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'Mouse-Tronaughts' to Test Low-Gravity in Space

RandBlade writes "The Telegraph has an article about plans to launch mice into space with simulated low-gravity for five weeks, to test the effects of low-gravity on their bodies. This "will be the first time mammals of any kind have lived in partial gravity for an extended period." Hopes are that this will provide information useful for plans to launch men to Mars, which has one-third of the gravity of Earth."

12 of 276 comments (clear)

  1. Re:Pardon? by worst_name_ever · · Score: 5, Informative

    RTFA... partial gravity != microgravity.

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  2. Re:First time for mammals by hokanomono · · Score: 4, Informative

    The point is: there is no documented experiment of humans living in partial gravity for an extended period.

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  3. Re:They contradict themselves in the article by epiphani · · Score: 1, Informative

    *thwap*

    zero gravity != partial gravity

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  4. Greek for mouse by kyknos.org · · Score: 2, Informative

    greek for mouse is mys ... so it is an astromys :o)

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  5. Re:They contradict themselves in the article by Mwongozi · · Score: 4, Informative

    Or you could try reading the article. Partial gravity is not the same as a weightless environment.

  6. Re:They contradict themselves in the article by Anonymous Coward · · Score: 3, Informative

    *thwap*

    zero gravity != partial gravity


    Go thwap yourself then. Gravity is never equal to zero. Every object in the universe attracts every other. If you have a calculator, determine the force from gravity applied to a human on earth. Then, calculate again from 1,000 km away. It's a small difference.

    In orbit, you experience weightlessness. IE, if you are travelling at 20,000 km/hour around the earth, and everything else on your spaceship is travelling at the same velocity, from your point of view you experience weightlessness. From earth, watching the spacecraft, everything looks normal.

    Go read a high school physics book, will you? Pay attention to frames of reference.

  7. Re:You mean astromouse ? by NonSequor · · Score: 4, Informative

    Actually astrum (star), is Latin. And nauta is Latin for sailor. So you need the Latin word for mouse which is mus (pronounced like the English word moose).

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  8. you are wrong by kyknos.org · · Score: 2, Informative

    astronaut is of greek origin indeed. latin for star is stella

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  9. Re:How are the mice supposed.. by Anonymous Coward · · Score: 2, Informative

    1.) it's not floating, there's microgravity, so while it won't be firmly attached to the ground, it will get there eventually.

    2.) you can't swim in space. swimming requires pushing against something (usually water), and it would take one hell of a lot of paddling to push enough air to push yourself.

    3.) I'm sure they'll come up with some way of feeding the mice. They usually do think of almost everything for these missions, and I'm sure the article didn't want to talk about such a mundane and trivial topic.

  10. Re:isn't this pointless? by Anonymous Coward · · Score: 1, Informative

    You mean F = MA ?

    If by "the effect of gravity" you mean force then yes it is related to the mass (not weight) of the object.

    So, yes an animal with a smaller mass would have a smaller force directly proportional exerted on it.

    But, the acceleration of gravity due to the planet would be constant regardless of size of the object.

    ?? surface-oriented instead of gravity-oriented ??

  11. Re:One difference... by Artifakt · · Score: 2, Informative

    Foo still falls through wire screen floors in 1/3 G nicely. Even for older experiments in microgravity, a small downward air flow, adding only about a 0.5 cm/second/second to the forces acting on the mouse, was enough to keep things about as clean as most privately owned mice have it.

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  12. Re:You only added 1km, not 1000km. by NeoThermic · · Score: 2, Informative

    Ohh, right, good eyes...
    thats the effect of having the units in m in the equation and forgetting that your challenge has been set at 1000km

    g=GM/r^2
    = 6.7 * 10^-11 N m^2 kg^-2 * 6.0 * 10^24 kg/(7.4 * 10^6)^2
    = 7.341 N kg^-1


    So a diffrence of 2.473 N kg^-1

    NeoThermic

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