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I found this article here in which it is said that it would be incorrect to say that g is not acceleration due to gravity but local gravitational field as there is no acceleration on a block placed on a table. Please can you explain this as I am in a school and I have read only that g is acceleration due to gravity and textbooks say this too.

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  • $\begingroup$ A block on a table is stationary, so is not accelerating. The Earth's gravity still acts as a force on it equal to $mg$, but the table is providing another force to balance this out. $\endgroup$
    – Henry
    Sep 19, 2020 at 13:58
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    $\begingroup$ A related ancient question of mine: physics.stackexchange.com/questions/96020/… $\endgroup$
    – user87745
    Sep 19, 2020 at 14:27
  • $\begingroup$ @DvijD.C. thank you for the reference. $\endgroup$ Sep 21, 2020 at 17:31

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The term gravitational acceleration refers to the free fall acceleration of an object on the other hand gravitational force simply can be found from Newton's law of universal gravitation and even an object stands on a table force effects though there is opposite force acting on object based from the table.

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"Acceleration due to gravity" is by far the more commonly used wording. As @Butane says, one has to understand that it is the acceleration in freefall. There is validity to his argument.

We have another field, the electric field, defined $$\vec{E} = \frac{\vec{F}}{q}$$ That's sort of a "specific force", so to speak. Force per unit of charge, where charge is the property that "makes" the force.

Is there a gravitational analogy? A gravitational field? Let's call such a thing $\vec{X}$, and try to construct it. $$ \vec{X} = \frac{\vec{F}}{m} $$ force per unit mass, a sort of "specific force" for gravity. But we know that the local force of gravity is $mg$, so $$\vec{X} = \frac{\vec{F}}{m} = \frac{m\vec{g}}{m} = \vec{g} $$In that sense $g$ is the gravitational field.

One can argue that there is a logical disconnect here in that we start knowing that the force is $mg$, and deducing the field from that. In the case of the electric field, the force is ... whatever it is, and $\vec{F} = q\vec{E}$ follows from that. The opposite direction of the reasoning.

But let's take a practical, functional, point of view. If you refer to "the gravitational field strength at the surface of the Earth" no one will know that you are talking about $g$. No one calls it that.

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