Homework #4:

Off[Clear::wrsym]
Off[General::spell]
Off[General::spell1]

Setup:

colors = {red, green, blue, cyan, magenta, yellow, 
            black} = 
{RGBColor[1, 0, 
              0]
, RGBColor[0, 1, 0]
, 
            RGBColor[0, 0, 1]
, 
            CMYKColor[1, 0, 0, 0]
, 
            CMYKColor[0, 1, 0, 0]
, 
            CMYKColor[0, 0, 1, 0]
, 
            CMYKColor[0, 0, 0, 
              1]
};
Plot[Table[i x, {i, 1, 7}]  // Evaluate, {x, 0, 1}, 
        PlotStyle -> colors];

Protect[{colors, red, green, blue, cyan, magenta, yellow, 
          black}];

Problem #1:
Parallax for Pluto

Clear["Global`*"]

<< Miscellaneous`PhysicalConstants`
<< Miscellaneous`Units`

?Miscellaneous`PhysicalConstants`*

The formula for resolving power is: λ /a≃ 1.22 Sin[θ]≃ 1.22 θ, where a is the apperature. From the text, Fig.1, the parallax angle is: θ = d/D where d = 5.91 * (10^12)m, and D is the distance to the star.

eq1 =  λ/a  ==  1.22 θ
eq2 = θ  ==  d /D

sol1 =  Solve[{eq1, eq2}, {D, θ}][[1]]

For Earth:

\(distance = \ \(D\  /. sol1\)\  /. {a -> \ 200\ Inch, \ 
          d -> 1.5\ *\ 10^11\ Meter\ , \ λ\  -> \ 
            5000\ Angstrom}\)

distance = MKS[distance]

Convert[distance, LightYear]

Convert[distance, Parsec]

For Pluto:

\(distance = \ \(D\  /. sol1\)\  /. {a -> \ 200\ Inch, \ 
          d -> 5.91*10^12\ Meter\ , \ λ\  -> \ 
            5000\ Angstrom}\)

distance = MKS[distance]

Convert[distance, LightYear]

Convert[distance, Parsec]

Ratio:

\(\(\({5.91*10^12\ Meter\ /\((1.5\ *\ 10^11\ Meter)\), 
      1595  \ Parsec\ /\((40.48  \ Parsec)\)}\)\(\ \ \)\)\)

Problem #2:
The sun at 4 light-years

Clear["Global`*"]

eqs =  
{ℓ ==  
          L/(4 π (d^2)), 
m  ==  -2.5  Log[
              10. ,  ℓ/ℓ0], 
M ==  
          m  -  5  Log[10. ,   d/d0]}

sol1 =  Solve[eqs, {L, m, ℓ}][[1]]  // 
        Simplify

array = {d/parsec, L,  ℓ, M,  m}  /. sol1

parsec =  3.09 * (10^16);
lightYear =  parsec/3.26;
d0 =  10 parsec;
 ℓ0 =  2.52 * (10^(-8));
earthSunDistance = 1.5 * (10^11);

array  /. {M -> 4.72,  d  ->  4 lightYear}

Problem #3:
Table of a star at different distances

Clear["Global`*"]

parsec= Convert[1.0 Parsec, 
        Meter]

lightYear  =  Convert[1.0 LightYear, Meter]

d0 =  10 parsec;
 ℓ0 =  
        2.52 * (10^(-8 )) Watt/(Meter^2);

solL = Solve[ℓ ==  L/(4 π (d^2)), 
          L][[1]]

L0 = L  /. 
          solL  /. {d -> 
            d0,  ℓ -> ℓ0}

ℓ =  L/(4 π (d^2))
m  =  -2.5  Log[
          10. ,  ℓ/ℓ0]
M =  m  -  5  Log[10. ,   d/d0]

array[
        d_] = {d/parsec, L,  ℓ, M,  
          m} /.{L -> L0}

\(\(Table[\ array[\ 10\^n\ parsec], {n, \(-1\), 3, 1}]\  // 
        TableForm[#, 
            TableHeadings -> {None, {"d", "L", "ℓ", "M", "m"}}] &\)\ \  // Chop[#, 10\^\(-14\)] &\)

Problem #4
Berry Text, #6: Gravitational Red Shift

Clear["Global`*"]

<< Miscellaneous`PhysicalConstants`
<< Miscellaneous`Units`

energyLoss =  G M m /r

(*  Find effective mass for photon of energy :  
          h ν *)  
sol1= Solve[h ν == m c^2, m][[1]]

z =  energyLoss/ (h ν) /. sol1

z =  
      z  /. {G -> GravitationalConstant,  c ->  SpeedOfLight,  
          r  ->  1000 Parsec,  M  ->  (10^9) SolarMass,  
          Newton ->  Kilogram Meter/(Second^2)}

MKS[z]