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What would the shape of orbits of planets be if, hypothetically, the gravitational force was proportional to the inverse of cube of distance from the Sun?

Please ignore other effects caused due to the same although I would also like to know what those effects will be?

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  • $\begingroup$ Related and probably very informative for you. $\endgroup$ – J. Manuel Dec 13 '19 at 14:55
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    $\begingroup$ This is a typical intermediate classical mechanics exercise. You should do the work for yourself. The starting point can be found in any decent textbook such as Taylor or Symon. $\endgroup$ – Bill N Dec 13 '19 at 15:32
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    $\begingroup$ I'm not sure why this was closed as "non-mainstream". As @BillN points out, calculating orbits in various potentials is a perfectly valid application of Newtonian mechanics. $\endgroup$ – Michael Seifert Dec 14 '19 at 13:53
  • $\begingroup$ I agree @MichaelSeifert, but I don't think it's worth it reopening the question, for then to close it for another reason (homework-like question). $\endgroup$ – stafusa Dec 14 '19 at 22:46
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TL;DR: The planets would not maintain a closed orbit.

Bertrand's Theorem states that the only possible central force potentials that will give rise to bound & closed orbits are $F(r) \propto r^{+1}$ (simple harmonic oscillator) and $F(r) \propto r^{-2}$ (gravity, Coulomb force). The proof is slightly involved and I will not reproduce it here. Orbits under an inverse cube law could oscillate between a $(r_{min}, r_{max})$ or go off to infinity or spiral into the sun -- but they will not follow a stable path.

However, the most interesting question I believe that could come out of this is the following: What would happen if you added an inverse cube law, in addition to a gravitational potential? In fact, Newton showed two very cool results about this hybrid potential.

  1. Given any particle motion defined by $(r(t), \theta(t))$ under gravity alone, one can find a particle motion under our hybrid potential that matches the distance $r(t)$, but has a different angular velocity $\tilde{\theta}(t)$.
  2. For any attractive potential $F(r)$, the above statement holds!

John Carlos Baez has a fantastic blog post on this subject that covers the answer to this question much more elegantly than I have, and also provides a more rigorous statement and proof.

I hope this answers your question!

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    $\begingroup$ Nothing in Bertrand's theorem prohibits orbits that are bounded but not closed; and such orbits would not fly off to infinity or spiral into the center. For example, the orbits in a $V \propto r$ potential eventually fill out all of the space between some $r_\text{min}$ and $r_\text{max}$, but they never get to $r = 0$ or $r = \infty$. $\endgroup$ – Michael Seifert Dec 13 '19 at 15:18
  • $\begingroup$ Thanks for the correction! I've edited my post. $\endgroup$ – nilai Dec 13 '19 at 15:21
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    $\begingroup$ Ironically, your original headline was actually correct — you just had the wrong reasoning for it. Take a look at my answer for the details. $\endgroup$ – Michael Seifert Dec 13 '19 at 15:37
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The best way to think about this is in terms of the effective potential. For a central-force problem in a potential $V(r)$, it turns out that we can always look at the radial motion of a particle with angular momentum $\ell$ and mass $\mu$ as being equivalent to 1-D motion in an effective potential given by $$ V_\text{eff}(r) = V(r) + \frac{\ell^2}{2 \mu r^2} $$ For an attractive gravitational potential, for example, we have $V(r) = -k/r$ (with $k > 0$). This leads to a potential that has a minimum at some value of $r$, which means that stable circular orbits exist. Orbits with $E < 0$ are always bounded; since $V_\text{eff}(r) \to 0$ as $r \to \infty$ and $V_\text{eff}(r) \to +\infty$ as $r \to 0$, the radial motion corresponds to oscillations about the minimum of $V_\text{eff}$.

On the other hand, for an force that depends on an inverse-cube law, we would have $V(r) = - k/r^2$ for some $k > 0$. This means that the effective potential will be of the form $$ V_\text{eff}(r) = \frac{1}{r^2} \left( \frac{\ell^2}{2 \mu} - k \right) $$ which will never have a single local minimum. There are three cases we can distinguish, depending on how strong the force is compared to the particle's angular momentum:

  • If $\ell^2/2 \mu > k$, the effective potential is always positive. The radial distance of such a particle will increase without bound (as the particle moves to regions where $V_\text{eff}$ is lower), which corresponds to the particle spiraling out towards infinity.

  • If $\ell^2/2 \mu < k$, the effective potential is always negative. The radial distance of such a particle will tend towards zero, which corresponds to the particle spiraling into the center.

  • If $\ell^2/2 \mu = k$, the effective potential is flat (and equal to zero). It is therefore possible to have a particle orbiting the center at any value of $r$, so long as its angular momentum has this "magic" value. However, such an orbit is unstable: if we give the particle a slight nudge inwards, it will continue to move inwards until it hits the origin, and if we nudge it outwards, its distance will increase linearly with time, spiraling out to infinity.

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Kepler's Second Law is the same for all two-body systems with a potential energy that depends on the distance between them $V(r)$. The Orbital Equation is the differential equation that gives the possible orbits, which is $\frac{\mathrm{d}^2u}{\mathrm{d}\theta^2} + u = - \frac{\mu}{L_z^2} \frac{f(u)}{u^2}$, where $u = \frac{1}{r}$, $\mu$ is the reduced mass of both bodies, $L_z$ is the angular momentum perpendicular to the plane in which the bodies move and $f(u)=-\frac{\mathrm{d}V}{\mathrm{d}r}|_{r=\frac{1}{u}}$. If you use $V(r)=-\frac{V_0}{r}$ then you get the solutions to the orbits of two bodies due to gravity or of two charges due to (only) the electric field. You could try using $V(r) = -\frac{V_0}{2r^2}$ and see which orbits you get.

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  • $\begingroup$ Kepler's Third Law does not hold for any potentials other than $V \propto 1/r$, at least not in the form $T^2 \propto r^3$. $\endgroup$ – Michael Seifert Dec 13 '19 at 15:01
  • $\begingroup$ Whoops, you're right. Kepler's third law only applies to elliptical orbits anyways. $\endgroup$ – Kirtpole Dec 13 '19 at 15:03

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