TITLE: Fireworks under inverse-cube gravity
QUESTION [8 upvotes]: What is the path of a projectile under an inverse-cube gravity law?

Imagine that the law of gravity was changed overnight from
$F(r) = G m_1 m_2 / r^2$
to 
$F(r) = G' m_1 m_2 / r^3$.
To be specific, suppose 
$G' = G r_E$ where $r_E$ is the radius of the Earth,
so that the force at the Earth's surface is unchanged.
I am wondering how would the arc of a fireworks rocket
compare to the parabolic path it would follow under the inverse-square law.
(In the U.S. at this time of year, the evening sky is full of fireworks
as we approach the 4th of July.)
Presumably, the same rocket would travel a bit higher and cover
a bit more distance horizontally, but what is the precise path it would follow?
It is known that the solutions to a 
central force 
that is inverse-cubic is a 
Cotes' Spiral,
which comes in three varieties:
     

But I am uncertain which of the three would apply here, and how to compute the relevant constants.
Perhaps a piece of an
epispiral,
something like this?
     


It would be instructive to see the inverse-square parabola and the inverse-cubic Cotes's spiral,
for the same projectile initial conditions, plotted together...
Addendum. After retrieving Arnol'd's book as per Mark Dominus's recommendation,
I wanted to share one interesting fact (p.37):  The only central-force laws
in which all the bounded orbits are closed are the inverse-square and inverse-cubic laws!

REPLY [2 votes]: Inverse square and linear laws $(\Large\frac {a}{r^2}$ and $a*r)$ are the only one to have stable closed orbits, 
proved by Bertrand in 1873: 
Joseph Bertrand, Théorème relatif au mouvement d'un point attiré vers un centre xe, C. R. Acad. Sci. 77 (1873), 849-853., 
see also F.C. Santos, V. Soares, A.C. Tort, An English Translation of Bertrand's Theorem, Lat. Am. J. Phys. Educ. 5 2011, 694-695. 
-ilan