A particle mass m moves under the attractive force of intensity (ar+b^2)/r^3 per unit mass. How to show that this motion is planar? What does it mean "force of intensity (ar+b^2)/r^3 per unit mass"? The motion is in central field.

Question

anonymous · Answer

torque about origin is zero as it is central motion. So, the direction as well as magnitude of angular momentum will remain constant. Now,
$\overrightarrow{L}=m\overrightarrow{r} \times\overrightarrow{v}$
Now$\overrightarrow{r}$ is perpendicular to $\overrightarrow{L}$ and $\overrightarrow{L}$ is fixed in direction as i said above. This means $\overrightarrow{r}$ is confined to rotate in the plane perpendicular to $\overrightarrow{L}$.

gorica · Answer

Actually, I know how to show that motion in central field is planar. I thought that I have to use a given magnitude of force. Is there a way to show it for this motion under this force?

anonymous · Answer

Well, no matter what is the nature of a central force (attractive, repulsive, inverse squared or whatever), planar motion is its characteristic. Like u can see for this force. It can be written in this vector form.
$\huge{\overrightarrow{F}=-m\frac{(ar+b^{2})\hat{r}}{r^{3}}}$

$\overrightarrow{\tau}=\overrightarrow{r}\times\overrightarrow{F}$
So, you can see that $\overrightarrow{\tau}$ will always be zero and then the rest of the reasoning follows.

anonymous · Answer

Also your question is asking the meaning of "force per unit mass" which is clearly acceleration.

gorica · Answer

I noticed that, in problems, when they say "force of magnitude ______ per mass unit", in Binet's equation there's no m.