The hypothetical, non-existent crowdfunding site go-launch-me is so confident, they are offering a two-for-one special. For a limited time, if your project is funded they will foot the bill for a 2nd project of identical cost and complexity.

I've decided to propose a Gold Ball project, an $m=$ 1000 kg sphere with a tiny laser interferometer and small camera on one side (and an equal one on the other for symmetry) to be launched into space and put in a distant retrograde orbit around the Earth, or perhaps heliocentric at or moving towards Sun-Earth L4 or L5. Some place far from strong gravity gradients.

Per the site's terms, if funded, they will pay for a second identical ball, so I will have two optically polished gold balls in space.

I propose that they be put in a tight orbit around each other. I estimate that the radii of the balls is $r=0.23$ meters, if they are placed at $R=0.5$ meters center-to-center there will be a 4 cm average gap between them if they orbit their common center of mass, their orbital period should be:

$$ T^2 = \frac{4 \pi^2 R^3} {2GM} $$

or about 101 minutes.

Wikipedia gives the value of G as

$$6.67408(31) \times 10^{-11} m^3 kg^{-1} s^{-2}$$

and the 31 is the uncertainty in the last two decimal places, so its roughly 50 ppm. The article goes on to report two very recent measurements at 12 ppm but they differ by almost three times that.

Their two laser interferometers are at different wavelengths, and so you can more easily measure the absolute surface-to-surface distance without the integer-fringe ambiguity that you might get from a single wavelength.

The tiny cameras in each sphere image the reflection of the Sun in the convex, polished gold surface of the other sphere, so precise angular information versus time is measured along with precise separation information. The orbits will be slightly elliptical (nothing is perfect) but that's pretty easy to account for.

Effects of solar photonic and wind pressure on their mutual 101 minute orbit cancel, since the spheres are identical.

Question: What could go wrong? Why might these gold balls not be a great way to measure G? Are there secondary forces or torques that spacecraft experience that could interfere with the measurement, or aspects of spacecraft design that would complicate the design of the golden balls beyond practicality?

To address @RussellBorogove's concern that this is too hard and should be in Physics SE, in the first experiment we'll measure the sum of both the "normal" $-\frac{GM}{r}$ potential plus any yet-undiscovered short-range gravitational effects.

note: hat tip to @rob for mentioning their answer in Physics SE which discusses a real-world consideration of such an experiment!

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    $\begingroup$ It is probably important to account for spacecraft charging. With two giant conductors being barraged by charged solar particles, electrostatic effects probably can't be ignored. $\endgroup$
    – Lex
    Commented Jan 29, 2019 at 6:32
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    $\begingroup$ Check your ball size calculation. I get r = ~*0.231* m. The density of gold is 19,320 kg/m^3, so they have to be a lot less than a cubic meter. I'm also going to mention charging (thanks, @Lex, for bringing it up), but the primary mechanism is irradiation by solar UV (and shorter wavelength) photons that eject electrons. All is not lost: put the two gold spheres inside a uniform, opaque sphere with a thermally controlled interior surface (to balance radiation pressure). All the equipment outside of the thermal sphere must be mass-balanced to cancel their gravity effects. $\endgroup$ Commented Jan 29, 2019 at 6:41
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    $\begingroup$ Do you know about Gravity Probe B? They had similar concerns with their 'gyros'/balls. $\endgroup$ Commented Jan 29, 2019 at 13:25
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    $\begingroup$ Current price of gold is $1311.90/oz goldprice.org 2000 kg = 70,400 oz. So that's a bit over 92 million just for the metal. Good luck with funding that :-) As a practical matter, depleted uranium is d denser than gold, and.much cheaper ($5/lb, per Google). (Osmium is even denser, but it's expensive, and the total world production is something like 500 kg per year.) $\endgroup$
    – jamesqf
    Commented Jan 29, 2019 at 19:48
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    $\begingroup$ @jamesqf The gold isn't "used up" by the experiment, when its over one could go get it and bring it home and return it for a refund I suppose, but I guess that space pirates could be a real risk then. $\endgroup$
    – uhoh
    Commented Jan 29, 2019 at 22:15

3 Answers 3


Lifting mass into space is expensive. Taking mass down from space is also expensive. Lifting large masses of gold is even more expensive. Insuring that you can bring the gold back is another expense. Not having defined what the experiment is measuring, means we have an expensive experiment without a well known outcome. Without having investigated all of other secondary effects which may affect the experiment means we wouldn't now how to set it up; or know what we would be measuring even if we recorded a result. Not having eliminated other ways of measuring G constant seems problematic as well.

