..
So this was my starting point when asked question...
 
  • #95
John Mcrain said:
I don't understand well this combinations of words.I am not physicist.

I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...whatever that mean..
So this was my starting point when asked question...
If you are getting into fine details of centrifugal force and rotating frames of reference then fine details on the definition of weight matter.

You cannot have it both ways. You cannot ask for an understanding and then disavow any basis for understanding.
 
  • #96
John Mcrain said:
I just learned in school if object is flying in level flight(keep same altitude) then lift force at wings must be equal to aircraft weight...
Yes, but that's only to a first approximation.
Let's assume this refers to the true weight, i.e. mg.

In an inertial frame, the plane has to follow the curvature of the Earth, so it has a small downward acceleration. To provide the centripetal force necessary, the lift must be a little less than mg.

In the frame of reference of the plane, there is no acceleration, so no centripetal force. Instead, we should apply the centrifugal correction since we know we are in a rotating frame. This gives centrifugal (up) + lift = mg, so again, lift is a bit less than mg.

An observer on the ground observes a centripetal acceleration (but different from that measured in the inertial frame) and needs to apply both centrifugal and coriolis corrections. As in the other frames, lift will turn out to be a bit less than mg.
 
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  • #97
haruspex said:
In the frame of reference of the plane, there is no acceleration, so no centripetal force. Instead, we should apply the centrifugal correction since we know we are in a rotating frame. This gives centrifugal (up) + lift = mg, so again, lift is a bit less than mg.

Why in this frame don't exist coriolis force?

And what is difference with this frame and frame of observer from gorund?
 
  • #98
John Mcrain said:
Why in this frame don't exist coriolis force?
In the plane's own reference frame it is stationary, so no acceleration (no centripetal) and zero velocity (no Coriolis).
John Mcrain said:
And what is difference with this frame and frame of observer from gorund?
The observer on the ground sees the plane accelerating radially and as having a nonzero velocity.
 
  • #99
haruspex said:
In the plane's own reference frame it is stationary, so no acceleration (no centripetal) and zero velocity (no Coriolis).
I allways think that centripetal force exist in every frame.
Isnt centirpetal force in plane own ref. frame gravity=mg?in this picture write it is present in every ref.frame ?
pic.png
 
  • #100
John Mcrain said:
I allways think that centripetal force exist in every frame.
Isnt centirpetal force in plane own ref. frame gravity=mg?in this picture write it is present in every ref.frame ?
View attachment 270511
It's wrong - where is it from?
Edit: Hi @A.T., I see this graphic is from an old post of yours.

In an observer's frame, the centripetal force is that component of the net force which accounts for an observed curved path. In the astronaut's own frame, the astronaut is stationary. No curved path to account for. Moreover, the centrifugal force exactly balances the normal force from the space station wall, so the net force is zero. This fits with the astronaut's view of being stationary.
 
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