what would a planetary orbit look like if gravity had constant magnitude?

Solution 1:

This is a nice question, and I don't have a comprehensive answer for you, but perhaps this will help a bit. For conservative forces, it's generally convenient to define the scalar potential $V$. The force will satisfy $\vec{F} = \vec{\nabla}V$, and your potential $V = r$ (yay, system is conservative). Since the kinetic + potential energy is constant, and $V$ is unbounded above, this means that the particle will always be in a bounded orbit around the body.

The orbits you've found are bounded, but not closed (on themselves), resulting in the flower-pattern trajectory. I do know that you only get closed orbits for potentials that are $\tfrac 1r$ (like ours), and $\tfrac 1{r^2}$ (where the force goes as $\tfrac{1}{r^3} \large)$, but I never really understood why (phase-space apparently).

Solution 2:

Consider an orbit $$\gamma: \quad t\mapsto z(t)=r(t)e^{i\phi(t)}$$ of such a particle. We are interested in the polar representation $$\phi\mapsto r(\phi)\tag{1}$$ of the resulting curve $\hat\gamma\subset {\mathbb C}$. Denoting the differentiation with respect to $t$ by a $\cdot$ and the differentiation with respect to $\phi$ by a $'$ we have $\dot r=r'\>\dot\phi$.

The kinetic energy of the particle (of mass $1$) is given by $$T={1\over2}(\dot r^2+ r^2\dot\phi^2)={1\over2}(r'^2+r^2)\dot\phi^2\ ,$$ and its potential energy simply by $V(z)=r$. It follows that $$E={1\over2}(r'^2+r^2)\dot\phi^2 + r\tag{2}$$ is constant along $\gamma$. A second invariant is the angular momentum $$L=r^2\>\dot\phi\ ,$$ so that we can eliminate $\dot\phi^2$ in $(2)$ and obtain the following differential equation for the functions $(1)$: $${1\over2}(r'^2+r^2){L^2\over r^4} + r=E\ .$$ This can be rewritten as a standard first order $ODE$ in the following form: $$r'=\pm{r\over L}\sqrt{2Er^2-L^2-2r^3}\qquad(r>0)\ .\tag{3}$$ The ODE $(3)$ can be separated. In fact, one obtains the inverse functions $r\mapsto \phi(r)$ by a mere quadrature: $${d\phi\over dr}=\pm{L\over r}\bigl(2Er^2-L^2-2r^3\bigr)^{-1/2}\ .$$ Unfortunately the resulting integral is nonelementary for general values of $E$ and $L$.