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As we know, that in a vacuum, bodies with different masses fall with same acceleration, which is due to acceleration due to gravity.

But Weight of a body is dependent on mass as well.

So, is it so, that weight has no function in vacuum?

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  • $\begingroup$ What do you mean by “function “? $\endgroup$
    – Bob D
    Commented Aug 29, 2020 at 7:28
  • $\begingroup$ By 'function', do you mean 'role'? $\endgroup$
    – Tony Stark
    Commented Aug 29, 2020 at 7:39
  • $\begingroup$ Weight is the force associated with the acceleration. Also, why do you ask about the vacuum? Do you perhaps mean the absence of gravity? $\endgroup$
    – my2cts
    Commented Aug 29, 2020 at 10:56
  • $\begingroup$ It is not clear what you are asking. A vacuum has no bearing on the concept of weight. It seems as though you may be thinking of general relativity, in which gravity only has meaning as an inertial (or "fictional") force. Since weight depends on gravity, it only has meaning in the same sense. $\endgroup$ Commented Aug 29, 2020 at 18:55

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It is precisely because the weight (i.e. the force on the object in a Gravitational Field) depends on mass that two objects of different masses fall with the same acceleration.

From Newton's Second law, we know that the acceleration a body experiences when a force is applied to it is given by $$\mathbf{a} = \frac{\mathbf{F}}{m},$$

in other words if a constant force $\mathbf{F}$ is applied to two objects with masses $m_1$ and $m_2$, and $m_1 < m_2$, then $\mathbf{a}_1 > \mathbf{a}_2$: the heavier mass will accelerate less.

Of course, the force due to gravity is not constant for objects that have different masses. From Newton's law of gravitation, we know that the force on two different masses will be different (i.e., they have different "weights"), and that

\begin{aligned} \mathbf{F}_1 &= -m_1 \frac{G M_\text{earth}}{R_\text{earth}^2}\mathbf{\hat{r}}\\ \mathbf{F}_2 &= -m_2 \frac{G M_\text{earth}}{R_\text{earth}^2}\mathbf{\hat{r}} \end{aligned}

Using Newton's Second Law, this means that:

$$\mathbf{a}_1 = -\frac{G M_\text{earth}}{R_\text{earth}^2}\mathbf{\hat{r}}=\mathbf{a}_2.$$

This constant acceleration is what we call $g=G M_\text{earth}/R_\text{earth}^2$. In other words, it is because the weight is proportional to the mass of an object that the net acceleration it experiences on the surface of the Earth is independent of the mass!

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  • $\begingroup$ Note that weight is not well defined in physics. Some authors define weight as the force of gravity on an object, and that's the definition here. The more common definition is weight is the force needed to keep an object in equilibrium against the force of gravity. In the first definition, an astronaut on the ISS has weight. In the second, more common, the astronauts are weightless, and furthermore, weight in that definition is what a bathroom scale measures.. The Wikipedia page referenced is an example of why one should not take Wikipedia as definitive of anything. $\endgroup$
    – garyp
    Commented Aug 29, 2020 at 13:50
  • $\begingroup$ @garyp Very good point. I hadn't considered that. $\endgroup$
    – Philip
    Commented Aug 29, 2020 at 13:53
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Even in vaccum weight is a quantity. It tells us the force required to lift an object .

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