# Measuring Mass In a Weightless Environment. **URL:** \ **Category:** Factual Questions\ **Created:** [November 14, 2012, 8:31am UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969 "2012-11-14T08:31:45Z")\ **Posts on this page:** 8\ **Page:** 2
**Author:** ![Leo\_Bloom](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/leo_bloom/32/10377_2.png) [@Leo\_Bloom](https://boards.straightdope.com/u/Leo_Bloom)\ **Post date:** [December 6, 2012, 6:17pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/21 "2012-12-06T18:17:24Z")
This is related to the electromagnetic idea (or is it the same?): couldn’t you accelerate the mass and keep at it–presuming it stays in one piece–until it is reduced to its ultimate subatomic particles, and do a simple E=mc^2? ---
**Author:** ![Leo\_Bloom](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/leo_bloom/32/10377_2.png) [@Leo\_Bloom](https://boards.straightdope.com/u/Leo_Bloom)\ **Post date:** [December 6, 2012, 6:21pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/22 "2012-12-06T18:21:57Z")
Actually all of **am77494** , and all other mass measurements, are of course E=mc^2-able. Just getting the right hammer for the nail. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [December 6, 2012, 6:33pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/23 "2012-12-06T18:33:09Z")
> [@Leo\_Bloom](#): > > This is related to the electromagnetic idea (or is it the same?): couldn’t you accelerate the mass and keep at it–presuming it stays in one piece–until it is reduced to its ultimate subatomic particles, and do a simple E=mc^2? Your definitions of “stays in one piece” and “reduced to its ultimate subatomic particles” can’t both be the ones I’m used to. ---
**Author:** ![Leo\_Bloom](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/leo_bloom/32/10377_2.png) [@Leo\_Bloom](https://boards.straightdope.com/u/Leo_Bloom)\ **Post date:** [December 6, 2012, 6:47pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/24 "2012-12-06T18:47:34Z")
> [@naita](#): > > Your definitions of “stays in one piece” and “reduced to its ultimate subatomic particles” can’t both be the ones I’m used to. 🙂 Well, I meant before the crack(s) up. But couldn’t you do additive analyses? (Real question.) ---
**Author:** ![Bullitt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bullitt/32/5725_2.png) [@Bullitt](https://boards.straightdope.com/u/Bullitt)\ **Post date:** [December 7, 2012, 4:03am UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/25 "2012-12-07T04:03:41Z")
> [@Rachellelogram](#): > > Since density=mass/volume, could you use water displacement to calculate the mass? It’s more convoluted than the proposed methods, of course. That works if and only if the object floats. If the object sinks, the displaced water is only equal to the object’s volume, not the object’s mass. ---
**Author:** ![naita](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/naita/32/5862_2.png) [@naita](https://boards.straightdope.com/u/naita)\ **Post date:** [December 7, 2012, 2:34pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/26 "2012-12-07T14:34:22Z")
> [@echo7tango](#): > > That works if and only if the object floats. > > If the object sinks, the displaced water is only equal to the object’s volume, not the object’s mass. Since we’re discussing weightless environments, there’s no such thing as “floats”. ---
**Author:** ![HoneyBadgerDC](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/honeybadgerdc/32/1033_2.png) [@HoneyBadgerDC](https://boards.straightdope.com/u/HoneyBadgerDC)\ **Post date:** [December 7, 2012, 2:57pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/27 "2012-12-07T14:57:46Z")
> [@Malacandra](#): > > Take a spring of known stiffness, attach it to the mass, extend the spring and measure the acceleration of the mass. This assumes you have an “immovable” object to attach the spring to, so works best for smallish masses in a largish spacecraft. This is what i was thinking, something like a bow and arrow where you know the stored energy. ---
**Author:** ![Quercus](https://avatars.discourse-cdn.com/v4/letter/q/7ab992/32.png) [@Quercus](https://boards.straightdope.com/u/Quercus)\ **Post date:** [December 7, 2012, 5:45pm UTC](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969/28 "2012-12-07T17:45:54Z")
> [@Malacandra](#): > > Take a spring of known stiffness, attach it to the mass, extend the spring and measure the acceleration of the mass. This assumes you have an “immovable” object to attach the spring to, so works best for smallish masses in a largish spacecraft. This is basically what the Skylab device does (as described by Stranger). Only measuring acceleration directly is kind of hard to do accurately, so it’s easier to let the spring go back and forth, first pushing the mass and extending, then contracting back and pulling the mass, and measure the total time it takes to go through a cycle. You’re still measuring acceleration, but when you do it going both ways, it’s easier to measure accurately. Again, it’s only accurate as long as you’ve got a big “immovable” object to attach the other end of the spring to. [Previous page](https://boards.straightdope.com/t/measuring-mass-in-a-weightless-environment/640969.md?page=1)