If object is at rest relative to the Earth's surface on a frictionless surface. What is the effect of Earth rotation and orbiting on an object? Does object rotate and orbit with Earth and stays at same position due to gravitational force. Or object stays same location and earth rotate underneath it?

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    $\begingroup$ An "object is at rest on a frictionless surface". At rest with respect to what? W.r.t. the surface or something else? Said in another way, the initial zero velocity is with respect to which reference frame? $\endgroup$ Jul 30, 2023 at 13:54
  • $\begingroup$ Hello @GiorgioP-DoomsdayClockIsAt-90 At rest with respect to the surface of the Earth. $\endgroup$
    – 123
    Jul 30, 2023 at 13:56

4 Answers 4


I think what you are actually asking is this:

If the object sits on a frictionless table surface with the Earth first assumed as not rotating and then you start rotating the Earth, would the object slide off or not (otherwise your question would be a contradiction in terms)?

The answer to this would be that it will slide off, because without friction the rotation could not translate to the object so it would stay in its original position because of its inertia.

With an already initially rotating Earth it only stays in the same place because it has the same rotational velocity imparted already from the outset.

Just consider satellites orbiting the Earth for instance. Their orbital position would be unaffected by any changes in the rotation of the Earth.

  • $\begingroup$ Satellites are a great way to visualize this, the only difference being the normal force. $\endgroup$
    – Turksarama
    Jul 31, 2023 at 2:58

If object is at rest on a frictionless surface. What is the effect of Earth rotation and orbiting on an object?

The effect of the rotation is that the surface of the Earth is a non-inertial reference frame. In this frame there is a centrifugal force and a Coriolis force.

The Coriolis force is proportional to the velocity in the rotating reference frame. Since that velocity is given to be 0, the Coriolis force is also zero.

The centrifugal force points away from the axis of rotation. Note that is not the same direction as the center of the earth.

This produces an effective potential which is a combination of the gravitational potential and the centrifugal potential. The equipotential surfaces are oblate ellipsoids rather than spheres.

If the frictionless surface is level, then it follows this oblate ellipsoid and there is no force to accelerate the object horizontally. However, if the frictionless surface does not follow this ellipsoid then the surface is not level and the object will slide downhill.

Or object stays same location and earth rotate underneath it?

This is contradicted by the given statement that the object is initially at rest on the frictionless surface.

  • $\begingroup$ So let's say if we drop an object at an instantaneous time interval dt then earth almost has not rotated, so now assuming the object is fully perpendicular to surface and not inducing any component of force sideways , does the object move or not ? $\endgroup$
    – Razz
    Jul 30, 2023 at 15:35
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    $\begingroup$ The initial velocity is defined at the initial time, not over an interval whether finite or infinitesimal. $\endgroup$
    – Dale
    Jul 30, 2023 at 15:49

I assume that the object is already on a perfectly spherical Earth with perfectly spherically symmetrical mass distribution, rotating with it, and suddenly the Earth's surface becomes frictionless.

Unless the object is on the equator it will start to drift to the equator because its inertia prevents it from following a latitude that is not the great circle known as the equator. Following any other latitude would require some horizontal acceleration which cannot be performed since the perfectly frictionless surface only permits the transfer of vertical forces.

The object would follow some curve towards the equator, cross it and then, symmetrically, enter the opposite hemisphere and curve back to the equator, oscillating forever — since there is no friction, it's physicist's paradise! — between the two hemispheres.

An observer resting relative to the surface of the Earth would additionally see the object drift, with increasing speed, to the West on its way to the equator. For this Earth-bound observer the reason is the Coriolis force; a non-rotating observer, by contrast, would see that the object's speed is smaller than the speed of the rotating Earth's surface closer to the equator, making the object fall behind Earth's rotation as it approaches the equator.

To this observer, the object's path would lie on the the great circle which crosses its "starting point" in the direction of motion it had when the surface became frictionless, while the Earth is rotating along under it, without any interaction resulting from this rotation.

  • $\begingroup$ So if you put an object on a flat frictionless finite plate, it would move to the edge at the equator. Would it also start to rotate? $\endgroup$
    – Orbit
    Jul 31, 2023 at 13:58
  • $\begingroup$ @Orbit I'm not sure whether you are aiming at a superposition question (the Earth is rotating, and the plate is rotating on the rotating Earth) or whether you have a gedankenexperiment of a rotating plate oriented so that its rotational axis aligns with a homogeneous gravitational field. As with the OP's original Earth question, you should also make unambiguous whether the object is rotating with the plate or, instead, is resting and put on a plate that is rotating under it. Whatever your scenario is: The object will follow its path unless net forces act on it. $\endgroup$ Jul 31, 2023 at 14:08
  • $\begingroup$ No additional layers, just my interpretation of the original question. Theoretical assumptions: Frictionless, perfectly flat and horizontal exist. Someone puts an object on this surface and releases it, what happens. I completely agree with your answer, but I think it will also start to rotate. If a plug is pulled from a container with water, the water will rotate, and the hemisphere it is on determines the direction. I am guessing the object will do the same. $\endgroup$
    – Orbit
    Jul 31, 2023 at 14:17
  • $\begingroup$ @Orbit The forming of vortices in fluids (notably the atmosphere) is not directly caused by the Earth's rotation. (The direction of large-scale vortices like cyclones is, as we know, dependent on which hemisphere you are on, but not the formation as such.) Even on the equator you'll have vortices in emptying bathtubs, even though there is no coriolis force, and you can easily create clockwise vortices on the Northern hemisphere. $\endgroup$ Jul 31, 2023 at 14:26
  • $\begingroup$ @Orbit So if you have a frictionless surface it will not transfer any lateral force on the object, period. If the object sits still to a non-rotating observer on the frictionless plate it will continue to sit still. If it is moving, it will continue moving in a straight line (the two statements are equivalent anyway because all non-accelerated observers are created equal). $\endgroup$ Jul 31, 2023 at 14:29

The object rotates with the Earth without slipping/sliding.

Since the Earth's angular velocity is constant (angular acceleration is zero) static friction is not needed for the object to rotate together with the Earth. Static friction is only required to oppose a force and prevent relative motion (sliding). In this case since there is no angular acceleration of the Earth, there is no net torque applied to the Earth, and thus no horizontal force applied to the object by the Earth for friction to oppose and prevent sliding.

Hope this helps.

  • $\begingroup$ Unless one is at the equator or one of the poles, staying in place relative to the Earth's surface does require a small but finite, constant lateral acceleration. If you imagine the Earth rotating very fast, being 10 feet away from the pole is like being on a merry-go-round, which clearly accelerates its occupants to force them on a circular trajectory. Consequently, if the Earth suddenly became frictionless, we U.S. Americans and Europeans would all drift to and across the equator as described in my answer. $\endgroup$ Jul 31, 2023 at 15:06
  • $\begingroup$ @Peter-ReinstateMonica Point well taken. But for the most part we can ignore it. Not sure its worth a downvote though (if you did downvote it). $\endgroup$
    – Bob D
    Jul 31, 2023 at 15:11
  • $\begingroup$ Yes, my downvote: Because what you ignore is in my opinion at the heart of the question ant completely changes the answer. You are suggesting (if I read you correctly) that the object, once surface friction ceases, keeps its angular velocity and stays more or less close to that surface point. Both assertions are entirely wrong. $\endgroup$ Jul 31, 2023 at 15:40

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