Aeronautical Engineering/Mmm ? Sounds UN-likely

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Question
FORCE

Radius of earth 6,371,000 meters + 400,000 meters up to ISS orbit height = 6771000 meters distance from earth barycenter into space (weightlessness).

Fg =  G x mass earth x 1 kg  /  6771000^2

Fg = .0000000000667384 x 5972190000000000000000000  /  45846441000000  =  8.69368257 Newtons

8.69368257 Newtons  =  .886508904122242 Kilograms

I don't think a 1 kg ball up in space near the ISS feels like it weighs .88 kilograms, do you ? ?

;0)
Sherice

Answer
Sherice - Your calculation is good, but you are getting mass and weight mixed up. A 1 kg mass remains 1 kg on the surface of the earth or in deep space or at the ISS. The weight (gravitational force) of that mass is 9.8 N at the earth surface and 8.69 N at the ISS altitude as you calculated. In other words:  w = mg. m does not change, but the acceleration of gravity,g, changes with altitude as does weight, w.

The reason the mass feels weightless on the space station is that the station is moving at just the right speed so the centrifugal force on every body in the station is equal to the weight of that body. Or to think of it another way, as the body falls the space station is moving in a circular path at the right speed so it falls at the same rate as the body.

Paul


p.s. Calculations are easier if you use decimal notation.

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Paul Soderman

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Aeronautics, Aerodynamics, Fluid Mechanics, Aeroacoustics, Noise Control, Muffler Design, Wind Tunnel Research.... I know nothing about India - do not ask about schools, jobs, application requirements, career choices, etc. for India. Please, no text message verbiage; I prefer full words in full sentences. Thanks.

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38 years as research engineer at NASA

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AIAA, NASA

Education/Credentials
B.S. and M.S. Aeronautical Engineering - U. of Washington, Graduate work Standford U.

Awards and Honors
AIAA Associate Fellow (American Institute of Aeronautics and Astronautics)

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