Weight of an Astronaut Between Two Stars - Newton's Laws of Motion Explained

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An astronaut positioned at the midpoint between two equal-mass stars experiences weightlessness due to the equal gravitational forces acting in opposite directions, effectively canceling each other out. While the astronaut is indeed weightless, this state is not solely due to the gravitational balance; it also relates to the concept of free fall. In orbit around a single star, an astronaut would also feel weightless because they are in a continuous state of free fall towards the star, despite the presence of gravitational force. The discussion emphasizes that weight is defined by the force exerted on an object, which can be influenced by the surrounding gravitational environment. Understanding these principles is crucial for grasping the nuances of Newton's laws of motion in space.
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I'm back again, this time regarding questions relating to Newton's laws of motion. :)

Homework Statement


Imagine an astronaut in space at the midpoint between two stars of equal mass. If all other objects are infinitely far away, how much does the astronaut weigh? Explain your answer.

The Attempt at a Solution


I'm not sure how to answer this, my initial thought was weightless, because equal forces of gravity are acting in opposite directions.
 
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He would be weightless, yes, but not for the reason you mention.

If you think about weight of an object as how much force is applied to the surface of this object (i.e. how much force an astronaut feels on his feet when standing), then under what circumstances would there always be weightlessness? Would an astronaut in orbit around a single star be weightless despite that he is accelerated by gravity?
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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