# The distance square in the Newton's law of universal gravitation is really a square?

When I was in the university (in the late 90s, circa 1995) I was told there had been research investigating the $$2$$ (the square of distance) in the Newton's law of universal gravitation.

$$F=G\frac{m_1m_2}{r^2}.$$

Maybe a model like

$$F=G\frac{m_1m_2}{r^a}$$

with $$a$$ slightly different from $$2$$, let say $$1.999$$ or $$2.001$$, fits some experimental data better?

Is that really true? Or did I misunderstand something?

• Anyways, I doubt anyone would try to verify the inverse-square law in the late 90s (Unless you went to uni in 1890s..). I'm quite sure general relativity was sufficiently established, and it makes Netwon's law outdated. – Manishearth Mar 7 '12 at 16:10
• @Manishearth :-) It was 1990s and maybe the professor did not mention the year of the researches... I remember it was just an aside comment. – Alessandro Jacopson Mar 7 '12 at 16:25
• Asaph Hall lived from 1829-1907. It was in 1894, before even special relativity was born. At that time the anomaly of mercury was a problem that cried for a resolution. Einstein solved the problem much later. – Arnold Neumaier Mar 7 '12 at 16:32
• farside.ph.utexas.edu/teaching/336k/Newton/node116.html – Jim Graber Mar 7 '12 at 17:54
• You would have tested this in the 20th century if you were trying to detect extra dimensions in a braneworld type thing--then the EM forces are confined to a 3D brane and fall off like $\frac{1}{r^2}$, while the gravitational forces live in the bulk and fall off like $\frac{1}{r^{d-1}}$, where $d$ is the number of spatial dimensions in your theory. People were interested in seeing if there were short-distance deviations from the inverse-square law in order to give credence to extra dimensions. – Jerry Schirmer Mar 8 '12 at 18:28

## 7 Answers

This was suggested by Asaph Hall in 1894, in an attempt to explain the anomalies in the orbit of Mercury. I retrieved the original article in http://adsabs.harvard.edu/full/1894AJ.....14...49H

Interestingly, he mentions in the introduction that Newton himself had already considered in the Principia what happens if the exponent is not exactly 2, and had concluded that the observations available to him strongly supported the exact power 2!

The story is retold, e.g., on p.356 of N.R. Hanson, Isis 53 (1962), 359-378.

See also Section 2 of http://adsabs.harvard.edu/full/2005MNRAS.358.1273V

• Asaph Hall III (1829,1907) or Asaph Hall IV (1859–1930)? – Alessandro Jacopson Mar 8 '12 at 16:56
• I didn't know there were two of them (in fact father and son). Unfortunately, the publication (see the link in my edited answer) doesn't help decide your query, and I have no idea how to find out. – Arnold Neumaier Mar 8 '12 at 17:15

Let's first see why the inverse square form is special. Betrand's Theorem states that only two types of central potentials will produce stable orbits. The harmonic oscillator potential $V=\frac{1}{2}kr^2$ and the potential $V=-\frac{k}{r}$ that will produce an inverse square force law. Obviously the age of the universe is finite, so the fact that planet's orbits survived until now need not imply it will continue to be so in the future.

Another argument why this type of potential is so common is that, when doing quantum field theory, the propagator (details depend on whether the particle is a (gauge) boson, fermion or scalar, i will stick with scalars for now) has form

$$\frac{1}{q^2+m^2}$$

Thus if this particle where the force carrier of your force with coupling $g$ the potential is basically the fourier transform of the propagator

$$V(r) =-g^2\frac{1}{(2\pi)^3}\int d^3k\frac{4\pi}{q^2+m^2}e^{i\vec{k}\cdot \vec{r}} = -g^2\frac{1}{r}e^{-mr}$$

This is the famous Yukawa potential. For massless force carriers the damping term goes to 1 and the force becomes long range with a inverse square force law. Upto small details this is analogous to the gauge boson case, e.g. the masslessness of the photon makes the EM force long range, where as the massiveness of W,Z bosons make weak forces short-ranged.

Above derivations use the three space dimensions. Theories with extra dimensions have suggested that large extra dimensions will alter the inverse square law at some not-so-short distances (sub-mm range). Published experimental results are to be found e.g. from the Eöt-Wash group ( http://www.npl.washington.edu/eotwash/experiments/shortRange/sr.html ) and are available on the arXiv.

