# Orbital motion

1. Mar 5, 2006

### andrevdh

I'm a bit unclear about the description of orbital motion in a plane by using the polar coordinates $(r,\theta)$. This coordinate system changes its orientation in the inertial reference frame, that it is rotating as the orbiting object moves along its path. In the derivation of the equations of motion the radial part comes to
$$F(r)=m(\ddot r\ -\ r\dot \theta^2)$$
My problem is that in a rotating reference system it is normally necessary to introduce a centrifugal force $F_C$, which sorts of explains the second term in the equation above since the centrifugal force is given as
$$F_C=mr\dot \theta^2$$
Why is the term then negative in the top equation?

2. Mar 5, 2006

### Hootenanny

Staff Emeritus
Doesn't the centrifugal force act in the opposite direction to the centripital force?

3. Mar 5, 2006

### phucnv87

Centrifugal force is in an opposite direction to the centripental force $\vec{F_C}=-\vec{F_r}$

4. Mar 6, 2006

### andrevdh

That is exactly my problem since for orbital motion the gravitational attractive force is in the negative $\vec r$ direction amounting to
$$F(r)=-G\frac{Mm}{r^2}$$
which is a bit confusing, unless the term in the equation of motion should not be interpreted as arising from the centrifugal force.

5. Mar 6, 2006

### vaishakh

The negative force is to keep the body in equilibrium in the rotating frame. The centrifugal force exerted away from the centre by the body is taken as positive direction. Unless an internal attraction the body cannot continue uniform orbital motion.

6. Mar 6, 2006

### andrevdh

Here is how I think the equation might be interpreted
$$-F_G=m\ddot r\ -\ F_C$$
thus
$$m\ddot r\ =\ F_C\ -\ F_G$$
the term $m\ddot r$ is the acceleration an observer in the rotating frame of reference observes. He explains it as a result of two forces the centrifugal $F_C$ and the attractive gravitational $F_G$?