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If an object would be accelerated to the escape velocity would it's weight be equal to zero?

Say I throw a ball from Earths surface (assuming frictionless atmosphere) upwards at 11.186km/s is it's weight equal zero? I assume mass would be greater because energy is greater but what about weight?

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    $\begingroup$ The ball is weightless when it is freely falling regardless of whether it has escape velocity or not. As soon as the ball leaves your hand, it is in free fall and so has zero weight. $\endgroup$ – Alfred Centauri Nov 27 '17 at 0:55
  • $\begingroup$ Is it's weight unchanged then? $\endgroup$ – user2820052 Nov 27 '17 at 0:55
  • $\begingroup$ Are all gas molecules weightles as they swing around? $\endgroup$ – user2820052 Nov 27 '17 at 0:58
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Say I throw a ball from Earths surface (assuming frictionless atmosphere) upwards at 11.186km/s is it's weight equal zero?

The ball is weightless when it is freely falling regardless of whether it has escape velocity or not. In your example, as soon as the ball leaves your hand, it is in free fall and so has zero weight.

From the Wikipedia article Weight:

Thus, in a state of free fall, the weight would be zero. In this second sense of weight, terrestrial objects can be weightless. Ignoring air resistance, the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, is weightless.

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The weight is not zero in free fall, it is just not in equillibrum (ignoring air-friction) with another force.

The weight may vary with the gravitational potential and not with the acceleration of the object (given for other forces then the gravitational), nor with its velocity.

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  • $\begingroup$ Is that a yes to my question? :) $\endgroup$ – user2820052 Nov 27 '17 at 11:29
  • $\begingroup$ It sounds (correctly) like a NO to me. $\endgroup$ – Chet Miller Nov 27 '17 at 13:14
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    $\begingroup$ Quick question for the Rodrigo: after the ball is released, what is the acceleration of the ball as read by an accelerometer attached to it? $\endgroup$ – Alfred Centauri Nov 27 '17 at 13:37
  • $\begingroup$ The thing is: are you thinking in the newtonian picture, or in einstein picture? The accelerometer reads g. From a point of view of Newton, it is not an inertial observer, so their readings are not trustable to comply with newtonians law... From an Einstein point of view there is no such thing as weight... The weigth doesnt vary with velocity... (Already stated above), so no. $\endgroup$ – Rodrigo Fontana Nov 27 '17 at 14:38
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    $\begingroup$ Rodrigo, I suspect that you misread my comment because the accelerometer attached the ball reads zero after the ball is released. $\endgroup$ – Alfred Centauri Nov 27 '17 at 16:31
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According to the Newtonian model of mechanics, massive bodies exert forces on other massive bodies, even if one of the bodies is in free fall. We refer to this as the weight of the body. So, bodies, even in free fall, have weight.

According to the General Relativity model of mechanics, massive bodies do not exert forces on other massive bodies (unless the bodies are in contact). There is no such thing as gravitational force, so bodies do not have weight.

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  • $\begingroup$ "So, bodies, even in free fall, have weight." - not according to the operational definition of weight. See, e.g., THE IMPACT OF AN OPERATIONAL DEFINITION OF THE WEIGHT CONCEPT ON STUDENTS’ UNDERSTANDING $\endgroup$ – Alfred Centauri Nov 27 '17 at 15:44
  • $\begingroup$ A quote: "Weight is distinguished from mass and is often defined as the force of gravitation exerted on a particular object (“gravitational weight” approach). The complexity of this pedagogy can be appreciated if one pays attention to the conceptual break: the heaviness of objects is directly associated with weighing results and so corresponds to the operational definition. Instead, when weight is equated with the gravitational force, weighing results become unreliable since the interpretation of weighing is not univocal." $\endgroup$ – Alfred Centauri Nov 27 '17 at 15:48
  • $\begingroup$ This is all correct within the framework of Newtonian mechanics. Within the framework of Relativistic mechanics, none of it is correct. So it all depends on which model one is using, and on its applicability/limitations to your particular situation. Most people are comfortable working, where appropriate, with either approach. Under typical practical conditions, the error incurred by applying Newtonian mechanics is acceptable. But don't try to use it to predict the precession of the elliptical orbit of Mercury. $\endgroup$ – Chet Miller Nov 27 '17 at 16:29

protected by Qmechanic Nov 27 '17 at 13:38

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