# Evidence that stationary masses in space actually attract each other

I'm finding it rather difficult to find experimental evidence that two stationary masses in space (unaffected by external massive bodies or gravities) actually attract one another. For moving masses, this is abundantly clear (planets, asteroids, etc.), but who has actually tried to measure forces of attraction between objects stationary in space with respect to the Sun, and has found through experimentation that the hypothesis of gravity being proportional to motionless masses is true?

I'm aware of the Cavendish experiment, however, this experiment is not what I'm looking for because the two balls are moving with the Earth, so they are not completely without motion with respect to the Sun. Relative to each other, the balls are stationary, but I am looking for an experiment conducted where there is no motion in the massive objects relative to the Sun.

• There is no absolute reference frame, so one cannot speak of things that are not moving in space. Everything is moving in some reference frame and stationary in some other. In each reference frame, the laws of physics must be the same. – Anthony Jan 20 '15 at 4:18
• The universe has no definable center. You might try to argue for the center of the observable universe, but this does not help us because all observations have shown the universe to be isotropic and homogeneous at the largest distance scales. – Anthony Jan 20 '15 at 4:24
• If you sent a rocket up in space to avoid moving "with" the Earth, then someone could object that you are still orbiting the center of the galaxy. If you sent a super deep space probe to leave the galaxy that would take a very very long time, and you'd still be in the thrall of the nearest supercluster. Eventually you realize there isn't much basis for saying whether you are at rest or moving, only that there is relative motion or relative rest. – Timaeus Jan 20 '15 at 4:25
• @Alexandru We don't "fail" to take motion into account, what we do is predict how relative motion changes in time based on what it current is. Because that's what we can measure and what we can predict. It is unfair to say we failed to take into account something that can't be measured, when what we did is used the available information and made detailed, testable predictions. – Timaeus Jan 20 '15 at 5:08
• @qweilun, et. al. Actually, I don't think that you can meaningfully talk about motion "relative to the observable universe". My understanding is that we are always at the center of the observable universe, because we are the observer. If so, then motion relative to the observable universe, is just motion relative to ourselves. – RBarryYoung Jan 20 '15 at 20:54

Your concern seems to be that the law of gravity might depend on whether the two bodies are moving relative to a third, external reference frame...

For example, you could do the Cavendish experiment in a lab in a University basement. Then you could load all the apparatus on a train and repeat the experiment while the train was rolling along (assuming no vibration and bends and jolts). You wonder if the results might be different, am I right?

I think you'd find there would be no difference and here's why:

Take a brick and count up all the protons and neutrons in it. Using the mass of the particles from particle physics, calculate the mass of the brick. Keep it stationary on the Earth and weigh it. This gives you the gravitational force due to the mass of the Earth.

Now watch the Moon as it orbits the Earth. Calculate the gravitational force that is required for the Moon to orbit as it does.

Taking account of the fact that the Moon is further away than the brick, check if the two results for the strength of the gravitational field are the same.

If they are, then the gravitational strength doesn't care if the object being attracted is moving (Moon) or stationary (brick).

• This is actually the best answer I think, because it proposes a very simple mechanism that experimentally refutes the proposal. You look at the relative difference of "Cavendishish" attractions for two lab-stationary balls and two lab-moving balls. – BjornW Jan 20 '15 at 11:13
• Isn't that essentially the Michelson-Morley experiment, except they used the Earth's rotation around both its axis and the sun in place of a train. – Pieter Geerkens Jan 20 '15 at 23:22
• Now that you mention it, I can't help but wonder - could the gravitational effects of the moon have affected the Cavendish experiment? – Alexandru Jan 21 '15 at 3:44
• What about the gravitational effects of the Sun? We're talking about very small forces here. Can a torsion balance actually rule those out? If not then, it would mean that G is not a valid gravitational constant. It may be valid on or around Earth, but what experiments have proven that it is valid everywhere? On top of this, are all the masses of other planets that we have calculated based on G? Because if so, their calculated surface gravity might not be the same as their experimental surface gravity might be. – Alexandru Jan 21 '15 at 14:15

