# Meaning of acceleration due to gravity

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.

• 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. Commented Sep 19, 2020 at 13:58
• A related ancient question of mine: physics.stackexchange.com/questions/96020/…
– user87745
Commented Sep 19, 2020 at 14:27
• @DvijD.C. thank you for the reference. Commented Sep 21, 2020 at 17:31

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.