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What if the gravitational force is

  1. Aug 24, 2010 #1
    Hi people,

    we all know that the equation for gravitational force is F= (G)(M1)(M2)/R^2

    which means that G.Force is inversely proportional to 1/R^2.

    I was wondering what if the gravitational force is given as F= (G)(M1)(M2)/R

    OR F= (G)(M1)(M2)/R^3 .. (i.e. G.Force is inversely proportional to 1/R or 1/R^3 instead of 1/R^2.. )

    How will this affect our solar system and universe at large???
     
  2. jcsd
  3. Aug 24, 2010 #2
    One thing I remember that's interesting is that there is something called Bertrand's theorem, which specifies the conditions under which you can have closed orbits (i.e. orbits that are bounded and which trace themselves), and this can only happen if the potential is proportional to -R-1 or to R2. So you could have orbits, but they would be chaotic!
     
  4. Aug 25, 2010 #3
    It's an interesting speculation. (Sailing close to the edge of the forum rules - but IMHO just inside)

    There's good reason to believe that the inverse square law is so ubiquitous in physics that it is a mathematical necessity for the universe itself - it leads to conservation of charge, mass, energy, and related things like the vectorial nature of electromagnetic and gravitational fields.

    It's reasonable to suppose that if the rule for gravity were different, so would be the same rules for everything else. The entire universe would unravel.

    Unfortunately we don't know enough about this universe to speculate about the behaviour of a different one.
     
  5. Aug 25, 2010 #4
    One simple conclusion would be that the force of gravity would be much stronger by a factor of r or much weaker by a factor of r. Likely we would not be here having this discussion.
     
  6. Aug 25, 2010 #5

    DaveC426913

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    Well, magnetism operates on 1/R^3. Try getting a metal object to orbit a magnet.
     
  7. Aug 25, 2010 #6
    Isn't it what it is because the force spreads out as the surface area of a sphere?
     
  8. Aug 25, 2010 #7

    DaveC426913

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    You mean like gravity?
     
  9. Aug 25, 2010 #8

    Cleonis

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    As darkSun remarks, only the harmonic force and the inverse square force have the property that they give rise to repetitive orbits. The orbits of other force laws look like spirograph drawings, so to speak.

    A force that is proportional to distance to the center of attraction is called a harmonic force (since in the 1-dimensional case such a force gives rise to harmonic oscillation.) It's also referred to as Hooke's law.

    If I remember correctly somewhere around 1/R^4 or 1/R^5 there is a runaway effect. An object will just keep spiralling in.

    I found a http://www.physics.sjsu.edu/tomley/Veff.html". Unfortunately the method of input is rather unclear.

    Now, there is software that I highly recommend (I'm using it myself) for generating Java simulations. It's called http://www.um.es/fem/EjsWiki/Main/WhatIsEJS?", it's open source and it's freely available. (Author: Francisco Esquembre)
    You can watch http://swampfox.fmarion.edu/sites/swampfox.fmarion.edu.engelhardt/files/pdfs/EJS_Tutorial/EJS_Tutorial.html" [Broken]

    What you need to provide as input is the equation of motion. EJS generates the java code for computing the orbit. Try it! It will be great fun for you to see the answer to your question in an applet that you have created yourself.

    I'm thinking now of creating one too, for my own website. So keep watching this thread. If you're going to try EJS, and something is unclear to you, then feel free to contact me.

    Cleonis
    www.cleonis.nl
     
    Last edited by a moderator: May 4, 2017
  10. Aug 25, 2010 #9
    Close. In the 2-body case (and a few specific setups with more bodies), any force function will allow exactly circular orbits...because the distance doesn't change, so the force may as well be constant. These orbits are not stable, however: for anything but the inverse square law or harmonic oscillator potential, any perturbation, no matter how small, will lead to a trajectory that diverges further and further from a circular orbit.

    The real world has a huge number of other bodies to perturb things, and no perfectly circular orbits (something that seems to get lost in some explanations, leading to the idea that an orbit is a precarious and fragile balance). Instead, the real world has elliptical orbits, which respond to small perturbations by becoming slightly different, equally-stable elliptical orbits, and a few more complex systems. A simple Newtonian inverse square law doesn't describe things exactly, but it comes extremely close.
     
