# Is it possible for a particle to slide on a frictionless sphere and orbit the sphere? [closed]

We are in an isolated system consisting of a large massive smooth sphere/planetoid with radius $$R$$ and due to its mass an nearly constant acceleration $$g$$ exists in its vicinity. A small particle is placed on top of it and given a shove. For simplicity let's assume the large sphere doesn't move.

I was curious that is it possible for the small particle to slide all the way and hence orbit the sphere?

I can write the equation in such a case as (if it is possible):

$$\frac{m v^{2}}{R}=m g-N.$$

But how do I know if it indeed is possible?

Edit: This question doesn't call for any computation, it is just a theoretical query if something can happen.

• Your equation implies that the sphere is in outer space, and the g is the acceleration caused by its gravity. – R.W. Bird Feb 19 at 14:49
• Yes, as I've mentioned that we've an isolated system. – Kashmiri Feb 19 at 15:12
• John Rennie added the "homework and exercises" tag and subsequently this question was closed for being a "homework and exercises" type question. Just reading this question it appears that it is not HAE. Voted to re-open and removed the "homework and exercises" tag as OP has also claimed that it is not - nor does this question even look HAE type. – joseph h Mar 3 at 4:59
• @Kashmiri FYI see this on meta.physics. – joseph h Mar 3 at 9:12
• Thank you for standing up. :) – Kashmiri Mar 3 at 10:35

In this case, if the shove is not too big to put the particle in orbit, the only possible arbitrary small displacement is: $$\boldsymbol \delta = R\boldsymbol \delta \theta$$.
The total virtual work must be equal to the virtual work of the resultant (that is mass x acceleration): $$m \boldsymbol{a.\delta} = m\boldsymbol {g.\delta} + \boldsymbol {N.\delta} + \boldsymbol {F.\delta}$$ where F is some tangential force as friction. For $$m\boldsymbol g$$ and $$\boldsymbol N$$, the force is perpendicular to the admissible displacement, so the virtual work is zero. And if additionally there is no friction: $$m \boldsymbol{a.\delta} = 0$$
As it must hold for any arbitrary displacement: $$m \boldsymbol a = 0$$
$$\boldsymbol a = R\frac{\partial^2 \theta}{\partial t^2} \implies \omega = \frac{\partial \theta}{\partial t} = cte$$