A particle travelling on the inside surface of a smooth sphere will just leave the surface when the reaction between the particle and the inner surface is zero.
Suppose a particle of mass
at the bottom A of a sphere of radius r is given a horizontal speed
It will move along the internal surface of the sphere. Suppose at some point B it has a speed![]()

If the line from the centre of the circle to the point B makes an angle
with the upwards vertical then the force of gravity will have a component
towards the centre of the circle.

Applying
towards the centre of the circle gives
(1)
Since the inside of the sphere is smooth, no work is done against friction, and energy is conserved in going from A to B. The potential energy at B is![]()
Hence![]()
Cancelling
gives
(2)
Put
in equation (1) and cancel
from both sides, obtaining
then
Substitute this into (2) to obtain![]()
Rearranging this formula for
gives
Since
we must have
Obviously![]()
The particle will leave the inside surface at some point if
This corresponds to the particle not having enough initial kinetic energy to reach the top of the sphere.