The first measurement of the gravitational constant was done by Henry Cavendish in a lab, in which the gravitational force between two lead balls was measured. They weren't moving. http://en.wikipedia.org/wiki/Cavendish_experiment

• Yes they were! They just weren't moving relative to one another. – Alexandru Jan 20 '15 at 3:46
• Every object in the universe is moving with respect to something else. They were attracted to each other in their own "stationary" reference frame. – philn Jan 20 '15 at 3:54
• "What are some of the theories that make a claim that no single object in the universe is stationary relative to, well, the universe?" Newtonian physics (really starting with Galileo) and Einsteinian physics. Ernst Mach had some very interesting things to say on these matters around the turn of the 20th century, which influenced Einstein very much. – dmckee --- ex-moderator kitten Jan 20 '15 at 4:08
• @Alexandru There is no such thing as "motion relative to the universe", it doesn't mean anything in physics. So if you use that term, you need to explain what you mean by it in physics terms. Perhaps relative to CMBR, as measured with red/blue shift? – hyde Jan 20 '15 at 6:19
• @Alexandru: Brian Greene's "Fabric of the Cosmos" has a very well-written chapter (Chapter 3, to be exact) on this aspect of and the development of our understanding of spacetime. Perhaps it can help you grasp the concepts involved better than a physics-oriented answer. – rubenvb Jan 20 '15 at 9:18

The concept of a body being "completely without motion in space" is thoroughly discredited, and you really should take the other responders seriously on this point.

According to Einstein's General Theory of Relativity, the nearest we can come to this state is by being in gravitational free fall. But now the Solar System provides the perfect test that you are looking for: every non-artificial body in the Solar System is in gravitational free fall. (OK, there are exceptions: for instance, when a small comet approaches the Sun, it might emit enough gases to measurably change its trajectory. But this is a side issue.)

In short, what you are looking for does not exist. It's like asking whether anybody has travelled to the edge of the world to find out whether the world has an edge or not. "But...but nobody's ever gone there! How can we know it doesn't exist if nobody's ever gone there?"

The biggest point is of course that all motion is relative, as many have pointed out. Even barring that, pretending that we live in Copernican times and assuming that the Sun is the center of the universe, the question posits that the force of gravity might be proportional to mass and velocity. This would mean something like $$\vec F\left( \vec v, \hat r, \frac{Mm}{r^2} \right)$$ Terms like $\vec v \cdot \hat r$ and $\vec v \times \hat r$ are ruled out by the observation that $\vec F$ is parallel to $\hat r$. So, we have $$\vec F = -G' \frac{v}{c} \frac{Mm}{r^2} \hat r$$ where $G'$ isn't necessarily the $G$ measured in the Cavendish experiment (constraints will be derived). Since the sun and the orbiting object might be moving in our Copernican/Newtonian "universe", for the solar system this becomes $$\vec F = -G' \frac{\left| \vec v + \vec v_{orb} \right|}{c} \frac{Mm}{r^2} \hat r$$ There are two simplifying limits, where the sun's velocity is 0 or much greater than the planet's orbital velocity.

Case 1: Sun's "universe" velocity is zero

Then, for an orbiting planet, the acceleration is $$\vec F = - \frac{G'Mm}{c} \frac{|\vec v|}{r^2} \hat r$$ This is still a centripetal force (with no torque, $\vec r \times \vec F = 0$). So, it works out that angular momentum $L$ is conserved. So, the effective one dimensional DE is, since $L = m r^2 \omega$ and $v = r \omega$, $$\frac{d^2 r}{dt^2} = - \frac{G'M}{c} \frac{r \omega}{r^2} + \frac{L^2}{mr^3} = \left(L^2 - \frac{G'ML}{mc} \right) \frac{1}{r^3}$$ This is directly integrable! Defining $$A = L^2 - \frac{G'ML}{mc},$$ the solution (starting from a point where dr/dt = 0) is $$r(t) = r_0 \sqrt{1 + \frac{At^2}{r_0^4}}$$ Three subclasses:

Subcase 1: $L > G'M/mc$ so that $A > 0$

This describes an orbit spiraling away to infinity, horribly contradicting observation.

