Let r = 1 / u {\displaystyle r=1/u} . Then the Binet equation for u ( φ ) {\displaystyle u(\varphi )} can be solved numerically for nearly any central force F ( 1 / u ) {\displaystyle F(1/u)} . However, only a handful of forces result in formulae for u {\displaystyle u} in terms of known functions. The solution for φ {\displaystyle \varphi } can be expressed as an integral over u {\displaystyle u}
A central-force problem is said to be "integrable" if this integration can be solved in terms of known functions.
If the force is a power law, i.e., if F ( r ) = a r n {\displaystyle F(r)=ar^{n}} , then u {\displaystyle u} can be expressed in terms of circular functions and/or elliptic functions if n {\displaystyle n} equals 1, -2, -3 (circular functions) and -7, -5, -4, 0, 3, 5, -3/2, -5/2, -1/3, -5/3 and -7/3 (elliptic functions).1
If the force is the sum of an inverse quadratic law and a linear term, i.e., if F ( r ) = a r 2 + c r {\displaystyle F(r)={\frac {a}{r^{2}}}+cr} , the problem also is solved explicitly in terms of Weierstrass elliptic functions.2
Whittaker, pp. 80–95. ↩
Izzo and Biscani ↩