# Does the integral of the vis-viva equation have any meaning?

If orbital speed of an elliptical orbit can be described by

$$v(r) =\sqrt{GM\left(\frac{2}{r}-\frac{1}{a}\right)},$$

then what would the meaning of its ($$\mathrm{d}r$$) integral be?

• How is asking for the physical meaning of an operation not a conceptual question, but instead a homework question? OP doesn't anywhere state they need help with the integral... Sep 20 at 19:57

Given any $$v(r)$$, the integral over space can be transformed as follows, $$\int v\,\mathrm{d}r=\int v\,\frac{\mathrm{d}r}{\mathrm{d}t}\mathrm{d}t=\int v^2\,\mathrm{d}t.$$ Recalling that kinetic energy is $$\sim mv^2$$, then the right-most integral is proportional to the integral of kinetic energy over time, which yields an action. So the original integral is something like an action per unit mass, I suppose.
$$\oint v_r \mathrm dr,$$ i.e. the radial component of the velocity integrated around a radial orbit, is a useful quantity, sometimes called the radial action. For example, it is an adiabatic invariant, meaning that it is preserved under (among other things) a change to the gravitational potential, as long as the change happens over many orbits. Also, it is used in the construction of action-angle coordinates.
However, I can't think of a context where it makes sense to put the total speed $$v\equiv |\vec v|$$ (and not just $$v_r$$) as the integrand. Putting the total speed in the above integral would actually just yield 0, i.e. $$\oint |\vec v| \mathrm dr =0,$$ since the ascending and descending contributions would cancel.