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
that is inverse-cubic is a
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!