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A change in the gravitational law

+ 6 like - 0 dislike
28 views

What would happen if the force of gravitation suddenly starts varying as $1$$/$$r^3$ instead of $1/r^2$ ? Would the symmetry of universe now seen be disrupted?

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user ARITRA
asked Oct 18, 2012 in Theoretical Physics by ARITRA (30 points) [ no revision ]
If you're asking physics questions to satisfy your own curiosity at 16 years old, you don't need to apologize for how it sounds. Good luck on your continuing education.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user AdamRedwine
Not a direct answer but might be of interest. There are three cases of a central power law where all the orbits can be written down explicitly using elementary functions---inverse square, direct proportionality (Hooke's law) and the $\frac 1{r^3}$ of your question. In the latter case, the orbits are the so-called Cotes' spirals about which the internet is full of information. By the way, all three were known to Newton and are dealt with in his "Principia".

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user mathuser4891

3 Answers

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It depends on what you mean by symmetry, But the force law $1/r^3$ does not have stable orbits. Small perturbations will de stabilize your orbits to fall inward or outward.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Prathyush
answered Oct 18, 2012 by Prathyush (690 points) [ no revision ]
As addition to this answer: http://en.wikipedia.org/wiki/Bertrand%27s_theorem.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user NiftyKitty95
Also the iron sphere theorem would not hold as it relies on area of a sphere increasing with $r^2$. Relativity gets around this by changing the geometry.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user SMeznaric
@NickKidman Wow! this and Bertrand's theorem are amazing! Thanks for the links. See also Dimension10's point 2 below and the Wiki page linked therein.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user WetSavannaAnimal aka Rod Vance
+ 5 like - 0 dislike

Such a change can only occur if space becomes 4 dimensional instead of 3 dimensional (see for example, my answer here). That will have quite profound implications. For example,

  1. The Riemann Curvature Tensor will have one more part (but this "part" means many more components) along with the Ricci Curvature Tensor and the Weyl Curvature Tensor. This actually related to the Electromagnetic force so in toher words, Gravity and EM will behave in the same way.

  2. The 4+1=5 dimensional spacetime is actually unstable (agreedly, I don't know why, but whatever), at least according to Wikipedia's "privileged character of 3+1 dimensional spacetime" article.

  3. The string theory landscape maybe a bit smaller (I think.).

  4. The Ricci curvature in a vacuum on an Einstein Manifold would no longer be exactly $\Lambda g_{ab}$. There will be an ugly coefficient of $\frac{2}{3} $

  5. The magnetic field could not be written as a vector, unlike the electric field. This is because it would have 6 components whereas the spatial dimension is only 4. So, perhaps humans would be more familiar with exterior algebras, earlier than us who live in 3+1=4 dimensions. Either that or we would be trying to find out how magnetism works. Or we would just die out, due to Implication (2).

  6. Cross Products would be harder to evaluate. Another reason why we would be more familiar with exterior algebras!

  7. Ok, so I'll come back to your question now:

So in 4D space the Newtonian gravitational force is: $$\vec F=G_5 \frac{m_1m_2}{r^3}$$ Here, $G_5$ is the gravitational constant in 4 spatial dimensions (5 spacetime dimensions, thus the 5 subscript.). It has to be different from the ordinary gravitational constant because it needs different units (since the $\frac{1}{r^3}$ immediately induces an extra units of $\operatorname{m}^{-1}$, so the Gravitational constant needs to be adjusted to have a units with extra $\operatorname{m}$, i.e $\frac{\operatorname{m}^4}{\operatorname{kg} \operatorname{s^2}}$. According to string theory, the gravitational constant gets multiplied by the string length for every extra dimension added: $$G_5=\ell_s G_4$$

The exact string length is obviously unknown (a free parameter of string theory, as far as I know!) but let us just say it is $1.2374713757$ multiplied by the 3+1=4-dimensional (i.e ordinary) planck length (it would be a coincidence if it were exact!). The 3+1=4-dimensional planck length is approximately $1.616199\times 10^{-35}$. Then, for the estimate, the string length would be approximately $2\times 10^{-35}$ (now you know why I chose 1.2374713757 !:) (not a factorial sign )). Then, $G_5\approx 2\times 10^{-35}\times6.673\times 10^{-11}=13.346\times 10^{-46}$

That is very weak gravity! Now, how heavy is an apple? It has a mass of say $0.1\operatorname{kg}$. Then, the gravitational force between that and the earth (in 5-dimensional spacetime,/4-dimensional space), is approximately around (after applying both the inverse cube proportion and the reduced, very weak, small, tiny, gravitational constant.),:: $$F\approx 4.82\times10^{-32} \operatorname{Newtons}$$$$ And in daily life, it is like 1 Newton!..

Finally, in summary, gravity, would be much weaker!.

P.S. Also see this recent paper: http://arxiv.org/abs/hep-ph/0011014.

answered Jun 17, 2013 by dimension10 (1,950 points) [ revision history ]
Minor comment to the answer (v2): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/hep-ph/0011014

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Qmechanic
@Qmechanic: Thanks for telling me. I edited my answer accordingly.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Dimensio1n0
+ 0 like - 2 dislike

some change will occur on earth as

  • gravity on earth will decrease.
  • earth will complete a cycle around earth in larger time.

and most importantly earth orbit will not be stable

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Rahul kumar walia
answered Aug 2, 2013 by Rahul kumar walia (-40 points) [ no revision ]
Instead of just putting these statements down, support them with valid physics arguments.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Dimensio1n0
we can show it by F= GMm/r^3 =mg, That's show that gravitational accelaration will decrese on earth .and same can be shown for the orbital motion for earth. Another explanation is that because 1/r^3 type of force do not support a stable closed orbit we can show it by mathematically.

This post imported from StackExchange Physics at 2014-03-30 15:52 (UCT), posted by SE-user Rahul kumar walia

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