Subcase 2: $L < G'M/mc$ so that $A < 0$

This describes an orbit spiraling in. Also bad.

Subcase 3: $L = G'M/mc$ so that $A = 0$

Fine, a falling straight trajectory.

So, this horribly fails.

Case 2: The Sun's "universe" velocity is large

The orbital velocity is much less than the speed of light, but the sun's might not beso expanding in series $$\frac{\left| \vec v_{sun} + \vec v_{orb} \right|}{c} \approx \frac{v_{sun}}{c} \left(1 + \frac{\hat v_{sun} \cdot \vec v_{orb}}{v_{sun}} \right)$$ This leads to two terms, the first of which is just Newton's laws (absorbing $G'v_{sun}/c$ into $G$) and the second of which is a perturbation: $$\vec F = - \left(1 + \frac{\hat v_{sun} \cdot \vec v_{orb}}{v_{sun}} \right) \frac{GMm}{r^2}$$ I don't intend on solving this more complicated equation, but I'll point out that if you did so (analytically or numerically), you could use observational data to put constraints on $v_{sun}$, which I believe would be severe.

Edit: This one sneaks through to lowest order if you assume that the sun's motion is normal to the solar system's plane; you'd have to look at nasty second order terms.

• What if its not orbital velocity that I mean to take into account, but if I were to postulate that a massive object's spin on its own axis is what defines its gravity, and that orbital velocity is a byproduct of competition for space due to two definining gravities of massive objects trying to compete for a remote space? – Alexandru Jan 20 '15 at 14:47
• I'd point out to you that Mercury and the Moon are tidally locked, with their periods of rotation equal to their periods of revolution, and Venus as well has a very long day. – jwimberley Jan 20 '15 at 15:00
• That's a great counter-point. Although Mercury is not tidally locked to the Sun (it was originally thought that it was, but I looked it up and it turns out that its not)...what if it is just a coincidence that the Moon happens to spin at a rate such that it's considered to be tidally locked? Can we really know for sure that this disproves my postulation? – Alexandru Jan 20 '15 at 15:08
• This concept could explain quantum entanglement, whereby an electron's spin is what tries to define its space, and competes with an entangled electron that is on the same axis, such that it transmits this energy for space competition through light in such a way to make the other, tangled electron, spin the opposite way. Thereby spin defines gravity and light is a measure of space competition (or quantum entanglement, depending on how you want to look at it). – Alexandru Jan 20 '15 at 15:22
• @Alexandru TBH, it seems that you really aren't paying attention to anything anyone is telling you. Wondering if something is possible (gravity dependence on velocity) and being curious is good, but refusing to listen and be critical of your own ideas is not. Your new postulate about orbital speeds is absurd. If you are serious about your proposal, why don't you compare the rotational speeds of planets with the force of gravity that binds them to the sun? You'll see that there is no relationship. It's easy for things to make sense in your head by ignoring reality. – jwimberley Jan 20 '15 at 16:16

I think Cavendish demonstrated this:

http://en.wikipedia.org/wiki/Cavendish_experiment

• I know Sofia. A good number of people on this site are vitriolic if you don't see things their way. There is much arrogance here as well. – Inquisitive Jan 20 '15 at 3:39
• Interesting, thank you for your answer, but I will update my question because this isn't what I am looking for, and I will explain why... – Alexandru Jan 20 '15 at 3:42
• It wasn't me that downvoted, but that seems to happen to me all the time on Stack Overflow. ;) I updated my question by the way. – Alexandru Jan 20 '15 at 3:46
• This is, in fact, and example of the very bad category of answer known as "link only". It could be improved considerable by describing the basics here. – dmckee --- ex-moderator kitten Jan 20 '15 at 4:11
• There is a way to convey truth. Insulting people is not the way. Guaranteed. – Inquisitive Jan 20 '15 at 16:54