Gravitational potential energy (general form) Knowledge-Based Study

Study notes

1. Beyond Uniform Gravity

   Near Earth's surface, where the gravitational field is approximately uniform, gravitational potential energy is expressed simply as $U = mgh$, with an arbitrary reference point such as the ground ($h = 0$). However, over astronomical distances where the separation distance $r$ varies significantly, Newton's law of universal gravitation reveals that the gravitational field diminishes inversely with the square of the distance.

   Because the gravitational force between two point masses $M$ and $m$ is attractive and given by:

   $$F_g = \frac{G M m}{r^2}$$

   the force approaches zero only as the separation distance approaches infinity. Consequently, the standard reference point for zero gravitational potential energy is chosen at an infinite separation ($r \to \infty$), where the interacting bodies exert no force on each other.

   Defining the General Formula

   Gravitational potential energy $U(r)$ is defined as the work done by an external agent to bring two masses from infinite separation to a distance $r$ at constant velocity, or equivalently, the negative of the work done by the conservative gravitational force:

   $$U(r) - U(\infty) = -\int_{\infty}^{r} F_g \, dr'$$

   Setting $U(\infty) = 0$ and evaluating the integral yields the general expression for gravitational potential energy:

   $$U(r) = -\frac{G M m}{r}$$

   Why is Gravitational Potential Energy Negative?

   • Zero at Infinity: When two objects are infinitely far apart, their mutual gravitational interaction ceases, so $U = 0$.
   • Attractive Interaction: Because gravity is purely attractive, an external force must do positive work to pull the objects apart, increasing $U$ toward zero.
   • Bound States: Bringing masses closer together (decreasing $r$) causes $U$ to become more negative, indicating that the system is bound and requires an input of energy to separate completely.

   Mechanical Energy of a Bound System

   For an object of mass $m$ moving in the gravitational field of a central mass $M$ (where $M \gg m$), total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$:

   $$E = K + U = \frac{1}{2}mv^2 - \frac{G M m}{r}$$

   Circular Orbits

   For a circular orbit of radius $r$, the gravitational force provides the centripetal acceleration:

   $$\frac{G M m}{r^2} = \frac{m v^2}{r} \implies K = \frac{1}{2}mv^2 = \frac{G M m}{2r}$$

   Substituting this result into the total mechanical energy equation reveals a crucial relationship:

   $$E = \frac{G M m}{2r} - \frac{G M m}{r} = -\frac{G M m}{2r}$$

   Notice that for circular orbits, $K = -\frac{1}{2}U$ and $E = \frac{1}{2}U = -K$. A negative total mechanical energy confirms that the orbiting body is gravitationally bound to the central body.

   Overcoming Gravitational Binding

   Binding energy is defined as the minimum energy required to completely separate a mass $m$ from a gravitational system to infinity, bringing its total mechanical energy to zero:

   $$E_{\text{binding}} = -E = \frac{G M m}{2r} \quad \text{(for a circular orbit)}$$

   Deriving Escape Speed

   To launch a projectile from the surface of a spherical planet of mass $M$ and radius $R$ such that it escapes the planet's gravitational field and reaches infinity with zero remaining kinetic energy, we apply conservation of energy:

   $$E_i = E_f$$

   $$\frac{1}{2}mv_{\text{esc}}^2 - \frac{G M m}{R} = 0 + 0$$

   Solving for the escape speed $v_{\text{esc}}$ yields:

   $$v_{\text{esc}} = \sqrt{\frac{2GM}{R}}$$

   Notice that the escape speed is independent of the mass of the escaping object and is $\sqrt{2}$ times the orbital speed at that same radius.