Question

in the lecture of 13(orbits and escape velocity). why the force F is the derivative of potential energy U ..thanks a lot

Answer 1

Lets suppose we have a force that depends on the position, x, of an object. The force may vary in any arbitrary way but it must have a definite value for each value of x. The work done by this force in moving an object from x1 to x2 is given by:\[W=\int\limits_{x_{1}}^{x_{2}}f(x)dx\]According to the work-energy theorem we may equate this with the change in the object's kinetic energy:\[K_{2}-K_{1}=\int\limits_{x_{1}}^{x_{2}}f(x)dx\]Now, let's re-write this equation by introducing an arbitrary reference point x0, between the points x1 and x2. We get:\[K_{2}-K_{1}=\int\limits_{x_{0}}^{x_{2}}f(x)dx - \int\limits_{x_{0}}^{x_1}f(x)dx\]Now we re-organize the terms and get:\[K_{2}-\int\limits_{x_{0}}^{x_{2}}f(x)dx=K_{1}-\int\limits_{x_{0}}^{x_{1}}f(x)dx\] This has the form of a conservation statement. We DEFINE the potential energy function of a particle to be:\[U(x)-U(x_{0})=-\int\limits_{x_{0}}^{x}f(x)dx\]This states that the potential energy at a point, relative to the reference point, is always defined as the negative of the work done by the force as the object moves from the reference point to the point considered. We are free to set U(x0)=0 if we please, since we are concerned only with changes of potential energy in any real problem. Now, we see that our re-arranged equation is just the conservation of mechanical energy:\[K_{2}+U_{2}=K_{1}+U_{1}\]Since \[U(x)-U(x_{0})=-\int\limits_{x_{0}}^{x}f(x)dx\]After differentiating, we get:\[\frac{dU}{dx}=-f(x)\]