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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:
     Mathworld: Cotes Spirals
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?
      epispiral piece
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!

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The validity of Gauss's law depends on the force being inverse square, so if gravity were inverse-cube it wouldn't be valid to approximate the external field of the Earth with a field of a point mass located at its center. The actual gravitational field right above the surface of the Earth would not behave like a straight inverse-cube field. – Henning Makholm Jun 28 '12 at 17:37
up vote 12 down vote accepted

The path of a ballistic projectile in an inverse-square field is not a parabola, but an ellipse, assuming that the projectile does not reach escape velocity, which is certainly true for fireworks rockets.

But fireworks don't go high enough for the $r$ in the denominator to vary significantly, so we can approximate this elliptical segment very closely by a parabolic segment instead.

Even if you change the $r^2$ to an $r^3$, the same will be true: the rocket doesn't go far enough from the earth for the $r$ distance to vary significantly, so the path will still look like a parabolic segment. Moreover, the parabola it looks like will be almost exactly the same as the parabola it would have been with inverse-square gravity.

(To put it another way, the path is a parabola if the Earth is an infinite flat plane rather than a sphere. So you only have to distinguish between elliptical and parabolic paths if the rocket is sufficiently far away that it can notice the curvature of the Earth. Fireworks don't get anything like that far away.)

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If you want to look into this in greater detail, the book Mathematical Methods of Classical Mechanics by V.I. Arnol'd is superb. By the end of chapter 2 you will have at your disposal all the mathematical tools necessary to analyze this kind of problem. – MJD Jun 28 '12 at 17:19
Great answer, Mark! I will retrieve Arnol'd's book from the library. Thanks! – Joseph O'Rourke Jun 28 '12 at 17:25
@Joseph I just noticed that you are the person who gave me the superb answer to my question about hexahedra with faces of areas $1,2,3,4,5,6$. Thanks again, and I am glad to have been able to return the favor. – MJD Dec 5 '14 at 22:02

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.


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While interesting and important, how does this answer the question given in the original post? – Willie Wong Feb 5 '13 at 15:01

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