Gravity as a Conservative Force

An example of a conservative force is gravity. An object of mass $m$ near the surface of the earth has the gravitational potential energy

$$U = mgz ~~~ \mbox{(gravity near earth's surface)}$$ (9.12)

where $z$ is the height of the object and $g = 9.8 \mbox{ m} \mbox{ s}^{-2}$ is the local value of the gravitational field near the earth's surface. Notice that the gravitational potential energy increases upward. The speed of the object in this case is $\vert u_g \vert = [2(E - mgz)/m]^{1/2}$. If $\vert u_g \vert$ is known to equal the constant value $u_0$ at elevation $z = 0$, then equations (8.11) and (8.12) tell us that $u_0 = (2E/m)^{1/2}$ and $\vert u_g \vert = (u_0^2 - 2gz)^{1/2}$.

There are certain types of questions which energy conservation cannot directly answer. For instance if an object is released at elevation $h$ with zero velocity at $t = 0$, at what time will it reach $z = 0$ under the influence of gravity? In such cases it is often easiest to return to Newton's second law. Since the force on the object is $F = -dU/dz = -mg$ in this case, we find that the acceleration is $a = F/m = - mg/m = -g$. However, $a = du/dt = d^2 z/dt^2$, so

$$u = -gt + C_1 ~~~~~ z = -gt^2 /2 + C_1 t + C_2 ~~~ \mbox{(constant gravity)} ,$$ (9.13)

where $C_1$ and $C_2$ are constants to be determined by the initial conditions. These results can be verified by differentiating to see if the original acceleration is recovered. Since $u = 0$ and $z = h$ at $t = 0$, we have $C_1 = 0$ and $C_2 = h$. With these results it is easy to show that the object reaches $z = 0$ when $t = (2h/g)^{1/2}$.