  11. Aug 26, 2010 #10
    I Actually meant gravity, As in, how can it then be anything else than 1/R^2. Mathematics are used to describe nature, not the other way around right? :tongue:
     
  12. Aug 26, 2010 #11

    Cleonis

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    I didn't remember correctly; I'm creating a Java applet (working as described in https://www.physicsforums.com/showpost.php?p=2852936&postcount=8" of this thread), and i tried R^-3, with the velocity at .99 of what would be needed for a circular orbit (which itself wouldn't be stable.)
    The object spirals in, and then all of a sudden it breaks away along a trajectory that looks like an escape trajectory. (I'll need to add zoom out capability to follow that trajectory.)

    By the way, it's not necessary to work with integers only. R^-2.5 is a force law too; the Java software for calculating powers handles that just as well.
     
    Last edited by a moderator: Apr 25, 2017
  13. Aug 26, 2010 #12

    DaveC426913

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    Don't bother. This is an artifact of your calculations, not a bona fide effect. I have programmed several orbital simulators in my lifetime and they all have this same quirk. (It is independent of the force used - I was using r^2.) At near 0,0, the orbiting object gains a vast boost in velocity and shoots off-screen.

    What's happening is that, unlike the real world, computer simulators have to operate in discrete increments; the precision of your positions, velocities and time intervals have a graininess. As your orbiting object nears the centre, these begin to manifest themselves. For example, the object can go from a reasonable distance to .000000000000001 distance in one interval, Then, when you calculate the force at this range, it is a vastly large value. So next time interval you update the velocity based on the force, both of which are now huge. Now you update the distance based on the velocity.

    The upshot is: Your object has effectively "teleported" from near 0,0 to 0,1 without having to experience any slowing effect by its passage in the interim. i.e. it is now at a quite safe distance from the centre to not be slowed, but it still has that huge velocity. And away it goes.


    The lazy way to deal with this is to put a governor on the force at small distances:
    if (d(x) < .1 OR d(y) < .1) {
    a(x) = a(x)/10 ;
    a(y) = a(y)/10 ;
    }
     
    Last edited: Aug 26, 2010
  14. Aug 26, 2010 #13
    If a central force [itex]F = F(r)[/itex] acts between two bodies, then the angular momentum L is still conserved and the trajectory lies in one plane and satisfies the equation (in polar form [itex]r = r(\phi)[/itex]):

    [tex]
    \frac{d^{2}}{d\phi^{2}}\left(\frac{1}{r}\right) + \frac{1}{r} + \frac{\mu \, r^{2} \, f(r)}{L^{2}} = 0
    [/tex]

    I know this as Binet formula, but it does not give me a link on Google. Here, [itex]\mu[/itex] is the reduced mass of the two--body system.

    For an attractive force that drops as the [itex]n[/itex]-th power of the distance:

    [tex]
    f(r) = -\frac{\alpha_{n}}{r^{n}}
    [/tex]

    we have the equation:

    [tex]
    \frac{d^{2}}{d\phi^{2}}\left(\frac{1}{r}\right) + \frac{1}{r} - \frac{\mu \, \alpha_{n}}{L^{2}} \, \left(\frac{1}{r}\right)^{n - 2} = 0
    [/tex]

    For the cases [itex]n = 2[/itex] and [itex]n = 3[/itex], the equation is linear in [itex]x \equiv 1/r[/itex]. Let's consider the case [itex]n = 3[/itex]. Then we have:

    [tex]
    \ddot{x} + \left(1 - \frac{\mu \, \alpha_{3}}{L^{2}}\right) x = 0
    [/tex]

    Subject to the condition [itex]\dot{x}(0) = 0[/itex], this equation has the solution:

    [tex]
    \frac{1}{r} = \frac{1}{r_{0}} \, \cos{\left(\sqrt{1 - \frac{\mu \, \alpha_{3}}{L^{2}}} \, \phi\right)}, \; \frac{\mu \, \alpha_{3}}{L^{2}} < 1
    [/tex]

    [tex]
    \frac{1}{r} = \frac{1}{r_{0}}, \; \frac{\mu \, \alpha_{3}}{L^{2}} = 1
    [/tex]

