Question 10.14: Gravitational potential energy. Determine the total gravitat...

The constant gravitational force per unit mass is a conservative force distribution given by

\mathbf{f} \equiv \mathbf{g}=-\nabla \psi,                (10.129a)

in which  \psi  is the gravitational potential energy density per unit mass σ. Form the scalar product  \mathbf{g} \cdot d \mathbf{x}=-\nabla \psi \cdot d \mathbf{x}=-d \psi,  which is equivalent to (10.125), and integrate this equation over the path traversed by the body point at x to obtain the potential energy   \psi(\mathbf{x})  per unit mass,

\mathbf{v} \cdot \mathbf{f}=-\nabla \psi \cdot \mathbf{v}=-\frac{\partial \psi}{\partial \mathbf{x}} \cdot \frac{d \mathbf{x}}{d t}=-\frac{d \psi(\mathbf{x})}{d t}                       (10.125)

\psi(\mathbf{x})=-\mathbf{g} \cdot \Delta \mathbf{x}                        (10.129b)

where   \Delta \mathbf{x} \equiv \mathbf{x}(P, t)  –  \mathbf{x}\left(P, t_0\right).  Then for a uniform gravitational field strength g and with  \sigma \equiv m  in (10.127) , we obtain

V(\mathscr{B}) \equiv \int_{\mathscr{B}} \psi(\mathbf{x}) d \sigma(P),                (10.127)

V(\mathscr{B})=\int_{\mathscr{B}} \psi(\mathbf{x}) d m=-\mathbf{g} \cdot \int_{\mathscr{B}} \Delta \mathbf{x} d m=-\mathbf{g} \cdot m \Delta \mathbf{x}^*,                      (10.129c)

where, from (5.12),  \Delta \mathbf{x}^*  is the displacement vector of the center of mass of   \mathscr{B}.  With  \mathbf{g}=-g \mathbf{k},  this delivers V = mgh in which  h \equiv \mathbf{k} \cdot \Delta \mathbf{x}^*=\Delta z^*  is the vertical displacement of the center of mass, a rule similar to the familiar particle relation (7.60). Thus, the total gravitational potential energy of a rigid body in a uniform gravitational field is equal to the gravitational potential energy of its center of mass particle.

m(\mathscr{B}) \mathbf{x}^*(\mathscr{B}, t)=\int_{\mathscr{B}} \mathbf{x}(P, t) d m(P)                  (5.12)

V_g=m g h               (7.60)