# "Gravitational Charge" in Newtonian Gravity in Analogy to the Electric Field

I am currently reading Carroll's GR book p. 48. There he says that

If you like, $$m_g/m_i$$ can be thought of as the "gravitational charge" of the body

with the gravitational mass $$m_g$$ and the inertial mass $$m_i$$. Why? Wouldn't it make more sense to say that just $$m_g$$ is the gravitational charge? After all, we don't call $$q/m_i$$ is the electric charge of a body.

• $q/m$ is called the specific charge. Sep 21, 2021 at 8:02

Imagine there is a uniform electric field $$\vec{E}$$ then for a particle with charge $$q$$ the force acting on that particle is $$\vec{F}=q\vec{E}$$. So the acceleration will be $$\vec{F}=\frac{q\vec{E}}{m}$$ Notice that if we double the charge the acceleration will also be doubled.
Imagine there is a uniform newtonian gravitaional field $$\vec{g}$$ then for a particle with inertial mass $$m_i$$ and gravitational mass $$m_g$$ the force acting on that particle is $$\vec{F}=m_g\vec{g}$$. So the acceleration will be $$\vec{F}=\frac{m_g\vec{g}}{m_i}=\vec{g}\text{ since }m_i=m_g$$ Notice that if we double the $$m_g$$ the acceleration will not be doubled.
In that sense, all particles have the same gravitational charge which is +1. The main difference compared to the electric field case is that q and m are independent, but $$m_i$$ and $$m_g$$ are not independent.