17
$\begingroup$

We know that the orbit of the earth is elliptical considering the force of gravity is inversely proportional to the square of the distance. But assume that, gravity does not depend on distance. Considering that situation, what will be the shape of the orbit? Will it be elliptical, circular, hyperbolic or what?

$\endgroup$
2
  • 4
    $\begingroup$ Interestingly, you can simulate those orbits at home just by trowing a marble in a funnel - a cone shaped funnel, like those common in kitchens and garages. $\endgroup$
    – Pere
    Commented Jul 18, 2023 at 15:30
  • 6
    $\begingroup$ I'm troubled by this. If gravitational attraction does not depend on distance then to all intents and purposes a body will feel the gravitational attraction of all bodies in the universe equally. The result will either be that there is zero net attraction resulting in all paths being along a straight line, or infinite attraction in every direction simultaneously. $\endgroup$ Commented Jul 18, 2023 at 15:51

4 Answers 4

43
$\begingroup$

Circular orbits are always possible for any central force law, but noncircular orbits would resemble rosettes. Here's a specific example for the case where the force is constant with distance: closed rosette orbit

By chance, I selected parameters that happen to make a (at least approximately) closed orbit. That won't happen in general. When I change the angular momentum slightly, I get the following. non-closed rosette orbit Here the red points show the starting and ending positions of the orbiting object. The positions and velocities don't quite match up, and if I continue the orbit indefinitely, it fills up the entire annulus between the minimum and maximum radii: just an annulus

Click here for an animation.

These kinds of orbits are common in galactic contexts, where things are orbiting inside extended mass distributions. The particular case of a force that is constant with distance corresponds to an orbit inside a density distribution that scales inversely with radius, i.e. $\rho\propto r^{-1}$. That's precisely the structure of the deep interior of the Navarro-Frenk-White density profile.

$\endgroup$
4
  • $\begingroup$ What software did you use for simulation / animations? $\endgroup$ Commented Jul 18, 2023 at 9:26
  • 1
    $\begingroup$ @VioletGiraffe Nothing straightforward to use, unfortunately! Python (with numpy and scipy) for calculations, Matplotlib for images, ffmpeg to make individual frames into a movie, and the top Google result for "mp4 to gif" to make the movie into a gif. For the specific problem, I modified some code I already had prepared for research purposes; basically I use quadrature integration to construct what one pericenter-to-apocenter orbital segment looks like and then stitch arbitrarily many of those together. (Happy to give further detail as an answer to a new question, if you are interested!) $\endgroup$
    – Sten
    Commented Jul 18, 2023 at 9:41
  • $\begingroup$ I see, thanks for the detailed explanation! Quite an elaborate effort. Have you tried manim? I haven't, but I see many people using it so I assume it's pretty good. github.com/3b1b/manim $\endgroup$ Commented Jul 18, 2023 at 9:46
  • $\begingroup$ @VioletGiraffe I was not aware of that. Just had a look, could be useful! $\endgroup$
    – Sten
    Commented Jul 18, 2023 at 9:58
8
$\begingroup$

In addition to Sten's great answer, it should be noted that under constant gravity all orbits are bounded. Therefore, there aren't any orbit like the hyperbolic ones mentioned in the question.

That can be seen from the fact that potential energy is unbounded. Every meter a particle moves away from the centre, it gains the same amount of potential energy. Therefore, for whatever kinetic energy the particle may have, there is a maximum distance it can reach even when all its kinetic energy is converted into potential energy.

Additionally, first and third Kepler's law wouldn't hold. However, second Kepler's law would hold, because it's equivalent to the conservation of angular momentum, and angular momentum is always conserved under a central force, no matter its magnitude.

$\endgroup$
1
  • $\begingroup$ @ProfRob - Fixed. Thanks for pointing. $\endgroup$
    – Pere
    Commented Jul 21, 2023 at 16:32
4
$\begingroup$

There wouldn't be standard orbits at all.

The Earth orbits the Sun because the Sun's mass dominates the local spacetime. By removing distance from the gravitational equation, the Earth would, with the same magnitude, now be attracted to every other object in the Universe, without the tempering effect of distance. It would be equally attracted to all similarly sized stars, and will be more attracted to larger stars, no matter how far away they are.

Everything would likely disintegrate.

The only reason the Earth is a single item is that its collective mass dominates its local spacetime enough to keep itself together. Without that local dominance, the Earth's internal pressure would blast its component pieces outward.

$\endgroup$
6
  • $\begingroup$ Try to imagine all life as you know it stopping instantaneously and every molecule in your body exploding at the speed of light. $\endgroup$
    – Mazura
    Commented Jul 20, 2023 at 1:36
  • 1
    $\begingroup$ Influence of distant objects would still be suppressed because they exert nearly the same gravitational pull on all elements of your system. That is, the Earth could still orbit the Sun because more distant things are pulling the Earth and Sun together, not the Earth separately; likewise the Earth could remain intact. Formally, the influence of distant things is suppressed by the extra $1/r$ dependence in the tidal force. (Also, the tidal force in this case is purely compressive! It wouldn't break things apart.) $\endgroup$
    – Sten
    Commented Jul 20, 2023 at 5:26
  • $\begingroup$ @Sten Distance is no longer taken into account. There is no difference in attraction between the 'nearside' and 'farside', regardless of distance between objects. The common image of spacetime, with each item making an indent marking its local influence is now absolutely flat, and Hill spheres and LaGrange points no longer exist. $\endgroup$ Commented Jul 20, 2023 at 14:40
  • $\begingroup$ Again, distance matters because distant objects exert the same gravitational pull on the Earth as on the Sun -- so they do not influence the dynamics of the Earth-Sun system. Within the freely falling frame, they have no net influence. (Aside from the tidal force.) $\endgroup$
    – Sten
    Commented Jul 20, 2023 at 15:48
  • $\begingroup$ (And you can explicitly show that Lagrange points and Hill spheres still exist. Their distances from the smaller object simply scale as the first power, instead of the $1/3$ power, of the mass ratio.) $\endgroup$
    – Sten
    Commented Jul 20, 2023 at 16:09
1
$\begingroup$

