# Coriolis effect causes in motion of a free falling object

• I
I don't understand what are the causes of the Coriolis effect for objects moving with respect to Earth. For istance consider an object free falling on the earth from an height $h$. Its tangential velocity its greater than the velocity of an object on the surface of Earth, hence it moves eastward (in the direction of rotation) while falling. But does it end up here? Is a kinematic consideration sufficient to explain completely the Coriolis effect in this case? Or there is something else, like conservation of angular momentum? If so how it is influent in this case?

sophiecentaur
Gold Member
2020 Award
For an object in free fall, the path taken will be along a part of an ellipse, as in any orbit, until it hits the ground. The Coriolis effect would not be 'in addition' to this but could be regarded as a part of it.

Soren4
andrewkirk
Homework Helper
Gold Member
Is a kinematic consideration sufficient to explain completely the Coriolis effect in this case?
Yes, the explanation you have given is sufficient to explain the effect. Angular Momentum does not add anything to the analysis. The analysis could be expressed differently, using angular momentum instead of Newton's equations, but that would be an alternative explanation, not a necessary additional explanation.

Soren4
For an object in free fall, the path taken will be along a part of an ellipse, as in any orbit, until it hits the ground. The Coriolis effect would not be 'in addition' to this but could be regarded as a part of it.
Yes, the explanation you have given is sufficient to explain the effect. Angular Momentum does not add anything to the analysis. The analysis could be expressed differently, using angular momentum instead of Newton's equations, but that would be an alternative explanation, not a necessary additional explanation.
Thanks for your answers! Would anyone be so kind as to give some further explanations about how angular momentum affects this phenomenon (deflection of free falling object as seen from Earth)? I get confused between the kinematic explanation and the conservation of angular momentum one, maybe this would help to make things clearer. Thanks in advice for your help.

sophiecentaur
Gold Member
2020 Award
There is no "deflection'. What happens is that the place the object arrives is not where you'd expect, because the Earth is spinning underneath. The Coriolis "force" is not a real force. it is just a fiction that's used for convenience. The path of the falling object in the inertial frame would be the same with or without the Earth's rotation.

Soren4
andrewkirk
Homework Helper
Gold Member
Would anyone be so kind as to give some further explanations about how angular momentum affects this phenomenon (deflection of free falling object as seen from Earth)? I get confused between the kinematic explanation and the conservation of angular momentum one, maybe this would help to make things clearer.
The angular momentum of the object is ##I\omega## where ##I=mr^2## is its moment of inertia about the Earth's rotation axis, ##r## is the object's distance from the rotation axis and ##\omega## is its angular velocity of rotation around the world. Conservation of angular momentum says that this amount, which is ##mr^2\omega##, must remain constant, in the absence of an applied torque. As the object falls, ##r## reduces, so ##\omega## has to increase to compensate, which causes the angular acceleration of the object Eastwards. It's a bit like when a spinning skater has their arms out and then brings them in close, thereby increasing their rate of spin.

Soren4
The angular momentum of the object is ##I\omega## where ##I=mr^2## is its moment of inertia about the Earth's rotation axis, ##r## is the object's distance from the rotation axis and ##\omega## is its angular velocity of rotation around the world. Conservation of angular momentum says that this amount, which is ##mr^2\omega##, must remain constant, in the absence of an applied torque. As the object falls, ##r## reduces, so ##\omega## has to increase to compensate, which causes the angular acceleration of the object Eastwards. It's a bit like when a spinning skater has their arms out and then brings them in close, thereby increasing their rate of spin.

Thanks so much! It's here my confusion: before, just looking at kinematics considerations we said that the velocity of the object is greater than the one of a point on the Earth, and that's why it seems to be deflected, if watched from the Earth. The tangential velocity of the object stays constant while falling, if observed in a steady inertial reference.

Angular momentum is conserved "with respect to" a steady inertial frame, right? Because in Earth system the point at height ##h##, before falling has zero angular momentum, so its angular momentum does not keep constant in Earth system. Nevertheless, as you showed, since ##r## gets smaller the velocity of the object must increase.

Isn't this in contrast with what I said before (i.e. the tangential velocity of the object stays constant while falling)? In other words, in a steady inertial reference frame, is the (tangential) velocity of the object increased when it almost reached the ground, or not?

andrewkirk
Homework Helper
Gold Member
The tangential velocity of the object stays constant while falling, if observed in a steady inertial reference.
That's not quite right. The tangential velocity is ##\omega r##. Since the angular momentum of ##m\omega r^2## is constant, that means that the tangential velocity of the falling object is proportional to ##1/r##. So the tangential velocity also increases as the object falls, but not as fast as the angular velocity increases.

So we conclude that the answer to this question
In other words, in a steady inertial reference frame, is the (tangential) velocity of the object increased when it almost reached the ground, or not?
is Yes.

Soren4