    [tex]
    \frac{1}{r} = \frac{1}{r_{0}} \, \cosh{\left(\sqrt{\frac{\mu \, \alpha_{3}}{L^{2}} - 1} \, \phi\right)}, \; \frac{\mu \, \alpha_{3}}{L^{2}} > 1
    [/tex]

    The first one describes a complicated curve that has asymptotes for the angles where the cosine becomes zero, i.e.:

    [tex]
    \sqrt{1 = \frac{\mu \, \alpha_{3}}{L^{2}}} \phi_{0} = \pm \, \frac{\pi}{2}
    [/tex]f

    and corresponds to an infinite motion, although we may have a loop around the center of force.

    The second one describes a circle, but only for a particular value of the angular momentum.

    The third always spirals toward the center of force because the hyperbolic cosine tends exponentially to infinity as the angle tends to both positive and negative infinity. The time it takes to fall is given by:

    [tex]
    t_{f} = \int_{0}^{\infty}{\frac{d\phi}{\dot{\phi}}} = \int_{0}^{\infty}{\frac{\mu \, r^{2} \, d\phi}{L}}
    [/tex]
    [tex]
    = \int_{0}^{\infty}{\frac{\mu \, r^{2} \, d\phi}{L}} = \frac{\mu \, r^{2}_{0}}{L} \, \int_{0}^{\infty}{\frac{d\phi}{cosh^{2}{\left(\sqrt{\frac{\mu \, \alpha_{3}}{L^{2}} - 1} \, \phi\right)}}
    [/tex]
    [tex]
    = \frac{\mu \, r^{2}_{0}}{L \, \sqrt{\frac{\mu \, \alpha_{3}}{L^{2}} - 1}}
    [/tex]

    i.e. it is finite. The particle is also predicted to come out of the center of force, but this is an unphysical solution.
     
    Last edited: Aug 26, 2010
  15. Aug 27, 2010 #14

    Cleonis

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    Yeah, that's what I should have concluded.

    Actually, the software I use, http://www.um.es/fem/EjsWiki/Main/WhatIsEJS", offers various solvers for differential equations. The default solver issued a warning to me: something like 'the numerical analysis has encountered a problem'. I then switched to another, less sophisticated solver. Sure enough the trajectory flew out of control.

    EJS also offers adaptive solvers, the time increment is automatically decreased if artifacts are detected. In addition, I can add a condition that the simulation automatically halts when the distance drops beneath a particular value
     
    Last edited by a moderator: Apr 25, 2017
  16. Aug 27, 2010 #15
    You don't need solving DEs. As I said, the angular momentum is conerved in central force field. Therefore, the trajectory is in a single plane (or, specially a line) determined by the initial postion and velocity (if those are collinear, then the trajectory is line). Therefore, we may use only two polar coordinates [itex](r, \phi)[/itex] to describe the motion. The Lagrangian of the system is:

    [tex]
    L = \frac{\mu}{2} \, \left(\dot{r}^{2} + r^{2} \, \dot{\phi}^{2} \right) - U(r)
    [/tex]

    where the potential energy is determined by [itex]F(r) = -U'(r)[/itex]. There are two conserved quantities:

    1) The angular momentum:

    [tex]
    l = p_{\phi} = \frac{\partial L}{\partial \phi} = \mu \, r^{2} \, \dot{\phi}
    [/tex]

    and

    2) The total energy:

    [tex]
    E = \dot{r} \, \frac{\partial L}{\partial \dot{r}} + \dot{\phi} \, \frac{\partial L}{\partial \dot{\phi}} - L
    [/tex]
    [tex]
    = \frac{\mu}{2} \, \left(\dot{r}^{2} + r^{2} \, \dot{\phi}^{2} \right) + U(r)
    [/tex]
    [tex]
    = \frac{\mu \, \dot{r}^{2}}{2} + \frac{l^{2}}{2 \, \mu \, r^{2}} + U(r)
    [/tex]

    But, this is just like the Law of Conservation of energy for a 1D motion in an effective potential energy:

    [tex]
    U_{l}(r) = U(r) + \frac{l^{2}}{2 \, \mu \, r^{2}}
    [/tex]