I agree with the answer of @MichaelRichardson that a constant, i.e. distance-independent, gravitational force throughout the universe would make any structures in it unstable. This follows directly from the The Virial Theorem, which tells us that in general the average kinetic energy $\bar{T}$ is related to the average potential energy $\bar{U}$ by the equation $2\bar{T}=k\bar{U}$ where $k$ is the power index of the interaction potential. For a constant force field we would have $k=1$ ( as $U=\alpha x$). Now if we let $x\rightarrow \infty$ we have in this case $\bar{U}\rightarrow \infty$ and per the Virial Theorem $\bar{T} \rightarrow \infty$. This is obviously not possible without any structure flying apart. In order to have localized structures the potential energy requires a power index $k<0$ (for gravity $k=-1$)

For clarification, with the usual law of gravity ($U=-\alpha /x$), an object is bound when the sum of kinetic energy and potential energy is less than zero ($T+U<0$)). For a circular orbit we have in this case $T=-1/2\cdot U$. For the Earth the orbital velocity $v=\sqrt{2T/m} $ is about $30 km/sec$. The nearest star has about a distance of $3\cdot 10^5 AU$ from the Sun, which would make its orbital velocity $1/\sqrt{3\cdot 10^5}$ times smaller, i.e. it could only move at about $0.05 km/sec$ to stay in orbit. However, stars in our galaxy have typically proper motions of the order of $100 km/sec$, so they are about 3 orders of magnitudes too fast to be bound. This is the reason why our solar system can be treated as a localized structure within but dynamically unrelated to the rest of the galaxy (never mind other galaxies).

Consider in contrast a hypothetical world where masses interact with a constant (distance independent) force of gravity. In this case the required equality between centrifugal and gravitational force would read

$$m\frac{v^2}{r} = const.$$

that is $v \propto{\sqrt{r}}$ . This means the orbital velocity would increase rather than decrease with increasing $r$. The orbital velocity of the star would now be $\sqrt{3\cdot 10^5}\cdot 30 km/sec = 1.6\cdot 10^4 km/sec$. So with such a force law the whole galaxy (and beyond) would be part of our solar system, or rather everything in our universe would be just one big system of randomly moving objects, without any kind of local structures being able to exist at all in the long term.

Of course, these absurd consequences of a hypothetical constant gravitational force not only contradict reality, but are merely the result of a conceptual inconsistency by ignoring that point masses (or masses that can be approximated as such) by definition imply a $1/r^2$ force law. This is because outside the mass distribution, the divergence of the force field must by definition vanish , i.e. (considering only the radial component here)

$$ div(\vec{F})=\frac{1}{r^2}\cdot \frac{\partial}{\partial{r}}(r^2\cdot F) = 0$$

For $F=const$, this would however result in a divergence $\propto{1/r}$ which contradicts the initial assumption that there are no mass sources outside the mass. It is immediately obvious that the above equation is only satisfied for $F=1/r^2$.

$\endgroup$
13
  • $\begingroup$ While I won't claim that cosmological dynamics with such a force law are sensible (although it's a super interesting question), I don't find this argument compelling. In our own universe, potential and kinetic energy diverge at large distances, just due to the homogeneous mass distribution. That doesn't prevent localized structures. Perhaps the key thing is that the virial theorem only applies to structures that are already localized. $\endgroup$
    – Sten
    Commented Jul 21, 2023 at 7:29
  • $\begingroup$ @Sten Please see my edited answer $\endgroup$
    – Thomas
    Commented Jul 22, 2023 at 16:51
  • $\begingroup$ I don't think it's a concern that circular velocities diverge with distance. Again that's true in our universe, too, due to the mass distribution! One feature of this force law is that the universe must eventually collapse (per @Pere's answer) -- but while we don't think that will happen to our universe, it's certainly been regarded as a possibility and not anything absurd. $\endgroup$
    – Sten
    Commented Jul 22, 2023 at 20:53
  • $\begingroup$ The most scary thing to me is that there's no shell theorem. But I'd have to think about whether that ends up being problematic. $\endgroup$
    – Sten
    Commented Jul 22, 2023 at 21:18
  • $\begingroup$ @Sten Pere's answer also implies that for a distance independent force law each mass is bound to every other individual mass in the universe regardless of their kinetic energies. This rules out the existence of any dynamically decoupled local structures. You can of course mimic a net constant gravitational force inside a continuous $\propto{1/r}$ mass distribution, but this would still be assuming a $\propto{1/r^2}$ force with regard to the individual mass elements (which is not what the OP was referring to). $\endgroup$
    – Thomas
    Commented Jul 23, 2023 at 8:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .