Fictitious forces arise because of the acceleration of the frame of reference itself. No real forces act on the object, still the object accelerates. So why it is said that if the earth were to rotate with more angular velocity we would fly off to the sky due to centrifugal force? Why I feel centrifugal force towards the sky if it just a fictitious force? Who is pushing me towards the sky?

  • $\begingroup$ Related: physics.stackexchange.com/q/109500/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Feb 15, 2016 at 8:22
  • 1
    $\begingroup$ Obligatory xkcd.com/123 $\endgroup$
    – Qmechanic
    Commented Feb 15, 2016 at 8:23
  • $\begingroup$ Nothing is pushing you towards the sky, the Earth's surface is constantly accelerating away from you... $\endgroup$ Commented Feb 15, 2016 at 12:18

2 Answers 2


Let's start with an easy Gendankenexperiment:

Imagine there was no gravity and you would be hovering a few feet above the Earth's surface and at a constant distance from the Earth's center of mass. What would you see? You would see the Earth rotating below you at hundreds of meters per second, right? After 24 hours the same spot would swoosh by underneath you, again.

In practice, of course, it would not be advisable to do this because you would have to watch out for those lamp posts and bridges and homes and mountain ranges that are sticking out from the surface coming at you at that speed, so you would have to make sure you are high enough to avoid all these obstacles!

Now let's make this a little more interesting:

Imagine you would be moving tangential to Earth with the same velocity as the surface below you to cancel out that fast rotation to first order. You are moving on a straight line (just like Newton ordered) while the point that was below you just a few moments ago, is moving on a circle (in geographic terms the parallel of your location). What are you going to see now? You are going to see that point falling away below you, first slowly and then ever more rapidly! A person observing you from that point will see you magically rise and fly ever further away, eventually disappearing behind the horizon only to be seen again some 12 hours later, flying off into space at the tangential velocity, i.e. you would be many thousands of miles above the surface by the time they see you, again. What did that? We call "that" a fictitious force.

We can, of course, not do this. We are being pulled down by gravity and then we stick to the surface due to friction, both of which are real forces. These are the forces that keep us on the surface and in the same place on the surface. Now we are not moving on straight lines but on circles, ourselves. This means, also by Newton's laws, that the total forces acting on us are the force of gravity and the force of the surface friction minus the inertial force that needs to be overcome to make us stick to our circular movement. If we could speed up the Earth's rotation, the inertial forces would become ever larger, eventually nulling out the gravitational force at the equator. The planet would bulge out and the atmosphere would stream into space and we, the lone surviver in our space suit could hop ever higher... until we reach escape velocity... at which moment we would seize to be bound and eventually we would fly off. So that's what one can feel (getting "lighter"), although it's not a particularly large correction to gravity on this planet.


An observer is in a laboratory standing on the floor of the laboratory which also has a horizontal turntable with a frictionless surface resting on the floor.
An object of mass $m$ is constrained to move in a horizontal circle on the turntable by a spring balance of length $r$ attached to the object and the centre of the turntable.
There is no relative motion between the object and the turntable.

The observer measures the radius $r$ and the constant angular speed $\omega$ of the object relative to the floor of the laboratory.

The observer standing on the laboratory floor would reason thus:
The mass is undergoing a centripetal acceleration of magnitude $r \omega^2$ and the force to do that is provided by the spring balance.
The reading on the spring balance is $S$ and indicates a force on the object towards the centre of the turntable.

Applying Newton’s second law $S = mr\omega^2$.

The observer now stands on the turntable and makes all measurements relative to the turntable.
The observer sees that the object is not moving and the spring balance is showing a reading of $S$ which indicates a force on the object towards the centre of the turntable.

The observer on the turntable reasons as follows.
No acceleration means that there is no net force on the object so there must also be a force of magnitude $S$ acting outwards on the object.
This is the centrifugal force.

The observer measures $r$ and the angular speed of laboratory relative to the turntable $\omega$ which turn out to be the same as the measurements made relative to the ground.
The observer on the turntable assigns a magnitude of $m r \omega^2$ to the force $S$.

So $S - m r \omega^2$ = 0$

In all that is written below the force that you feel is the force on you due to the spring balance.
It does not matter which frame of reference you use the force on you due to the spring balance will always be the same.

You are on the Equator of a spherical spinning Earth with radius $r$.
A spring balance is connected between you and the Earth.

Observing you from space the reading on the spring balance is $- S$, the negative sign indicating that the force on you due to the spring balance is away from the centre of the turntable.
The force $-S$ is what is usually called the normal reaction on you due to the ground.

Applying Newton’s second law gives $\frac {GM_EM}{r^2} +(- S) = m r \omega^2$.

If the observer is standing on the Earth next to you then the gravitational attractive force is the same as before as is the reading on the spring balance but now you are seen to be stationary relative to the Earth.

The same reasoning as that used for the rotating turntable will give

$\frac {GM_EM}{r^2} +(- S) +(- m r \omega^2) = 0$.

If the Earth was now made to rotate faster then $\omega$ increases.
The result is that the reading on the spring balance and hence the force exerted by the spring balance on you decreases.

Increasing the speed further there will come a time when the reading on the spring balance is zero.
You are now in a geostationary orbit about the Earth and feel weightless.
The spring balnce is not exerting a force on you.

The only force acting on you is the gravitational attraction and that produces the centripetal acceleration or the gravitational force is equal in magnitude but opposite in direction to the centrifugal force.

Now make the Earth rotate even faster.
If you are to have no motion relative to the Earth then the spring balance must now exert a force on you towards the centre of the Earth.
This is equivalent to saying that you must cling on to the Earth thus applying an inward force otherwise you will continue orbiting the Earth but at an angular speed which is lower than that of the Earth.

If you did cling on and attained the rotational speed of the Earth and then released your grip, you would then start moving away from the Earth in an elliptical or hyperbolic trajectory.

In practice things are much more complicated, for example one would have to consider the fact that the matter which makes up the Earth is under compression and is fairly weak when subjected to tensile stresses.


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