Path of a free particle over a sphere

[gordunox]

Homework Statement

Consider a particle of mass m whose motion is restricted to occur on the surface of a sphere of radius R. There are no other forces acting on the particle. Demonstrate that the motion occurs along a circle on the sphere.

Relevant Equations

$$\phi=-\frac{c_{2}}{c_{1}}\cot\theta+c_{4}$$

 

[haruspex]

Try working from the other end, find the relationship between $\theta$ and $\phi$ on a great circle.
E.g. let $\vec P$ be a fixed point on a unit sphere and $\vec Q$ be $\pi/2$ away from it. Write each in Cartesian form but using spherical polar coordinates. The dot product is zero. This gives a relationship between the latitude and longitude of Q, the locus being a great circle.
(But the mix of angle and trig in your equation looks very unlikely to me. Maybe do a sanity check on special cases first; $\theta=0$, for example.)

 

[vela]

Check your equation for $\dot{p}_\theta$.

 

[Frabjous]

Couldn’t a free particle move along a line of latitude?

 

[haruspex]

Couldn’t a free particle move along a line of latitude?

Only at the equator. Elsewhere it would require a force with a component tangential to the surface.

 

[Frabjous]

Only at the equator. Elsewhere it would require a force with a component tangential to the surface.

How does a particle with an initial velocity not pointed in a great circle direction move then. It cannot be on a great circle.

 

[gordunox]

Check your equation for $\dot{p}_\theta$.

You are right, that's the mistake. I will change the equation and see what I get.

 

[vela]

How does a particle with an initial velocity not pointed in a great circle direction move then. It cannot be on a great circle.

How can the velocity not be in a direction of a great circle?

 

[Frabjous]

How can the velocity not be in a direction of a great circle?

You’re correct. I had the wrong picture in my head.

 

[Orodruin]

I mean, the easy way of solving this is to arrange your coordinates such that the particle is at $\theta = \pi/2$ and $\dot\theta(0) = 0$. This is always possible and the solution is an affinely parametrised equator. You can always transform back to any other coordinate system should you really really want to.