In terms of payload the sphere encloses roughly half the volume of a sphere of the same size (side and diameter) and the cost per volumetric unit comes out the same. In that context the usage is a wash.
A sphere is stronger than a cube, however. Where the sides and corners meet on a cube are weak points and require reinforcement. This adds to the cost and weight (with weight being the more important vector).
Advances in construction technology and materials has largely made this a moot point though. The only forces a satellite has to contend with are launch stresses (pack it well) and gravity (shape is irrelevant).
Not all satellites are spheres. The Iridium, by Motorola, is a rectangle.
2006-07-28 23:51:04
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answer #1
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answered by Anonymous
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Orbiting satellites, or Deep Space Satellites.
Orbiting Sat's. have to use Earths gravitational pull, therefore they must be the same shape as Earth. And even then corrections need to made once in a while, they have little rockets for that, too.
Deep space Sat's Can be any shape.
2006-07-29 06:37:18
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answer #2
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answered by Anonymous
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External forces will go tangential, it wont bounce back. Cubes receive horizontal and vertical force that will just reflect.
2006-07-29 06:33:20
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answer #3
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answered by Anonymous
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I think it might be to to reduce the chances of static electricity.Also a sphearical shape can hold maximum energy in it.
2006-07-29 06:42:05
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answer #4
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answered by Anonymous
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spheres present far less resistance to friction when they are sent through our atmoshere out into space.
2006-07-29 06:34:01
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answer #5
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answered by ha_mer 4
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http://en.wikipedia.org/wiki/Satellites
2006-07-29 06:32:02
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answer #6
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answered by Nickname 5
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becase of there shape they can easily move in space
2006-07-29 06:38:22
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answer #7
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answered by rajesh b 1
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all r not spheres
2006-07-29 08:59:00
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answer #8
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answered by jalaj 2
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they are spherical to collect signals and reflect them to the same point
2006-07-29 18:27:55
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answer #9
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answered by koki83 4
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i think its because of gravitational force
2006-07-29 06:31:41
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answer #10
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answered by Anonymous
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