3
$\begingroup$

If two people are in a box (no windows) One is on the earth at 1G gravity and the other one is Accelerating in space with only the acceleration (inertial equal to 1G gravity) is there a way for the man inside the box to tell if he's on the earth or accelerating in space

$\endgroup$
  • $\begingroup$ Why do you ask? $\endgroup$ – Russell Borogove Aug 10 '17 at 23:50
4
$\begingroup$

@OrganicMarble's answer is correct. An absolutely uniform gravity field would require infinite mass at infinite distance and so realistic gravity fields will usually have local gradients that can be measured with sufficiently sensitive equipment. There might be some pathological situations where higher terms also cancel and gravity appears extremely uniform, but that's best left as another question.


Although it is not named explicitly, the question refers to the Equivalence Principle:

In the theory of general relativity, the equivalence principle is any of several related concepts dealing with the equivalence of gravitational (mass) and inertial mass, and to Albert Einstein's observation that the gravitational "force" as experienced locally while standing on a massive body (such as the Earth) is the same as the [pseudo-force][] experienced by an observer in a non-inertial (accelerated) frame of reference.

The problem here is that the Earth is rotating, and so the Earth-box is not in an inertial frame to begin with.

One way this would be noticeable would be to look for non-inertial effects such as the precession of a Foucault pendulum. However, at the equator that precession goes to zero.

Another way to notice it would be in the motion of a simple gyroscope. For example, at the equator, a gyroscope initially spinning with a vertical axis would rotate in a vertical plane once a day. If it pointed "up" at noon, it would point "west" about six hours later, and "down" another six hours after that. Of course if you were not on the equator, or the initial direction were not "up", the motion would be more complex, but still indicate clearly that you were not in an inertial frame, except in the particular situation that you were at the North Pole and the initial direction were "up", or that it were at some orientation that coincided with the Earth's axis no matter where you were.

So to be absolutely sure, you probably should use the gyroscope with several initial directions, or combine the gyroscope with the Foucault pendulum.

enter image description here

above: Foucault Pendulum animation, from here.

enter image description here

above: Gyroscope animation, from here.

| improve this answer | |
$\endgroup$
  • $\begingroup$ But neither the Gyroscope nor the Foucault Pendulm could tell us a difference between the slowly rotating Earth and a rocket in space rotating with the same speed. $\endgroup$ – Uwe Aug 11 '17 at 17:08
  • $\begingroup$ @Uwe that is absolutely correct! The question is short, if it had a period at the end it would be one sentence. Since it talks about acceleration and not rotation, I've just pointed out that there is rotation on Earth weather you like it or not. But if the question did include rotation of the box in space, somehow simulating the rotation of the Earth, then maybe the two might be indistinguishable. But that sounds like a new question, and it sounds like you've got an answer already! $\endgroup$ – uhoh Aug 11 '17 at 17:28
  • 1
    $\begingroup$ To find out if you are in a_rotating_ rocket, you could try to change the rotation via a reaction wheel. If it works, you're in a rocket. If not, you're on earth, or the rocket has countermeasures. $\endgroup$ – sweber Aug 15 '17 at 4:43
  • $\begingroup$ @sweber i.stack.imgur.com/tjLD8.jpg I think measuring the mass or moment of inertia of the box is not intended to be part of the question, but it's a good idea! $\endgroup$ – uhoh Aug 15 '17 at 6:54
3
$\begingroup$

With sufficiently precise instruments* you can tell which box you are in because the Earth-based box gravity vectors will point to the center of the earth as you move about in the box (box on the left) whereas the space box will have parallel acceleration vectors (box on the right)

* probably not achievable in practice for a reasonably sized box but this is a thought experiment after all.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Don't forget that there is also acceleration due to Earth's rotation. The Foucault Pendulum would be a good experiment to do in a box as well. $\endgroup$ – uhoh Aug 11 '17 at 3:43
  • $\begingroup$ @uhoh The Foucault Pendulum should be very long to show the effect. Within a small box, the influence of the Earth's rotation is too small. The first pendulum made by Foucault was 67 m long. A vaccuum chamber may help to reduce the length of the pendulum, but I don't believe a length of only 2 m is possible. The pendulum should swing for about half an hour on its own to show a rotation. $\endgroup$ – Uwe Aug 11 '17 at 10:28
  • $\begingroup$ @Uwe Unfortunately there's no such thing as a Foucault torsional pendulum, they have to point straight down. However, unless you can prove this quantitatively, you are just guessing without basis in engineering. A heavy mass on a quartz fiber in low pressure can probably do this. And as OrganicMarble points out above, this is a Gedankenexperiment. $\endgroup$ – uhoh Aug 11 '17 at 10:36
  • 1
    $\begingroup$ Re probably not achievable in practice for a reasonably sized box but this is a thought experiment after all -- That's not true. This is exactly how GOCE worked. It used three pairs of highly sensitive 1D accelerometers to measure Earth's gravity gradient tensor. $\endgroup$ – David Hammen Aug 11 '17 at 11:25
  • $\begingroup$ The shortest Foucalt pendulums in this list en.wikipedia.org/wiki/List_of_Foucault_pendulums#cite_note-2 is 6.5 and 6 m long. But even with a pendulum that fits in the box, you don't know if the box is on earth or on a slowly rotating rocket. $\endgroup$ – Uwe Aug 11 '17 at 11:41
2
$\begingroup$

Both gravity fields are not totally equal. In the box on Earth, gravity at top is a little smaller than at the bottom. But in the other box gravity is the same at bottom and top.
There are ultra precise atomic clocks that show a very small difference in frequency when mounted at different heights and there are gravimeters with very high sensitivity. The relative time difference is -1,09 · 10-16 per meter of height.
On Earth the gravity decreases with every meter of increased height by 3.1 µm/s² (https://de.wikipedia.org/wiki/Schwerefeld and https://en.wikipedia.org/wiki/Gravity_of_Earth#Free_air_correction). The best superconducting gravimeters have a resoultion of 0.1 nm/s² (https://www.bkg.bund.de/DE/Observatorium-Wettzell/Messsysteme-Wettzell/Lokale-Messsysteme/Gravimeter/gravimeter_cont.html). Such gravimeters are also able to measure the influence of tidal effects of Moon and Sun to the Earth, but only in the box on earth tidal effects will be measureable.
A modern seismometer will show if there are the typical seismic waves of the Earth or completely different vibrations of a rocket in space.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.