PC1672 Advanced dynamics

3.3 Gravitational potential energy

The gravitational potential energy of two masses separated by a distance is

$\begin{displaymath}U(r)=-{GMm\over r}\end{displaymath}$

Using the chain rule for $\nabla$

we find that the corresponding force is

$\begin{displaymath}{\bf F}=-{{\rm d}U\over{\rm d}r}\hbox{\boldmath $\nabla$}r\end{displaymath}$

which agrees with the form above since

$\begin{displaymath}\hbox{\boldmath $\nabla$}r={\bf e}_r\end{displaymath}$

Note that a

potential corresponds to a

force.

For an object moving near the Earth, the total energy

$\begin{displaymath}E={1\over 2}mv^2-{GM_Em\over r}\end{displaymath}$

is constant. We can use this to show that an object with speed

$\begin{displaymath}v_{\rm esc}=\sqrt{2GM_E\over R_E}=11.2\ \hbox{\rm km s$^{-1}$}\end{displaymath}$

at the surface of the Earth is just able to escape from the Earth to infinity.

Textbook references

Mike Birse
17th May 2000