All in all, this proposal seems limited by many issues. However not is all hopeless.

Back to the drawing board:

  1. Define what we want to measure
  2. Investigate why we can't get the same results on earth/ why we need to measure this in space. (what Secondary effects are we eliminating by performing this in space?)
  3. Ok so we have now decided that it really has to be in space now
  4. Investigated all of the expected secondary effects that may occur which would messup the experiments (say ask on Physics SE)
  5. Ok now ere are cooking!
  6. (Given what was found in step 4), Investigate how we could build a space craft which will
    1. Deploy the experiment to space,
    2. Measure the primary effect we are investigating,
    3. Reduce the secondary effects,
    4. Measure how much those secondary effects are affecting the experiment
    5. and perform this as cheaply as possible. (You could ask that as a question on Space SE perhaps).
  7. Go back to step 1 and see if step 6 could answer the question.

So this is hard to make work well. Really hard. The main problems are:

  • Obtaining very accurate relative measurements of the positions of two objects without influencing them.

  • Eliminating external forces and creating two "perfectly free falling" objects.

These are (more or less) the two things that certain proposals for measuring gravitational waves run head long into. Some pretty good work to progress solutions to these problems has been made though. LISA pathfinder springs to mind. They did't use 1 Tonne gold balls, they didn't want the 2 masses to interact. But other than that the basic idea's pretty much the same. Scaling it up should be just a matter of cost. You said you had a two-for-one deal right...


Challenges to measuring G in a laboratory between two known masses include accounting for other forces and handling the huge force from Earth's gravity. Going to a distant Earth orbit or heliocentric orbit removes the later but lets see what other forces show up with two bare masses and the space environment as laboratory.

The gravitational force between the two identical masses $F_G = \frac{G m^2}{R^2}$ is about 2.67E-04 Newtons. Let's see how some other forces stack up against that.


Comments 1, 2 immediately pointed out that spacecraft charging is an issue.

Let's estimate how much charge would cause say a 1 ppm change in the force between the two spheres ($F_E$ six decimal places lower than $F_G$) and see how easy it would be to measure that to better precision.

The Coulomb force between two point charges $+q$ separated by $R$ is

$$F_{Coulomb} = \frac{q^2}{4 \pi \epsilon_0 R^2}$$

Charge on the two spheres will tend to move to the outside faces which means we can't use Newton's Shell theorem like we do for gravity, so that would be an upper limit. Luckily this is well studied. In Table 1 of On approximate formulas for the electrostatic force between two conducting spheres (viewable here) for $d/a$ of 2.2 we see that the numerically calculated (and series expanded) reduction is about 0.14 relative to the Coulomb point or shell force above. I'll just keep that factor.

$$F_{Electrostatic} \approx \text{0.14}\frac{q^2}{4 \pi \epsilon_0 R^2}$$

Setting $F_{Electrostatic} = 1 \times 10^{-06} F_{G}$ I get only 2E-03 Coulombs which is pretty tiny! Were it distributed evenly on a 0.23 m radius sphere, the electric field on the surface $E=\frac{q}{4 \pi \epsilon_0 r^2}$ (r=0.23 meters) would be about 40 Volts per meter, or 400 mV/cm. That's certainly measurable (deep space scientific spacecraft measure DC and LF fields far lower) but it becomes a serious undertaking. The field between the two spheres would be lower as discussed previously, and would be the relevant field to measure.

Comments mentioned positive charging by photoelectron generation. There might be some negative charging from the solar wind as well. Either way, a charging management system will be necessary, and that complicates the design, the smooth shape, and the mass distribution.


For even 1 ppm you need to manage the mass drift and uncertainty of the 1,000 kg to 1 gram. For eight digits; thats 10 milligrams. There's no easy way to measure the sputter rate of the outer gold surface directly, though it can be done at certain locations using the resonant piezo monitors used for film thickness measurement in evaporators.

What about meteorite impacts? They can add or remove mass and are difficult to quantify.

The electronics and batteries could be subject to outgassing over extended periods of times, that would have to be managed by sealing the entire system and containing all outgassing products. Pressure rise could make internal arcing of even low voltages a problem.

All in all, the answer to

Two 1000 kg gold spheres orbit their CM in near-contact, great way to measure G or limited by spaceflight issues?

is probably:

No, not a great way, and yes indeed, probably limited by spaceflight issues.

As @TomSpilker mentions, building some kind of shield around the pair could go a long way, but that's beyond the scope of the question.


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