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Suppose you are kidnapped and you don't know who did it. You have no memory of the trip that brought you to the building where you are now being held. You have reason, however, to suspect that you are not on earth but rather in a large (5km diameter possibly larger) space habitat that uses gravity simulated through rotation. There are many typical objects in the room, possibly even some devices for making measurements if needed.

So, trapped inside of a room without windows, is there any way to prove you are not on earth using simple found objects and your understanding of physics?

Would the answer to this question change if you are on a non-spinning spaceship that accelerates at 9.8 m/s*s?

Artistic rendering of spinning space habitat.

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    $\begingroup$ Re "There are many typical objects in the room ..." Does this include a smart phone with an accelerometer, gyroscope, magnetometer, and GPS? $\endgroup$ – David Hammen Dec 29 '17 at 16:43
  • $\begingroup$ About the second question. It depends on the room shape and axes of rotation. You can have a floor closing in itself and this will point to a thorus rotating about its outside axes $\endgroup$ – Alchimista Dec 29 '17 at 17:15
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    $\begingroup$ This question (v2) is just Einstein's elevator gedankenexperiment. $\endgroup$ – Qmechanic Dec 30 '17 at 19:31
  • $\begingroup$ Find an apple tree. $\endgroup$ – Mitchell Dec 30 '17 at 19:47
  • $\begingroup$ This appears to be a list-based question, which is considered off-topic as too broad. $\endgroup$ – Kyle Kanos Jan 3 '18 at 11:03
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I think it would be worth picking up on Ben's point about measuring non-uniformities, which I assume means measuring the variation in the gravitational acceleration with height.

Assuming you have a ruler and some sort of timing device you can measure the gravitational acceleration by making a pendulum and measuring its period. Since the period is given by:

$$ \tau = 2\pi \sqrt{\frac{\ell}{g}} $$

You can use the pendulum to measure $g$, and you can measure how $g$ changes with height. On the Earth's surface the value of $g$ is given by:

$$ g = \frac{GM}{r^2} $$

with $r$ equal to the radius of the Earth, $r_e$. Differentiating this gives us the equation describing how $g$ varies with height:

$$ \frac{da}{dr} = -\frac{2GM}{r^3} $$

At the Earth's surface this has the value $3.07 \times 10^{-6}$ s$^{-2}$.

In a uniformly accelerating spaceship the variation of $a$ with height would be zero, so that's easily distinguishable.

In the stereotypical rotating torus the acceleration is given by:

$$ a = r\omega^2 $$

with $r$ being the radius of the torus and $\omega$ the angular velocity, so we get:

$$ \frac{da}{dr} = \omega^2 $$

So this time we find $da/dr$ is a constant rather than proportional to $r^{-3}$. We'd also measure a different value. Suppose we have a space station with a radius of $100$m, so the angular velocity has to be about $0.31$ radians per second to get an acceleration equal to $g$. In that case when we measured $dg/dr$ we'd get the value $0.0981$ s$^{-2}$. And again this is easily distinuishable from the value on Earth.

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    $\begingroup$ I think it would be worth picking up on Ben's point about measuring non-uniformities, which I assume means measuring the variation in the gravitational acceleration with height. The terrestrial field is also nonuniform in direction, even at a fixed height. $\endgroup$ – user4552 Dec 30 '17 at 19:54
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If you want to check whether you're rotating this is pretty easy. (1) Measure the Coriolis force. (2) Measure nonuniformity of the centrifugal force (gradients indicating variations in strength and direction). This is how we can prove that the earth is spinning, using a Foucault pendulum.

Would the answer to this question change if you are on a non-spinning spaceship that accelerates at 9.8 m/s*s?

The equivalence principle says that you can't tell the difference between this situation and a uniform gravitational field. However, you could check for the nonuniformities that you expect if you're near a spherical gravitating body.

