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 $E$?

  • $\begingroup$ Didn't they just give the formulae for the force (probably just the modulus, and using words to describe directions, which are nowadays described by vector notation)? $\endgroup$ – Phoenix87 Dec 30 '14 at 22:11
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    $\begingroup$ Would History of Science and Mathematics be a better home for this question? $\endgroup$ – Qmechanic Dec 30 '14 at 22:16
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    $\begingroup$ I believe so. It's not obviously off topic here (in the same way that some other questions are), but I think it is off topic. $\endgroup$ – David Z Dec 30 '14 at 22:30
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    $\begingroup$ Newton did his work before even a solid notion of calculus had been developed. It is unlikely that he thought of gravity as a continuous function over a metric space (aka a "field"). He did, however, think about gravity as being "... caused by an Agent acting constantly according to certain laws; but whether this Agent be material or immaterial, I have left to the Consideration of my readers.". Today we call that agent "the real vacuum" and we think of it as immaterial. He was close... even though the next round of cigars had to wait until 1915 when Einstein collected them. $\endgroup$ – CuriousOne Dec 30 '14 at 22:31
  • $\begingroup$ Newton interpreted $g$ as a constant for all the inhabitants of the Earth. When we calculate the weight of objects on the Earth we don't need $G$ and the mass of the Earth, $g$ is practical. $\endgroup$ – Sofia Dec 30 '14 at 23:12

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