Eroica

2004-Nov-17, 08:56 AM

The Free Dictionary (http://encyclopedia.thefreedictionary.com/Absolute+magnitude) gives a formula for calculating the apparent magnitude of an asteroid from its absolute magnitude:

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 [ie full, like the full-moon]. This is a physical impossibility, but it is convenient for purposes of calculation....

Apparent Magnitude

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

app mag = H - 2.5*log10[p(X)*(a*b)²]

Where:

X = phase angle (0° for opposition)

p(X) = phase integral (integration of reflected light) (<1) = (⅔)*((1-χ/π)*cos(χ)+sin(χ)/π)

At opposition, an asteroid's phase integral is ⅔

a = distance in AUs betweeen Earth and asteroid

b = distance in AUs between Sun and asteroid

But when I tried using this formula on Vesta to confirm that it is a naked-eye object at favourable oppositions, I got an apparent magnitude of 2.59, instead of 6.1.

app mag = H - 2.5*log10[⅔*(a*b)²]

H = 3.2

a = 1.152 (distance between Earth and Vesta when Vesta is at perihelion)

b = 2.152 (Vesta's perihelion)

app mag = 3.2 - 2.5*log[⅔(1.152*2.152)²] = 2.59

What's wrong with this picture? ](*,) :-k

Vesta (http://encyclopedia.thefreedictionary.com/4%20Vesta)

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 [ie full, like the full-moon]. This is a physical impossibility, but it is convenient for purposes of calculation....

Apparent Magnitude

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

app mag = H - 2.5*log10[p(X)*(a*b)²]

Where:

X = phase angle (0° for opposition)

p(X) = phase integral (integration of reflected light) (<1) = (⅔)*((1-χ/π)*cos(χ)+sin(χ)/π)

At opposition, an asteroid's phase integral is ⅔

a = distance in AUs betweeen Earth and asteroid

b = distance in AUs between Sun and asteroid

But when I tried using this formula on Vesta to confirm that it is a naked-eye object at favourable oppositions, I got an apparent magnitude of 2.59, instead of 6.1.

app mag = H - 2.5*log10[⅔*(a*b)²]

H = 3.2

a = 1.152 (distance between Earth and Vesta when Vesta is at perihelion)

b = 2.152 (Vesta's perihelion)

app mag = 3.2 - 2.5*log[⅔(1.152*2.152)²] = 2.59

What's wrong with this picture? ](*,) :-k

Vesta (http://encyclopedia.thefreedictionary.com/4%20Vesta)