If you are free-floating in space you have NO weight. Weight is based on gravity. Your weight on every planet would be different. Less on smaller planets or the moon and more on the bigger ones. There are many web sites that will show you your exact weight on any planet.
2006-06-25 21:47:25
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answer #1
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answered by Smart Dude 6
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It is depended on the gravity. If we near to any planet or any space thing, we have weight but we are alone in the space so no weight of the man in this place.
2006-06-26 08:43:38
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answer #2
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answered by rav 2
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If you are free-floating in space you have NO weight. Weight is based on gravity. Your weight on every planet would be different. Less on smaller planets or the moon and more on the bigger ones. There are many web sites that will show you your exact weight on any planet.
2006-06-26 07:16:36
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answer #3
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answered by Anonymous
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In the space - depends how far he's located from any planets...
If he's located in space without any plantets, moons, etc. his weight is 0 :) great for who needs to do a diet :)
2006-06-26 04:54:29
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answer #4
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answered by Arcady 3
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w = mg, if in the space where he is, has no gravitation, then there would not be any weight, only mass would be there.
2006-06-26 04:47:51
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answer #5
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answered by kalyan_panda 2
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Depends on how close they are to something with gravity and how strong the gravity is. If they are out away from everything they are weightless.
2006-06-26 04:46:25
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answer #6
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answered by draciron 7
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His weight is dependent on his proximity to another mass.
2006-06-26 04:46:49
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answer #7
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answered by slagathor238 5
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WHEN U R IN SPACE, THERE IS NO FORCE OF GRAVITY.IF THERE IS NO FORCE OF GRAVITY, THERE WUD NOT BE ANY WEIGHT IN THE SPACE.THE ONLY THING WHICH WUD BE THERE IS MASS.
2006-06-26 04:57:51
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answer #8
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answered by poppins_renu 1
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zero as
w=m * g
and g in space is zero.
yes his mass will remain same as it is independent of g.
2006-06-26 07:26:35
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answer #9
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answered by Garima M 1
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The weight of man in space is zero.......................................
2006-07-02 18:02:02
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answer #10
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answered by Anonymous
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