Gravitational Potential Energy

Suppose that a spherically symmetric object of mass $M$ is located at the origin of our coordinate system. The gravitational force, due to the gravitational attraction of mass $M$, experienced by a point object of mass $m$ and displacement is ${\bf r}$ (located outside the former mass) is

$${\bf f}= -G\,M\,m\,\frac{{\bf r}}{r^3}.$$ (1.257)

[See Equation (1.241).] Now, $r=(x^2+y^2+z^2)^{1/2}$. It is easily demonstrated that

$$\frac{\partial r}{\partial x} = \frac{x}{r},$$ (1.258)

$$\frac{\partial r}{\partial y} = \frac{y}{r},$$ (1.259)

$$\frac{\partial r}{\partial z} = \frac{z}{r}.$$ (1.260)

Consider

$$\nabla\!\left(\frac{1}{r}\right) \equiv \frac{\partial (1/r)}{\pa... ...al (1/r)}{\partial y}\,{\bf e}_y+ \frac{\partial (1/r)}{\partial z}\,{\bf e}_z.$$ (1.261)

(See Section A.19.) It follows that

$$\nabla\!\left(\frac{1}{r}\right) = -\frac{1}{r^2}\,\frac{\partial r}{\partial x}\,{\bf e}_x -\frac... ...{\partial y}\,{\bf e}_y-\frac{1}{r^2}\,\frac{\partial r}{\partial z}\,{\bf e}_z$$

$$= -\frac{x}{r^3}\,{\bf e}_x- -\frac{y}{r^3}\,{\bf e}_y-\frac{z}{r^3}\,{\bf e}_y$$

$$= - \frac{{\bf r}}{r^3},$$ (1.262)

where use has been made of Equations (1.258)–(1.260). The previous equation can be combined with Equation (1.257) to give

$${\bf f} = \nabla\!\left(\frac{G\,M\,m}{r}\right).$$ (1.263)

A comparison with Equation (1.47) reveals that the gravitational force field of our spherical object is a conservative field with the associated potential energy

$$U({\bf r})= - \frac{G\,M\,m}{r}.$$ (1.264)

Note that, by convention, the potential energy at infinity is zero.

A particle of mass $m$ moving in the gravitational field of our spherical object has a conserved energy

$$E= K+ U = \frac{1}{2}\,m\,v^2 - \frac{G\,M\,m}{r},$$ (1.265)

where $v$ is the particle's instantaneous speed. (See Sections 1.3.2 and 1.3.5.)

Let us again model the Earth as a sphere of mass $M$ and radius $R$ that is centered at the origin. Consider an object that is launched from the surface of the Earth, in an arbitrary outward direction, with speed $v_{\rm escape}$. Suppose that the object only just manages to escape from the Earth's gravitational field. It follows that the object's speed at infinity (i.e., $1/r=0$) is zero. Thus, it is clear from the previous equation that the object's conserved energy, $E$, is also zero. Hence,

$$0=\frac{1}{2}\,m\,v_{\rm escape}^2 - \frac{G\,M\,m}{R}$$ (1.266)

at the surface of the Earth, which implies that

$$v_{\rm escape} = \left(\frac{2\,G\,M}{R}\right)^{1/2}= \left[\fra... ....972\times 10^{24})}{6.371 \times 10^{6}}\right]^{1/2}=11.19\,{\rm km\,s^{-1}}.$$ (1.267)

The speed $v_{\rm escape}$, which is known as the escape speed, is the minimum speed at which an object must be launched from the Earth's surface if it is to reach outer space.