# Do you experience no centrifugal force when in orbit?

I think that the answer is no, but in some sense it feels counter intuitive so I just wanted to check here.

So if I understand it correctly, the General theory of relativity tells us that any sufficiently small object in orbit around Earth (that does not "experience" any forces) essentially follows a straight line through the curved space time, that its experience would be essentially indistinguishable from traveling through deep space in a straight line. Is that the case?

If so, this feels a bit freaky, since from our point of view here on Earth we can clearly see that objects in orbit are turning, yet they experience no lateral force.

• Re, "no centrifugal force." Centrifugal force from what? Do you mean, the "force" due to the orbit itself? In the Newtonian model, that force is exactly equal and opposite to the force of gravity, that otherwise would cause the spacecraft and passengers to fall to the ground. Commented Oct 12, 2020 at 13:34
• I recommend changing ".. any object in orbit..." to "... any sufficiently small object ...". A large enough object will 'feel' the stretching and squeezing due to the inhomogeneous field around a central body. Commented Oct 12, 2020 at 13:35
• My previous mental model (of a stable orbit) was that the object in the orbit has some velocity in the direction of the tangent of the orbit and the gravity is what curves the path. Essentially a rotating ball on a tether. In this model the object experiences centrifugal force (I believe equal and opposite to gravity). But clearly, this model is wrong and that's not what happens in reality.
Commented Oct 12, 2020 at 15:02
• New Veritasium video: Why Gravity is NOT a Force Commented Oct 18, 2020 at 2:26

You are essentially right. In free fall, you are in an inertial frame of reference, there are no forces acting on you. You follow a "straight" path through space-time, though it might be better to say "least curved" path. You can read more about "straight" paths in curved space-time in my answer to If gravity is curvature of space why are more massive objects “heavier?”.

When you stand on the ground, you are accelerated away from this path by the force of the ground pushing up on you.

We often treat the surface of the Earth as an unaccelerated inertial frame of reference, particularly when thinking of horizontal directions. We want $$F = ma$$ to work. But there is the force from the ground pushing upward, and $$a= 0$$. So we add a downward force that acts on everything to the left side of the equation. This is the force of gravity. The force of gravity is a pseudo-force.

There are other pseudo forces, such as centrifugal force and Coriolis force. You can read about them in Pseudo Force and Inertial and Non-Inertial frames and Coriolis Force: Direction Perpendicular to Rotation Axis Visualization.

Gravity is a counter-intuitive one because it isn't obvious that free fall is an inertial frame of reference. After all, it is obviously accelerating toward Earth.

• Your linked answer is very interesting, I am curious about the origin of the force from the ground pushing upward. You say that objects in spacetime follow the geodesic, also that essentially everything in our universe is traveling by the speed 1 second/s into the future which is sort of equivalent to 1 light-second of distance in the spatial dimensions. Is this "constant" time velocity the reason why we have to be accelerated upwards to stay at rest on Earth? That we always have this velocity in the time direction and the geodesic of the spacetime simply leads to the center of Earth?