Gravity

Gravity

$$F:\eq G\frac{m_1m_2}{r^2}$$

Newtonian definition

Let there be two objects, A and B such that:

• A's centre of mass acts at position $x_A$, with mass $m_A$
• B's centre of mass acts at position $x_B$, with mass $m_B$

Additionally:

• One may use $r$, which is the distance between $x_A$ and $x_B$
• $G$ (the universal gravitational constant, which has units: $[G]$$\eq ML^3T^{-2}$ or $[G]\eq[F]L^2M^{-2}$
  • I prefer $[G]\eq[F]\cdot(ML^{-1})^{-1}\cdot(ML^{-1})^{-1}$

Then the magnitude of the force acting on each due to the other's presence is:

• Force, $F$, or $F_{A,B}$, is defined as follows: $$F:\eq G\frac{m_Am_B}{\Vert x_A-x_B\Vert}$$ or $$F:\eq G\frac{m_Am_B}{r^2}$$
  • where $\Vert\cdot\Vert$ here represents a norm, which in standard cases of $x_{A,B}\in\mathbb{R}^3$ or $\in\mathbb{R}^2$ would be the Euclidean norm
  • Specifically the forces act as follows:
    1. On A, the force due to gravity from B has magnitude $F$ as defined above in direction towards $x_B$ from $x_A$
    2. On B, the force is simply minus the force on A from B, or the force of magnitude $F$ in direction $x_A$ from $x_B$

Dimensions and Units

In what follows we use the standard physical dimensions, $L$ for length, $M$ for mass, $T$ for time, and $[\alpha]$ to denote the dimensions of $\alpha$.

We start with the obvious, dimensions of our terms:

• $[r]\eq L$
  • Thus: $[r^2]\eq L^2$
• $[m_A]\eq M$ and $[m_B]\eq M$ also
• $[F]\eq MLT^{-2}$ from $f\eq ma$ (see: force), we may also write $[F]\eq\frac{ML}{T^2}$

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Deriving the units of $G$, we see $[G]\eq ML^3T^{-2}$, or $[G]\eq[F]L^2M^{-2}$ I prefer $[G]\eq[F]\cdot(ML^{-1})^{-1}\cdot(ML^{-1})^{-1}$

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To find the dimensions of $G$ we may solve as usual:

• $$\text{LHS}:\eq[F]\eq\left[G\frac{m_Am_B}{r^2}\right]\eq:\text{RHS}$$

  $$\implies\ \text{LHS}\eq MLT^{-2}\eq [G]M^2\cdot\frac{1}{L^2} \eq [G]M^2L^{-2}\eq\text{RHS}$$

  $$\implies\ MLT^{-2}\eq [G]M^2L^{-2}$$

  $$\implies\ [G]\eq\frac{MLT^{-2} }{M^2L^{-2} }\eq\frac{ML^3}{M^2T^2}$$ (we may then stop here with $[G]\eq \frac{L^3}{MT^2}\eq ML^3T^{-2}$ as our answer)

  $$\implies\ [G]\eq \underbrace{\frac{ML}{T^2} }_{\text{Force} }\cdot \frac{L^2}{M^2}$$

  $$\implies\ [G]\eq [F]\left(\frac{L}{M}\right)^2$$ - many would stop here, however

  $$\implies\ [G]\eq [F]\left(\frac{M}{L}\right)^{-2}$$

We arrive at:

• $$[G]\eq[F]\cdot\left(\frac{M}{L}\right)^{-1}\cdot\left(\frac{M}{L}\right)^{-1}$$

As the reader may know I like to think of $\circ^{-1}$ as rate, this means (in SI units) that:

• $[G]$ is force per ($\text{ kg }$ per $\text{ meter }$) per ($\text{ kg }$ per $\text{ meter }$)

This may be said better as "(force per each (kg per meter away)) per each (kg per meter away)"