Gravity

Gravity

[math]F:\eq G\frac{m_1m_2}{r^2} [/math]

Newtonian definition

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

• A's centre of mass acts at position [ilmath]x_A [/ilmath], with mass [ilmath]m_A[/ilmath]
• B's centre of mass acts at position [ilmath]x_B [/ilmath], with mass [ilmath]m_B[/ilmath]

Additionally:

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

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

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

Dimensions and Units

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

We start with the obvious, dimensions of our terms:

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

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

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To find the dimensions of [ilmath]G[/ilmath] we may solve as usual:

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

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

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

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

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

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

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

We arrive at:

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

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

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

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