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Question 3: Weightlessness Astronauts in orbit around the Earth are not so far away that they dont feel Earths gravity, ae that an identical 1 kg mass were resting on a scale sitting on the Equato, somewhere in Kenya. What would this scale read? Does the force of gravity alone provide the means for the centripetal force, is this why they are equal ? For explaining why there is no normal force, could you see it as the astronaut having the same acceleration as the ship and so the relative acceleration = 0, so it's under the illusion almost of having no acceleration so there's no force ? Please !! If you have different reasoning please share. I'm just trying to wrap my head around it so everything helps. Thank you. Also I am very lost on the second last part.

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Answer #1

Concept required- The question requires the concept of Gravity , mass and Weight of a body. Let's see them in our solution

Solution-

Part a-

The astronaut feels weightless in the space even under the influence of gravity because there is no external contact force which pushes or pulls their body. This situation is same as that of weightlessness felt during the ride of free fall ride

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Part b -

Here we have the free body diagram of an astronaut

Astronaut force due to gravity

We see that there is no contact which could provide a normal force to the astronaut and only gravity is acting

The weightless is just a feeling. The term just means that there is no contact forces acting on the object or person, which generally occurs in case of free-fall. An object always has a mass and hence can never be truly weightless. It's just a feeling, a sensation.

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part c-

the acceleration of the box at the equator will be. At the equator of earth , the weight is slightly different due to the centripetal force of the object. So some of the force of gravity is used to push us in the circular motion. Calculations show us that this centripetal force is about 0.03 m/s2.This force is "supplied" by the force of gravity, so the net acceleration due to gravity becomes 9.8-0.03 = 9.77 m/s2

Thus the weight of 1 kg mass at the equator will be

W = 1 X 9.77 = 9.77 N

Thus we can say that the value of 'g' is slightly lesser at equator than at other high latitude places on earth, but only just slightly

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Part d-

I think that there is no need to define the gravity in terms of the mass or normal force because as we saw earlier the variation in the acceleration due to gravity was due to the position of latitude and too was very minor. Moreover the normal force or mass has no effect on the value of 'g' . If we were to change the definition of 'g' , defining it in terms of position on earth would make more sense as we saw above

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Part e-

This sort of discussion will not be of any vale if you go to a field of engineering like computer science or electronics. If you go for civil engineering, or mechanical engineering, it might be of some value to you then

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