- #1
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This is Newton's law of universal gravitation.
$$F=G\frac{m_1.m_2}{r^2}$$
Gravitational field $$g$$ is derived from this formula
$$g=G\frac{m_1}{r^2}$$ This is named gravitational "field" strength.
If Newton knew nothing about "field concept" and formulated his formula in the form of "action at a distance", how did he interpret $$g$$?
My question is valid for Coulomb as well.
Wasn't this formula
$$
E = \frac{Q}{4\pi\varepsilon_{0}r^2}
$$
invented by Coulomb?
If so, what did he name for $$E$$?
$$F=G\frac{m_1.m_2}{r^2}$$
Gravitational field $$g$$ is derived from this formula
$$g=G\frac{m_1}{r^2}$$ This is named gravitational "field" strength.
If Newton knew nothing about "field concept" and formulated his formula in the form of "action at a distance", how did he interpret $$g$$?
My question is valid for Coulomb as well.
Wasn't this formula
$$
E = \frac{Q}{4\pi\varepsilon_{0}r^2}
$$
invented by Coulomb?
If so, what did he name for $$E$$?