The Absolute magnitude reference article from the English Wikipedia on 24-Jul-2004 (provided by Fixed Reference: snapshots of Wikipedia from wikipedia.org)

Absolute magnitude

In astronomy, absolute magnitude is the apparent magnitude, m, an object would have if it were at a standardized distance away.

It allows the overall brightnesses of objects to be compared without regards to distance.

Table of contents

1 Absolute Magnitude for stars (M) 1.1 See also 1.2 Computation 1.2.1 Example 1.3 Apparent magnitude 2 Absolute Magnitude for planets (H) 2.4 Calulations 2.4.2 Example 2.5 Apparent magnitude 2.5.3 Example

Absolute Magnitude for stars (M)

In stellar astronomy, the standard distance is 10 parsecs (about 32.616 light years, or 3×10¹⁴ kilometers). (A star at ten parsecs has a parallax of 0.1")

In defining absolute magnitude it is necessary to specify the type of electromagnetic radiation being measured. When referring to total energy output, the proper term is bolometric magnitude. The dimmer an object (at a distance of 10 parsecs) would appear, the higher its absolute magnitude. The lower an object's absolute magnitude, the higher its luminosity. A mathematical equation relates apparent magnitude with absolute magnitude, via parallax.

Many stars visible to the naked eye have an absolute magnitude which is capable of casting shadows from a distance of 10 parsecs; Rigel (-7.0), Deneb (-7.2), Naos (-7.3), and Betelgeuse (-5.6).

See also

• Hertzsprung-Russell diagram Relates absolute magnitude or luminosity versus spectral color or surface temperature.

For comparison, Sirius has an absolute magnitude of 1.4 and the Sun has a magnitude of 4.5/4.8. Absolute magnitudes generally range from -10 to +17.

Computation

You can compute the Absolute Magnitude of a star given its apparent magnitude and distance:

• M_star=m_star+2.5*log10((10/dist_star(in parsecs))²)

or

• M_star=m_star+5*(1+log10(parallax_star))

or

• M_star=m_star+2.5*log10((32.616/dist_star(in lyrs))²)

Example

• Rigel has a visual magnitude of m_v=0.18 and distance about 773 lyrs.
  • M_vRigel=0.18+2.5*log10*((32.616/773)²) = -6.7
• Vega has a parallax of 0.133", and an apparent magnitude of 0.03
  • M_vVega=0.03+5*(1+log(0.133)) = +0.65
• Alpha Centauri has a parallax of 0.750" and an apparent magnitude of -0.01
  • M_v*=-0.01+5*(1+log(0.750)) = +4.37

Apparent magnitude

Given the absolute magnitude, you can also calculate the apparent magnitude from any distance:

• m_star=M_star-2.5*log10((32.616/dist_star(in lyrs))²)

Absolute Magnitude for planets (H)

For planets, comets and asteroids a different definition of absolute magnitude is used which is more meaningful for nonstellar objects.

In this case, the absolute magnitude is defined as the apparent magnitude that the object would have if it were one astronomical unit (au) from both the Sun and the Earth and at a phase angle of zero degrees. This is a physical impossibility, but it is convenient for purposes of calculation.

Calulations

Formula for H: (Absolute Magnitude)

• H_body = m_sun-2.5*log10(a_body*d_body²)
• Where
  • m_sun = apparent magnitude of sun at 1 au = -26.73
  • a_body=geometric albedo of body (between 0 and 1)
  • d_body=diameter of spherical body (in au)

Or

• H_body ~= 2.5*log10(1329/(a_body*d_body²(in km)))
  • a_body=geometric albedo of body (between 0 and 1)
  • d_body=diameter of spherical body (in km)

Example

Moon: a_moon = 0.12, d_moon = 3476 km

• H_moon ~= 2.5*log10(1329/(0.12*3476²)) = +0.21

Apparent magnitude

The absolute magnitude can be used to help calculate the apparent magnitude of a body under different conditions.

• m_body = H_body -2.5*log10(p_body(χ)*(d_SunToBody*d_{DistanceToBody})²)
• Where
  • χ = phase angle
    • Law_of_cosines cos(χ) = ( ( d_{DistanceToBody}² + d_BodyToSun² - d_{DistanceToSun}²) / (2 x d_{DistanceToBody} x d_BodyToSun) ).
  • p_body(χ) = phase integral (integration of reflected light) (<1)
    • Example: (An ideal diffuse reflecting sphere) - A reasonable first approximation for plantary bodies
      • p(χ) = (2/3)*((1-χ/π)*cos(χ)+sin(χ)/π)
        • A full-phase diffuse sphere reflects 2/3 as much light as a diffuse disc of the same diameter
  • Distances: (MUST be in au)
    • d_{DistanceToBody} = distance betweeen observer and body
    • d_SunToBody = distance between sun and body
    • d_{DistanceToSun} = distance between obverser and sun

Example

Moon

• H_moon = 0.21
• d_SunToBody = d_{DistanceToSun} = 1 au
• d_{DistanceToBody} = 384500 km = 2.57x10^-3 au
• How bright is moon from earth?
  • Full moon: χ = 0, (p(χ) ~= 2/3)
    • m_moon = 0.21 -2.5*log10(2/3*(2.57x10^-3)²) = -12.4
      • (Actual -12.7) A full moon reflects 30% more light at full phase than a perfect diffuse reflector predicts.
  • Quarter moon: χ = 90, (p(χ)~=2/(3*π) if diffuse reflector)
    • m_moon = 0.21 -2.5*log10(2/(3*π)*(2.57x10^-3)²) = -11.1
      • (Actual ~ -11.0) The diffuse reflector formula does better for smaller phases.