    There can be a finite motion only around a point of stable equilibrium which is the point of a potential energy minimum. The necessary condition for a minimum is:

    [tex]
    U'_{l}(r) = U'(r) - \frac{l^{2}}{\mu \, r^{3}} = -F(r) - \frac{l^{2}}{\mu \, r^{3}} = 0
    [/tex]

    Using the force law [itex]F(r) = -\alpha_{n}/r^{n}, \; \alpha_{n} > 0[/itex] for an attractive force, we get:

    [tex]
    \frac{\alpha_{n}}{r^{n}_{0}} - \frac{l^{2}}{\mu \, r^{2}_{0}} = 0
    [/tex]

    [tex]
    r_{0} = \left(\frac{\mu \, \alpha_{n}}{l^{2}}\right)^{\frac{1}{n - 2}}
    [/tex]

    But, this is a point of extremum, to test whether the point is max or min, we need the second derivative:

    [tex]
    U''_{l}(r) = -F'(r) + \frac{3 \, l^{2}}{\mu \, r^{3}} = -\frac{n \, \alpha_{n}}{r^{n + 1}} + \frac{3 \, l^{2}}{\mu \, r^{3}}
    [/tex]

    [tex]
    U''(r_{0}) = -n \, \alpha_{n} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{n + 1}{n - 2}} + \frac{3 \, l^{2}}{\mu} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{3}{n - 2}} = (3 - n) \, \alpha_{n} \, \left(\frac{l^{2}}{\mu \, \alpha_{n}}\right)^{\frac{n + 1}{n - 2}}
    [/tex]

    We see that this point is a local minimum ([itex]U''(r_{0}) > 0[/itex]) if [itex]n < 3[/itex] and then we can have finite orbits. Nevertheless, these orbits need not be closed curves since the period in the radial direction is not necessarily equal to the period in the angular motion (as in the n = 2 case).

    The point is a local maximum ([itex]U''(r_{0}) < 0[/itex]) if [itex]n > 3[/itex]). In that case the motion is either infinite and the trajectory asymptotically approaches and recedes from infinity or, the particle simply falls in the center of force.

    The case [itex]n = 3[/itex] is marginal and was discussed in my previous post. It has the same behavior as the [itex]n > 3[/itex] case.

    Finally, if the force has that proprerty, then the potential energy is:

    [tex]
    U(r) = -\frac{\alpha_{n}}{n - 1} \, \frac{1}{r^{n - 1}} + C, \; n \neq 1
    [/tex]

    [tex]
    U(r) = \alpha_{1} \, \ln{r} + C, \; n = 1
    [/tex]

    If [itex]n > 1[/itex], then the potential energy vanishes at infinity and it is possible for a particle to free itself from the attractive force field and become free. If, on the other hand, [itex]n \le 1[/itex], then the potential energy at infinity is infinite and no particle can ever leave the force field.
     
  17. Aug 27, 2010 #16

    Cleonis

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    You are correct, and you demonstrate so impressively.

    Arguably, it's better to solve analytically what is solvable, and use numerical analysis only for cases that do not lend themselves to exhaustive analytic solution. Still, as confirmation I think it's worthwile to run the numerical analysis.
     
  18. Aug 27, 2010 #17
    Interesting! How about raising it to a 2/3 power? What happens then?
     
  19. Aug 28, 2010 #18

    Cleonis

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    Well, nothing spectacular.
    The general force law as a function of distance R is Rn. For n=1 and n=-2 you get orbits that loop back on themselves, for all other force laws with n>-2 you still get orbits, but they're not cyclic, and they all look pretty much the same.

    As Dickfore has shown, the case of n=-3 will crash the simulation. Taking smaller time increments only delays the inevitable.
     
  20. Aug 29, 2010 #19

    Cleonis

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    I have created the simulation for forces that vary with distance according to Rn, where n can be any number, not just integers. I called it 'Force laws simulation'. It's on my website now.

    You can find my website's URL on the contact tab of my physicsforums member profile.
     
    Last edited by a moderator: Apr 25, 2017
  21. Aug 29, 2010 #20

    DaveC426913

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    Cool! A force of 1/r^2.75 results in a double-four-leaf-clovered orbit.
     
    Last edited: Aug 29, 2010
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