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# 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.

 Contents

## Absolute Magnitude for stars and galaxies (M)

In stellar and galactic astronomy, the standard distance is 10 parsecs (about 32.616 light years, or 3×1014 kilometres). A star at ten parsecs has a parallax of 0.1" (100 milli arc seconds).

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).

For comparison, Sirius has an absolute magnitude of 1.4 and the Sun has an absolute visual magnitude of 4.83 (it actually serves as a reference point).

Absolute magnitudes for stars generally range from -10 to +17. The absolute magnitude for galaxies can be much lower (brighter). For example, the giant elliptical galaxy M87 has an absolute magnitude of -22.

### Computation

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

$M = m + 5 \log_{10}\frac{d_0}{d}\!\,$

where $d_0\!\,$ is 10 parsecs (≈ 32.616 light-years) and $d\!\,$ is the star's distance; or:

$M = m + 5 (1 + \log_{10}\frac{\pi}{\pi_0})\!\,$

where $\pi\!\,$ is the star's parallax and $\pi_0\!\,$ is 1 arcsec.

#### Example

Rigel has a visual magnitude of mV=0.18 and distance about 773 light-years.
MVRigel = 0.18 + 5*log10(32.616/773) = -6.7
Vega has a parallax of 0.133", and an apparent magnitude of +0.03
MVVega = 0.03 + 5*(1 + log10(0.133)) = +0.65
Alpha Centauri has a parallax of 0.750" and an apparent magnitude of -0.01
MVα Cen = -0.01 + 5*(1 + log10(0.750)) = +4.37

### Apparent magnitude

Given the absolute magnitude $M\!\,$, you can also calculate the apparent magnitude $m\!\,$ from any distance $d\!\,$:

$m = M - 5 \log_{10}\frac{d_0}{d}\!\,$

## 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.

### Calculations

Formula for H: (Absolute Magnitude)

$H = m_{Sun} - 5 \log_{10}\frac{ \sqrt { a } r}{d_0}\!\,$

where $m_{Sun}\!\,$ is the apparent magnitude of the Sun at 1 au (-26.73), $a\!\,$ is the geometric albedo of the body (a number between 0 and 1), $r\!\,$ is its radius and $d_0\!\,$ is 1 au (≈149.6 Gm).

#### Example

Moon: $a_{Moon}\!\,$ = 0.12, $r_{Moon}\!\,$ = 3476/2 km = 1738 km

$H_{Moon} = m_{Sun} - 5 \log_{10}\frac{ \sqrt { a_{Moon} } r_{Moon}}{d_0} = +0.25\!\,$

### Apparent magnitude

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

$m = H + 2.5 \log_{10}{(\frac{d_{BS}^2 d_{BO}^2}{p(\chi) d_0^4})}\!\,$

where

$d_0\!\,$ is 1 au, $\chi\!\,$ is the phase angle, the angle between the Sun-Body and Body-Observer lines; by the law of cosines, we have:

$\cos{\chi} = \frac{ d_{BO}^2 + d_{BS}^2 - d_{OS}^2 } {2 d_{BO} d_{BS}}\!\,$

$p(\chi)\!\,$ is the phase integral (integration of reflected light; a number in the 0 to 1 range)

Example: (An ideal diffuse reflecting sphere) - A reasonable first approximation for planetary bodies

$p(\chi) = \frac{2}{3} ( (1 - \frac{\chi}{\pi}) \cos{\chi} + (1/\pi) \sin{\chi} )\!\,$

A full-phase diffuse sphere reflects 2/3 as much light as a diffuse disc of the same diameter
Distances:
$d_{BO}\!\,$ is the distance between the observer and the body
$d_{BS}\!\,$ is the distance between the Sun and the body
$d_{OS}\!\,$ is the distance between the obverser and the Sun

#### Example

Moon

$H_{Moon}\!\,$ = +0.25
$d_{OS}\!\,$ = $d_{BS}\!\,$ = 1 au
$d_{BO}\!\,$ = 384.5 Mm = 2.57 mau
How bright is the Moon from Earth?
Full Moon: $\chi\!\,$ = 0, ($p(\chi)\!\,$ ≈ 2/3)
$m_{Moon} = 0.25 + 2.5 \log_{10}{(\frac{3}{2} 0.00257^2)} = -12.26\!\,$
(Actual -12.7) A full Moon reflects 30% more light at full phase than a perfect diffuse reflector predicts.
Quarter Moon: $\chi\!\,$ = 90°, $p(\chi) \approx \frac{2}{3\pi}\!\,$ (if diffuse reflector)
$m_{Moon} = 0.25 + 2.5 \log_{10}{(\frac{3\pi}{2} 0.00257^2)} = -11.02\!\,$
(Actual approximately -11.0) The diffuse reflector formula does better for smaller phases.