# Conservative forces intuition

Take a gravitational field (with all the field lines pointing inwards) and a perfectly circular curve as an object's trajectory. To find the work exerted by the force on the object, compute the line integral $\oint_C\overrightarrow{F} \cdot \overrightarrow{v}=Work=0$; which is to be expected as there is no tangential component to a conservative force, and $\overrightarrow{F} \cdot \overrightarrow{v}=0$ everywhere.

However, work is being done on the planet as it isn't moving in a straight line, and $\overrightarrow{F}$ is the only force around to provide this work.

Therfore, is perfect circular motion unphysical? Is the real motion more like Newton's 'infinitesimal pulling in' model pictured below, the downward pulls being the points where $\overrightarrow{F} \cdot \overrightarrow{v} \ne 0$, for instance? Have I misunderstood something?

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Yes, you have misunderstood something: there is no work being done. What makes a conservative force conservative is the fact that it can be written as the gradient of a scalar field, and thus we can assign a PE to every position the object occupies. And of course the PE in a gravitational field depends on the distance to the origin only.

So there is a force acting on the object orbiting in a circle, but it isn't doing any work, since there is no change in PE. When work is done there is a change in PE compensated by an equal but opposite change in KE. That doesn't happen with a circular orbit. Bt it does, on the other hand, if the object moves in a straight line.

It is the same situation as when you are carrying a heavy object: you have to do a lot of force to keep it from falling, but if you are moving on level terrain at constant speed, you are not doing any mechanical work, despite all the huffing and puffing...

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With regards to the straight line, my meaning was that without any force whatsoever present it would move in a straight line. –  Alyosha Dec 27 '12 at 16:15