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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.

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  • $\begingroup$ Your equation implies that the sphere is in outer space, and the g is the acceleration caused by its gravity. $\endgroup$ – R.W. Bird Feb 19 at 14:49
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    $\begingroup$ Yes, as I've mentioned that we've an isolated system. $\endgroup$ – Kashmiri Feb 19 at 15:12
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    $\begingroup$ 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. $\endgroup$ – joseph h Mar 3 at 4:59
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    $\begingroup$ @Kashmiri FYI see this on meta.physics. $\endgroup$ – joseph h Mar 3 at 9:12
  • $\begingroup$ Thank you for standing up. :) $\endgroup$ – Kashmiri Mar 3 at 10:35
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A useful tool in this case is the method of virtual displacements of D'Alambert. We use the geometrical constraints of the problem to define the admissible displacements.

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: $m \boldsymbol a = 0$

$$\boldsymbol a = R\frac{\partial^2 \theta}{\partial t^2} \implies \omega = \frac{\partial \theta}{\partial t} = cte$$

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This scenario is possible (with the exception of unmoving frictionless spheres not existing). The particle has kinetic energy and no way to convert it into another form, thus, by conservation of energy, the particle will keep moving forever.

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