# Shouldn't ball has to move in the circle even after cutting the rope by which it was attached?

The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.

Angular momentum is given by$$^*$$ $$\mathbf L=\mathbf r\times\mathbf p$$ Without loss of generality, let's assume after the rope is cut our object is moving along the line $$y=1$$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum $$L=rp\sin\theta$$ where $$\theta$$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).

Now, since there are no forces acting on our object, $$p$$ is constant. Also, $$r\sin\theta$$ is just the constant $$y=1$$ value given by the line the object is moving along. Therefore, it must be that $$L$$ is constant.

This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.

$$^*$$Note that this applies to any type of motion, not just circular motion.

The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $$L = \vec r \times \vec p$$. $$r \sin \theta$$ and $$mv$$ remain constant after such an event.

The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.