# 5.5: The acceleration due to gravity

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If you have studied some physics before reading this textbook, you may have been surprised by our choice of dimension for $$g$$ to be force per unit mass rather than acceleration. This is indeed an unconventional choice as $$g$$ is usually presented as “the acceleration due to Earth’s gravity” instead of the “strength of Earth’s gravitational field”. Our choice comes from the potential difference between inertial mass, $$m_I$$, and gravitational mass, $$m_G$$, which we distinguish in this section.

Consider the simple model of a mass falling freely near the surface of the Earth in the absence of air-resistance. The only force exerted on the mass is its weight, $$m_G\vec g$$, which is given in terms of gravitational mass (the mass that determines how an object experiences gravity). Both the weight and the acceleration of the object point downwards. The free-body diagram for the mass is shown in Figure $$\PageIndex{1}$$, where the $$y$$ axis was chosen to be vertically upwards (co-linear with the acceleration). Figure $$\PageIndex{1}$$: Free-body diagram for a mass that is free-falling in the absence of air resistance (drag).

Writing out the $$y$$ component of Newton’s Second Law, being careful to distinguish between inertial and gravitational mass, and noting that both the weight and the acceleration are in the negative $$y$$ direction: \begin{aligned} \sum F_y = -F_g &= -m_I a\\ \therefore m_Gg &= m_I a\end{aligned} This makes it clear that $$g$$ is not necessarily the acceleration due to gravity. It is only the acceleration due to gravity in the limit that the inertial and gravitational masses are the same. If $$m_G=m_I$$, then we have: \begin{aligned} a = g\end{aligned} and indeed, the acceleration of objects near the surface of the Earth has a magnitude of $$g$$. It is also clear that the dimensions of $$g$$ can also be written as an acceleration, and in most cases, one writes that, near the surface of the Earth, $$g=9.8\text{m/s}^{2}$$. You should however remember that this is only true when inertial and gravitational masses are the same, and that $$g$$ really should be thought of as the strength of the gravitational field, not as an acceleration.

This page titled 5.5: The acceleration due to gravity is shared under a CC BY-SA license and was authored, remixed, and/or curated by Howard Martin revised by Alan Ng.