One potential tested here is here $$V(r)=-G\frac{m_1m_2}{r}(1+\alpha\exp(-r/\lambda))$$

The below plot shows the exclusion limits for both parameters $\alpha$ and $\lambda$ But of course, Newton’s theory is not correct; instead Einstein’s theory is correct. If you use general relativity GR, you usually talk about curvature, etc. rather than forces.
Nevertheless, the results can be expressed in terms of an effective force.

This reference http://farside.ph.utexas.edu/teaching/336k/Newton/node116.html gives

$F = -GM/r^2 -3GMh^2/(c^2r^4)$

where h is momentum as the first order correction. Higher orders have been calculated by the PPN and gravitational wave people. This correction is very small except for very fast moving objects. In practise, it applies to bodies orbiting very near to a black hole or neutron star. Famously, it is also responsible for the precession of the perihelion of Mercury.

• How do you know Einstein's theory is correct? Newton's theory sure seemed correct at the time. – wim Mar 8 '12 at 0:34
• Of course we don't know Einstein's theory is absolutely correct. Experiment (from Mercury onwards) tells us that it is more correct than Newton's theory. But many scientists, particularly the quantum gravity people, expect a further correction is yet to come. – Jim Graber Mar 8 '12 at 22:20
• It seems like "your" expression gives the right value for the perihelion precession, however it is not identical to the first order correction effective force used by JPL/NASA and that is the post-Newtonian expansion to the first order. That is at least if you approximate $h^2=GMr$ as they do in your link. – Agerhell May 5 at 18:36

There was indeed some talk of the exponent on $r$ during the late 90's and the early years of the 21st century. The problem, as I recall, was dark matter, which can only be observed indirectly by looking at the anomalous rotation of galaxies. It was suggested that perhaps Newton's law broke down under certain conditions. Again, as I recall, that while a number of papers were published, nothing much came of the idea.

• Milgrom's modified gravity was a deviation from Newton's law at various acceleration scales. Modified gravity is still somewhat alive, but the models have grown very complex. – Jerry Schirmer Apr 27 '13 at 16:56

Building on Jim Graber's answer:

We can absorb the perturbation term into the correction to the power law. $$F = -GM/r^2 \frac1{(r/r_0)^{\delta(r)}} \approx -GM/r^2 \left( 1 - \delta(r)(r/r_0 - 1) \right)$$

and we get

$$\delta(r) = -\frac{3h}{r^2c^2}\frac1{r/r_0 - 1}$$

I'm not sure about the physical meaning of $r_0$ though (renormalization scale?).

If $r/r_0 > 1$ then $\delta(r)$ is negative and we have $\alpha$ like 1.9999...

• Still, the expansion in terms of Yukawa potential is more physical – pcr Mar 13 '12 at 3:20

There has indeed been such research about the Pioneer anomaly: Two spacecrafts launched in the 1970s into the outer solar system did not move quite as expected (as calculated due to gravity and solar wind) after ca. 1980. Only in/after the 2000-2010 decade the source of the discrepancy, accidental thrust by thermal radiation, became generally accepted consensus. Previously, it was at least conceivable to interprete the data as containing hints of subtle differences between actual observed gravity and our theoretical understanding of gravity.

JPL who calculates the orbits of celestial bodies to a high precision uses an expression for the acceleration of one body of negligible mass due to the gravitational force of one other body that looks like:

$$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}{(1-\frac{4GM}{rc^2}}+\frac{v^2}{c^2})\hat{r}+ \frac{4GM}{r^2}(\bar{v}\cdot{\hat{r}})\frac{v^2}{c^2}\hat{r}$$

Ignoring velocity dependent parts we have: $$\frac{d\bar{v}}{dt}=-\frac{GM}{r^2}\hat{r}+\frac{4(GM)^2}{c^2r^3}\hat{r}$$

So actually keeping $$a = 2$$ but adding a small "inverse cubic" part is actually done do fit experimental values better, even though the inverse cube term is not invented but comes from trying to approximate general relativistic effects using what is known as "the post-Newtonian expansion".

The reason for this is a bit complicated, but basically adding a small inverse cube part as well as velocity-dependent parts explains what it is known as the "anomalous precession of perihelion".

See for instance expression 4-61 in this paper, titled "Formulation for Observed and Computed Values of Deep Space Network Data Types for Navigation"