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  • $\begingroup$ Would the room need to be very tall to accommodate a Foucault pendulum? $\endgroup$ – futurebird Dec 29 '17 at 15:41
  • $\begingroup$ It is very easy to point out that the second part is not really an answer, just a general hint. What are non-uniformities? How to check them? None of this is described. $\endgroup$ – kpv Jan 1 '18 at 17:57
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General answer without getting into mathematics -

First Part -

  1. Set a pulley at pretty high place in the room.

  2. Put a weight directly below the pulley

  3. Tie a good quality string to the weight

  4. Pull the string to lift the weight so it goes significantly higher than its original position.

  5. Due to conservation of angular momentum, the weight will lean to forward direction of rotation. This way, you can also find direction of rotation. You may use a laser pointer at the bottom of the weight and note the shift in the laser projected on the floor as the weight goes from bottom to the top.

You may also drop something from a height that would act as a projectile and would not land exactly below it.

Second Part is tricky due to equivalence principle so, I will try and wait for others to comment -

Due to the practicality of imparting a constant force, it can be distinguished in following manner -

Lift something significantly heavy to the roof. Leave a balance with a weight on it at a distance. The balance should start recording the weight now. Then let the heavy weight fall freely to the floor. Suppose, it does not bounce at all post fall.

I will take four scenarios here -

  1. The space ship is accelerated by a constant force. In this case, the time taken by the fall will be smaller than the one predicted by the perceived g as the ship will accelerate more during the fall. The balance will record higher weight during the fall, and lower weight on impact and then correct weight. i.e. just one cycle of weight change - higher, lower, correct.

  2. The room is a blimp hovering in the atmosphere of a planet - In this case, the balance would record multiple cycles of higher and lower weight due to its movement back and forth before it balances again.

  3. The room is fitted on springs on the surface of a planet - In this case, the balance would record multiple cycles of higher and lower weight due to its movement back and forth before it balances again.

  4. The room is actually on the surface of a planet - In this case, the balance would record no cycles of higher and lower weight.

I will wait for others to point out if there are any problems with this explanation, or if there are other scenarios that need to be explained.

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  • $\begingroup$ I don't understand your pulley experiment. Do you really mean a pulley, or are you just talking about hanging a weight from a fixed point? If the former, then what's on the other side of the pulley? Due to conservation of angular momentum, the weight will lean to forward direction of rotation. In the rotating frame, there is a Coriolis force in the direction you claim, but it's velocity-dependent, $2\textbf{v}\times\Omega$. In the kind of rotating space station the OP has in mind, $\Omega$ is small, so an expt like this with small $v$ is not a very practical. A projectile is better. $\endgroup$ – user4552 Dec 30 '17 at 20:01
  • $\begingroup$ The experiment in the second part really can't work, because of the equivalence principle. Your idea is to detect the inertia of the spaceship. But a blimp hovering in the earth's gravitational field would also have some amount of inertia. $\endgroup$ – user4552 Dec 30 '17 at 20:03
  • $\begingroup$ @BenCrowell: I mean a pulley and the weight is pulled over the pulley. The conservation of angular momentum will cause a shift in position. In other words, if you drop something from a height, it would not land directly below it in a rotating frame because, during the free fall, the falling object is not undergoing acceleration. I also mentioned dropping something, which is a projectile. $\endgroup$ – kpv Dec 30 '17 at 20:18
  • $\begingroup$ @BenCrowell: Actually, I did not mention dropping in first case, but you may use dropping as well which would act as a projectile. $\endgroup$ – kpv Dec 30 '17 at 20:24
  • $\begingroup$ @BenCrowell: In second case, I somehow agree with you on the blimp example. In this case, the blimp/ship will move down first and then will move up again giving an upward acceleration that would give a fluctuation on higher side of g. It may actually oscillate for some time. You may also consider the room fitted on springs on earth's surface. In case of a rocket in space, that should not happen unless it is specially designed to do so, not sure how hard it would be to design it. $\endgroup$ – kpv Dec 30 '17 at 20:34

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