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Obviously a balance scale is not going to add value to the task in free-fall. Weight in space is more accurately described as mass, but the need to define it still exists. Regardless of if you're doing scientific experiments, mixing ingredients, or calculating mass to thrust values for delta-v, you need to know the mass or weight of things.

So how do you figure out the mass of an object, if you can't use a balance scale?

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    $\begingroup$ You don't weigh things as such, because things are weightless, but as you rightly said, they have mass which doesn't require gravity to have a value $\endgroup$ – RhysW Aug 1 '13 at 14:09
  • $\begingroup$ "Weight in space is more accurately described as mass, but the need to define it still exists." . . . what? Weight is the force a mass produces in a gravitational field. If you don't have a gravitational field (or you're in a reference frame which considers gravitational acceleration to be zero, such as Earth orbit), then the "weight" of an object is literally meaningless. $\endgroup$ – imallett Oct 31 '14 at 3:37
  • $\begingroup$ @imallett The OP wants to measure mass. Weight is literally zero, but mass can be measured as inertia, as the answers show. $\endgroup$ – Qsigma Mar 23 '15 at 12:21
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For pretty precise measurement you use linear acceleration of the body with fixed force (say, spring pulled until its force reaches nominal value) and then you measure its speed when launched.

Kinetic energy $ E={{1}\over{2}}mv^2 $ will be equal to potential energy of the "launcher" (which can be easily calibrated by launching an object of known mass and measuring its speed using the same assembly).

Now, given known energy and measurable speed we can calculate the mass: $$ m={{2E}\over{v^2}} $$

For example, assume the launch assembly is $ 5 kg $. You launch $5 kg$ weight out of the assembly and measure the distance it travels over the course of 1 second. It goes $ 10 m/s $;

$$ E= {{1}\over{2}}(5kg+5kg)(10m/s)^2 = 500J $$

that's the force of the assembly.

We launch a human, and the result of measurement is $ 4 m/s $.

$$ m = {{2 \cdot 500J}\over{(4m/s)^2}} = 62.5kg $$ Substract the known $5 kg$ of assembly, and the human mass is $57.5 kg$.

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    $\begingroup$ This is one more case, where we need proper formulas ... $\endgroup$ – s-m-e Aug 1 '13 at 11:40
  • $\begingroup$ This is essentially what astronauts have been using. Dr. William Edgar Thornton filed a patent for a device like this in 1968 while working for NASA studying the effects that zero G has on human bodies, which included the need to measure mass changes. Today, a nearly identical device called the SLAMMD (Space Linear Acceleration Mass Measurement Device) is used to do the same thing (to measure astronauts' mass during prolonged stays on the ISS). $\endgroup$ – Lèse majesté Aug 1 '13 at 16:58
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You can basically use all effects, that depend on the mass of an object. Spring pendulums, some magic with centrifugal forces, angular momentums .....

For determining the mass of for instance a human body, see Space Linear Acceleration Mass Measurement Device.

There is some really cool website by NASA, which provides some nice insight and an experiment, which demonstrates an option of how it can work: http://quest.nasa.gov/space/teachers/microgravity/4inert.html

SLAMMD

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A quick search through the space agencies' sites yielded this photo of the IM-01M mass measuring device in the Russian orbital segment:

Anatoly Ivanischin undertakes periodic weighing of his body on the "Mass measuring" device. Credits: Roscosmos. The likely author of the photo is Anton Shkaplerov. Ivanisin

Peggy Whitson: We weigh ourselves with a Russian-built device called the body mass measuring system. It calculates inertia as it moves a mass back and forth on a calibrated spring. It's very simple system, but seems to work very well. After we calibrate the spring, we climb on the device and release the spring. Our mass is calculated from the inertial forces on the spring.

Source: http://spaceflight.nasa.gov/feedback/expert/answer/isscrew/expedition5/index.html

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  • $\begingroup$ @Philipp: I think this is a good answer in that it supplies evidence of an actual device in use. The principle is that a mass on a spring will oscillate (when pushed away from the unloaded length) per the equation here: en.wikipedia.org/wiki/Harmonic_oscillator#Spring.2Fmass_system so if you calibrate the spring, you can measure the frequency of oscillation and thus the mass attached to the spring. The bonus here is that it is contained, so doesn't require the space of a linear acceleration system. And a spring is easy to calibrate. $\endgroup$ – Phil H Aug 11 '13 at 20:15
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Following up on Deer Hunter's answer showing the Russian device, we can make a mass on a spring oscillate; this is the equation of motion:

$$x \left( t \right) =A \cos \left( \sqrt{k \over m}t \right)$$

(from Wikipedia's harmonic oscillator page)

This gives the frequency to be:

$$f = \sqrt{k \over m}$$

So, if we move the mass away from the rest position, we can measure the frequency of its oscillation over a period of time. If we have carefully calibrated the spring and know the spring constant $k$ to high precision, we can calculate the mass $m$.

The benefits of this over a linear acceleration device are its simplicity, robustness and that it is contained; the mass oscillates in place, rather than moving in space, and you need only measure the frequency which can be done over a number of oscillations. By comparison, a linear acceleration device requires knowledge of the force involved, careful measurement of velocity and is single-shot operation. There's not a lot of space in a space station!

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    $\begingroup$ ...although I believe in case of linear accelerators the comfort for weighing people is somewhat higher. Also, for the frequency-based measurements the body must be somewhat rigid. Oscillating parts inside it might falsify the results. $\endgroup$ – SF. Aug 12 '13 at 1:42
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You can use rotation to make a balance scale work. Attach the object being weighed on one end of a rod attached to a pivot and a moveable counterweight to the other. Only one position on the rod for the counterweight will allow the system to be balanced for any given test mass.

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    $\begingroup$ This is okay for objects with uniform mass distribution or much smaller than the pivot. If the object is large and e.g. most of the mass is concentrated on its top, there will be a significant error. $\endgroup$ – SF. Aug 1 '13 at 11:09
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The following is only an example. As ernestopheles already said, one could use any effect that depends on mass.

To measure the mass of small solid objects, I'd put a small sliding carriage between 2 springs of known properties. Place the mass in the carriage, deflect the carriage a bit and let it go. Measure the frequency of oscillation.

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