Question 13.6: The Change in Potential Energy A particle of mass m is displ...
The Change in Potential Energy
A particle of mass m is displaced through a small vertical distance Δy near the Earth’s surface. Show that in this situation the general expression for the change in gravitational potential energy given by Equation 13.13 reduces to the familiar relationship relationship \Delta U_g=m g \Delta y.
U_f-U_i=-G M_E m\left(\frac{1}{r_f}-\frac{1}{r_i}\right) (13.13)
Conceptualize Compare the two different situations for which we have developed expressions for gravitational potential energy: (1) a planet and an object that are far apart for which the energy expression is Equation 13.14 and (2) a small object at the surface of a planet for which the energy expression is Equation 7.19. We wish to show that these two expressions are equivalent under the conditions described in the problem.
U_g(r)=-\frac{G M_E m}{r} (13.14)
U_g \equiv m g y (7.19)
Categorize This example is a substitution problem.
Combine the fractions in Equation 13.13:
(1) \Delta U_g=-G M_E m\left(\frac{1}{r_f}-\frac{1}{r_i}\right)=G M_E m\left(\frac{r_f-r_i}{r_i r_f}\right)
Evaluate r_f-r_i and r_i r_f if both the initial and final positions of the particle are close to the Earth’s surface:
r_f-r_i=\Delta y \quad r_i r_f \approx R_E{}^2Substitute these expressions into Equation (1):
\Delta U_g \approx G M_E m\left(\frac{\Delta y}{R_E{}^2}\right)=m\left(\frac{G M_E}{R_E{}^2}\right) \Delta y=m g \Delta ywhere g=G M_E / R_E{}^2 from Equation 13.5.
g=G \frac{M_E}{R_E{}^2} (13.5)