Difference between revisions of "Gravity"

	Gravity
	F:=Gm1m2r2

Latest revision as of 11:14, 3 October 2017

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: $[G]$ =ML3T−2 or [G]=[F]L2M−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 FA,B, is defined as follows: F:=GmAmB∥xA−xB∥ or F:=GmAmBr2
- 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]=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 $[F]\eq\frac{ML}{T^2}$

[Expand]

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

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)"

@@ Line 9: / Line 9: @@
 Additionally:
 * One may use {{M|r}}, which is the distance between {{M|x_A}} and {{M|x_B}}
-* {{M|G}} (the [[gravitational constant]], which has units:  [[Dimensions of|{{M|[G]}}]]{{M|\eq ML^3T^{-2} }} or {{M|[G]\eq[F]L^2M^{-2} }}
+* {{M|G}} (the [[universal gravitational constant]], which has units:  [[Dimensions of|{{M|[G]}}]]{{M|\eq ML^3T^{-2} }} or {{M|[G]\eq[F]L^2M^{-2} }}
 **  I prefer {{M|[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:

Difference between revisions of "Gravity"

Latest revision as of 11:14, 3 October 2017

Newtonian definition

Dimensions and Units

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