CONSTANTS

Needs["Notation`"]

Symbolize[(NotationTemplateTag[e])]; Symbolize[(NotationTemplateTag[G])]; Symbolize[(NotationTemplateTag[h])]; Symbolize[(NotationTemplateTag[ℏ])]; Symbolize[(NotationTemplateTag[σ])]; Symbolize[(NotationTemplateTag[(Subscript[k, B])])]; Symbolize[(NotationTemplateTag[(Subscript[m, p])])]; Symbolize[(NotationTemplateTag[(Subscript[m, e])])]; Symbolize[(NotationTemplateTag[c])]; Symbolize[(NotationTemplateTag[AU])]; Symbolize[(NotationTemplateTag[lyr])]; Symbolize[(NotationTemplateTag[pc])]; Symbolize[(NotationTemplateTag[(Subscript[M, ♁])])]; Symbolize[(NotationTemplateTag[(Subscript[R, ♁])])]; Symbolize[(NotationTemplateTag[(Subscript[M, ☉])])]; Symbolize[(NotationTemplateTag[(Subscript[L, ☉])])]; Symbolize[(NotationTemplateTag[(Subscript[R, ☉])])];Symbolize[(NotationTemplateTag[(Subscript[T, ☉])])]; Symbolize[(NotationTemplateTag[kpc])]; Symbolize[(NotationTemplateTag[Mpc])];Symbolize[(NotationTemplateTag[g])];Symbolize[(NotationTemplateTag[eV])];

e=4.8032×(10^(-10))(* esu *) (* 1.602*(10^(-19)) Coulombs *);
G=6.67408*(10^(-8)) (* (cm^3)(g^(-1))(s^(-2)) *) (* *(10^(-11)) N m^2 kg^-2*);
g=9.81 (* m (s^(-2)) *) (* *(10^(-11)) N m^2 kg^-2*);
h=6.626069*(10^(-27)) (* erg s *)(* *(10^(-34)) J s *);
ℏ=(h/(2*π));
σ=5.67*(10^(-5)) (* erg (s^(-1)) (cm^(-2)) (K^(-4)) *) (* *(10^(-8)) J  (s^(-1)) (m^(-2)) (K^(-4)) *);
(Subscript[k, B])=1.38065*(10^(-16))(* erg (K^(-1)) *) (* *(10^(-23)) J (K^(-1))*);
(Subscript[m, p])=1.672*(10^(-24)) (* g *) (*(10^(-27)) kg *);
(Subscript[m, e])=9.109*(10^(-28))(* g *) (*(10^(-31)) kg *);
c=29979245800  (* cm (s^(-1)) *) (* 299792458 m (s^(-1))*);
AU=14959787970000 (* cm *) (* *(10^11) m *);
lyr=9.4605*(10^17) (* cm *) (* *(10^15) m *);
pc=206265*AU; kpc = (10^3)*pc; Mpc = (10^3)*kpc;
(Subscript[M, ♁])=5.97*(10^27) (* g *) (* *(10^27) kg *);
(Subscript[R, ♁])=6.371*(10^8)(* cm *) (* *(10^6) m *);
(Subscript[M, ☉])=1.989*(10^33) (* g *) (* *(10^30) kg *);
(Subscript[L, ☉])=3.846*(10^33) (* erg (s^(-1)) *) (* *(10^26) J (s^(-1)) *);
(Subscript[R, ☉])=6.96*(10^10) (* cm *) (* *(10^8) m *);
(Subscript[T, ☉])=5772 (* K *);
eV = 1.60218*(10^(-12))(* erg *);

OwnValues[e];OwnValues[G];OwnValues[h];OwnValues[ℏ];OwnValues[σ]; OwnValues[(Subscript[k, B])];OwnValues[(Subscript[m, p])];OwnValues[(Subscript[m, e])];OwnValues[c];OwnValues[AU]; OwnValues[lyr];OwnValues[pc];OwnValues[(Subscript[M, ♁])];OwnValues[(Subscript[R, ♁])];OwnValues[(Subscript[M, ☉])]; OwnValues[(Subscript[L, ☉])];OwnValues[(Subscript[R, ☉])];OwnValues[(Subscript[T, ☉])];OwnValues[kpc];OwnValues[Mpc];OwnValues[g];OwnValues[eV];

keV=(10^3)*eV;MeV=(10^3)*keV;GeV=(10^3)*MeV;TeV=(10^3)*GeV;
γ[v_]:=1/(Sqrt[1-((v^2)/(c^2))])
sec=1;min=60*sec;hr=60*min;day=24*hr;year=365.2422*day;
arcsec=1/206265;

Scratchpad

(* Here is a simple calculation of the Earth's semimajor axis via Kepler's Law, given the units defined above *)

P=1*year;
a=((((((P^2)*G*(Subscript[M, ☉]))/(4*(π^2)))))^(1/3))

(* That's in centimeters, so lets convert to AU, then light-minutes *)
a/AU

a/lyr*(year/min)

(* How about some more constants. Here's an example calculation using the Saha equation to find the ratio of singly ionized Hydrogen to neutral hydrogen abundance in the Sun. You'll learn about the Saha equation in AST 232 and AST 241 *)

T=5880;
P=10;
χ=13.6*eV;

NIItoNI=(((Subscript[k, B])*T)/(2*P))*(((((2*π*(Subscript[m, e])*(Subscript[k, B])*T)/(h^2))))^(3/2))*Exp[-χ/((Subscript[k, B])*T)]

(* Thats 10,000 neutral hydrogens for every 1 ionized hydrogen! *)