# Centrifugal force affecting satellite

I'm trying to explain the behaviour of a geostationary satellite using different frames of reference.

1. Inertial frame: The satellite has a circular motion with angular velocity $\omega$. The centripetal force $F$ required for this motion is created by the gravitational pull of Earth. Earth itself rotates around its axis with $\omega$, but that is irrelevant. OK

2. Rotating frame ($\omega$): The frame of reference is fixed to Earth. Everything appears stationary. Gravity is still present, which still acts on the satellite with force $F$. Due to the acceleration of our frame of reference we introduce a centrifugal force, which acts on the satellite with $-F$. The forces cancel out, so the satellite's lack of acceleration is explained. OK

3. Rotating frame ($2\omega$): This frame of reference rotates around Earth's axis with angular velocity $2\omega$. The satellite appears to have angular velocity $-\omega$. The centripetal force $F$ is provided by gravity. However, we have not yet accounted for the acceleration of our frame of reference! There should be a centrifugal force of $-2F$, meaning that the satellite should be accelerating away from Earth!
Not OK

How do we explain case 3?

## 1 Answer

The key is the coriolis force.

The coriolis force is $F_c = -2m\Omega \times v$. Here $\Omega$ is the rotation of the frame of reference and $v$ is the linear speed of the satellite.

If you do the calculations, left as an exercies for the reader, you'll get the missing force.

In case 2 the coriolis force is 0, because the velocity $v$ has to be used in the local frame of reference. And there $v = 0$.

• Note - Coriolis force is a "fictional" force that appears on moving objects in rotating frames of reference - which is why you didn't need it for either case 1 or 2 (1 = stationary frame, and 2 = stationary satellite). – Floris Aug 22 '14 at 14:13
• @Floris: Like centrifugal force. But note that in case 2 (I've just added a note) the frame of reference is rotating; there is no coriolis force because $v=0$, – rodrigo Aug 22 '14 at 14:16
• It might be "cleaner" to state that the Coriolis force is always there - but since it involves the product of $v$ and $\Omega$ it will be zero if either of them is zero. This is why you can get the right answer for case 1 and 2 without knowing about it. But on the whole, good answer - this is the explanation. – Floris Aug 22 '14 at 14:50