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I just want to be sure I understand correctly.

They ask us to find the speed at which an object moves weightlessly in a tunnel around the center of the Earth. Assume Earth is homogeneous, and the part of Earth beyond the tunnel does not contribute to gravity.

I suppose the gravity and centrifugal force need to cancel each other. But I was wondering if we also need to take the Coriolis force (or any other force by the way) into account?

If the object moves weightlessly in the tunnel, is it as if the object was moving in orbit around Earth? Because it doesn't touch the walls of the tunnel, shouldn't the Coriolis force apply?

The Coriolis force only applies to an object that was in contact with Earth, and then not any more. But in this case, if the object doesn't touch the wall, this doesn't apply, do I understand this correctly?

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    $\begingroup$ You write: "The coriolis force only applies to an object that was in contact with Earth, and then not anymore." It would appear that you have been led to believe that coriolis effect is applicable only in ballistics. However, in meteorology the fact that the Earth is rotating is taken into account, referred to as 'coriolis effect'. Air mass is buoyant; motion of air mass is not ballistic motion. To see that air mass is buoyant: think about an airship (zeppelin). An airship maintains altitude because it has neutral buoyancy. So: you first need to sort out how you understand 'coriolis'. $\endgroup$
    – Cleonis
    Commented Oct 9, 2022 at 12:04
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    $\begingroup$ The setup that is described is odd. The setup has an object circumnavigating a gravitating mass such that the circumnavigating object is weightless. There is only one form of circumnavigating motion with that property: orbital motion. Having that orbital motion occur in an evacuated tunnel makes the setup more elaborate, but it will not change the motion. Take the International Space Station, its orbit around the Earth has a period of 90 minutes. The orbit of the ISS is not affected by the Earth's rotation underneath that orbit: there's no reason to attribute coriolis effect to ISS orbit. $\endgroup$
    – Cleonis
    Commented Oct 9, 2022 at 12:14
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    $\begingroup$ Continuing: if you would use a coordinate system that is co-rotating with the Earth to describe the motion of the ISS, then in terms of that rotating coordinate system you would attribute a coriolis force. But that is pointless; the orbit of the ISS is not affected by the Earth's rotation underneath it, so any coriolis force that is attributed does not tell you anything usefull about the ISS orbit. $\endgroup$
    – Cleonis
    Commented Oct 9, 2022 at 12:19
  • $\begingroup$ @Cleonis Someone else also commented a few minutes after the question was posted, I couldn’t see his name unfortunately. The setup they give is : six subways lines move in concentric circles around the center of Earth, each in its own tunnel with a radius from the Earth Center of 1000, 2000, ... 6000 km And as said they ask at which speed they move weightlessly in their tunnels $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 12:31
  • $\begingroup$ @Cleonis So if I understand correctly, this is just a case of orbital motion ? It’s in fact possible I understood the Coriolis Force differently $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 12:34

3 Answers 3

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But I was wondering if we also need to take the Coriolis Force (or any other force by the way) into account ?

If the object moves weightlessly in the tunnel, it's as if the object was moving in orbit around Earth ? Because it doesn't touch the walls of the tunnel, the Coriolis Force shoudn't apply ?

The Coriolis force is a fictitious force, meaning that it appears whenever we are doing our analysis of physics using a rotating reference frame. In that rotating reference frame it applies for all objects, regardless of whether they are in contact with the Earth or not.

This force is not due to contact with the Earth, it is due to the fact that we are using a non-inertial coordinate system to do our analysis. So everything in that analysis is affected. As a hint for your specific problem, it may be that you can build your tunnels in a specific location which will eliminate or simplify the Coriolis force.

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  • $\begingroup$ If the tunnel is around Earth Center and parallel to Equator, there shouldn’t be any Coriolis Force, is this the case ? So the tunnels are built in concentric circles around Center of Earth, and they are parallel to Equator, so the rotation speed will be the same everywhere in the tunnel because they all have the same distance to Center of Earth, and so no Coriolis Force happens. So the tunnels around the center are not north - south but west - east, did I see this correctly ? Thank you so much for your support $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 13:44
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    $\begingroup$ An east-west tunnel at the equator doesn’t eliminate the Coriolis force, but it does simplify it. The force becomes vertical so it adds simply with gravity and the centrifugal force $\endgroup$
    – Dale
    Commented Oct 9, 2022 at 14:10
  • $\begingroup$ Ah yes of course, it starts to make sense. It would point either toward Earth Center or toward outer space (through the remaining Earth Crust of course). You say we would need an inertial coordinate system to avoid it entirely, would this even be possible in such a situation ? If we view it from a fixed point in Outer Space for example, would it be possible to avoid it because the reference point isn’t part of Earth and its rotation ? $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 16:19
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    $\begingroup$ @wengen yes, you could avoid any Coriolis force by doing the analysis in an inertial frame, as you say. Then you would have to account for the movement of the Earth $\endgroup$
    – Dale
    Commented Oct 9, 2022 at 19:57
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If your tunnel goes south to north, you can not avoid to bump to the wall, since at the equator your initial velocity perpendicular to the tunnel is that of the earth rotating. Farther north, the velocity of the tunnel is slower, so you will bump into it, and this you could attribute to a Coriolis force in the rotating system.

If your tunnel goes around the equator you can have this weightless path.

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    $\begingroup$ If the tunnels go around the Equator, so have a west east path, the Coriolis Force would point toward the core of Earth or toward outer space, it would partially add or cancel Gravity and Centrifugal Force, isn’t it ? $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 16:52
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Coriolis force with a horizontal component occurs if (i) you are working in a frame of reference that is rotating with the Earth's surface and (ii) the object's motion has any north-south component.

Since the tunnel is attached to the Earth's surface, condition (i) applies. Unless the tunnel is constructed underneath the equator then condition (ii) applies.

This is why the groundtrack of a satellite such as the ISS does not close after one orbit. In space it returns to the same position after one orbit, but its groundtrack (projected onto the surface of the Earth) is not closed because the Earth has rotated underneath it. In the ground reference frame, the satellite experiences a sideways force which is the Coriolis force. Your tunnel scenario is exactly the same except at a negative altitude.

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  • $\begingroup$ If the tunnels are built around the Equator, the Coriolis force would be parallel to Gravity and Centrifugal Force, but it wouldn’t disappear ? The only way to avoid Coriolis is to watch the tunnels from outside the Earth, is that correct ? We would need to watch from a point in space. The Groundtrack example means that while the object in the tunnel may make a full rotation, it wouldn’t be in the same place as before because the tunnel would have rotated further with the Earth, do I see this correctly ? $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 16:37
  • $\begingroup$ If the tunnels are built below Equator, the force would just point toward Earth Center or toward outer space, isn’t it ? It wouldn’t disappear ? $\endgroup$
    – wengen
    Commented Oct 9, 2022 at 16:41
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    $\begingroup$ @wengen Yes, you are correct. Only the horizontal component disappears - but it is this component that you need to worry about in the tunnel scenario. $\endgroup$
    – gandalf61
    Commented Oct 9, 2022 at 17:27

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