# Angular acceleration of orbit

I have been spending some time examining the relationship between linear and angular quantities, especially in relation to orbits:

Consider a normal two body system (like earth-sun) with masses m and M (M>>m), separated by a distance R.

We know that $$v = R\dot{\theta}$$ (1)

Consequently, $$a=R\ddot{\theta}$$ (2)

Therefore, $$\frac{v}{\dot{\theta}}=\frac{a}{\ddot{\theta}}$$ (3)

Using the centripetal force, we can easily find that the linear acceleration is $$a = v^2/R$$. Substituting this into (3) for $$a$$, we get

$$\frac{v}{\dot{\theta}}=\frac{v^2}{R\ddot{\theta}}$$ (4)

Solving for $$\ddot{\theta}$$, we get $$\ddot{\theta}=\frac{v\dot{\theta}}{R}$$. (5)

Now, we know that $$v = R\dot{\theta}$$ from (1). Substituting this into (5) for $$v$$,

we get $$\ddot{\theta}=\frac{R\dot{\theta}^2}{R}$$, leaving us with

$$\ddot{\theta}=\dot{\theta}^2$$.

Is this correct? I thought the angular acceleration had to be 0 since any body in uniform circular motion (like earth around sun) moves at a constant angular speed. My head is spinning, I have no idea what's happened here. It also seems weird that $$\ddot{\theta}=\dot{\theta}^2$$. It just seems wrong...

• Your first two equations refer to tangential motion. The $v^2$/r is for centripetal. Nov 15, 2021 at 15:52

Starting from first principles, we can write the position vector of an object as

$$\vec r = R \hat r$$

where $$\hat r$$ is a unit radial vector. So the velocity vector is

$$\displaystyle \vec v = \frac {d \vec r}{dt} = \frac {dR}{dt} \hat r + R \frac {d \hat r}{dt}$$

If the object is moving in a circle about the origin then $$\frac{dR}{dt}=0$$ so

$$\displaystyle \vec v = R \frac {d \hat r}{dt}$$

If the angle between the object's position vector and the x-axix is $$\theta(t)$$ then in cartesian co-ordinates we have $$\hat r = (\cos \theta, \sin \theta)$$, so

$$\displaystyle \frac {d \hat r}{dt} = (-\dot \theta \sin \theta, \dot \theta \cos \theta) = \dot \theta \hat \omega$$

and

$$\vec v = R \dot \theta \hat \omega$$

where $$\hat \omega = (-\sin \theta, \cos \theta)$$ is a unit tangential vector. The acceleration of the object is

$$\displaystyle \vec a = R \ddot \theta \hat \omega + R \dot \theta \frac {d \hat \omega}{dt}$$

If the object is undergoing uniform circular motion then $$\theta = kt$$ for some constant $$k$$ so $$\ddot \theta=0$$. However, we can see that $$|\vec a|$$ is not zero because $$\frac {d \hat \omega}{dt} \ne 0$$. In fact

$$\displaystyle \frac {d \hat \omega}{dt} = \dot \theta (- \cos \theta, - \sin \theta) = -\dot \theta \hat r \\ \displaystyle \Rightarrow \vec a = R \dot \theta \frac {d \hat \omega}{dt} = - R (\dot \theta)^2 \hat r = - \frac {|\vec v|^2}{R} \hat r$$

which gives us the magnitude and the direction of the acceleration vector for an object in uniform circular motion.

• Thanks, this really cleared things up! Nov 15, 2021 at 14:27