Calculating Apparent Magnitude of The Moon

In summary, the apparent magnitude of the Moon can be calculated using the equation -2.5 log(I/I0), where I is the flux of light from the source and I0 is the flux of a reference source. However, the flux of light from the Sun that reaches the Earth is not the same as the flux of light that is reflected by the Moon, as only a small fraction of the reflected light reaches the Earth. Therefore, the calculated apparent magnitude of the Moon is not accurate and needs to be adjusted to account for this.
  • #1
luuurey
24
0
Hello everyone,
I've been wondering if I can calculate the apparent magnitude of The Moon since I know the apparent magnitude of The Sun(say -27mag) and reflection coefficient of Moon's surface(say 0.12).
We know the equation

[itex]
\Delta m = -2.5 log_{10}(I/I_0)
[/itex]

Where I is the flux of the light that comes from source and I_0 is flux of a reference source. I know the magnitude of The Sun, so I can say:

[itex]
-27 = -2.5 log_{10}(I_S/I_0)
[/itex]

Where I_S is the flux that I can measure on the sunlight. I know that the flux of the light from The Moon is just reflected sunlight and hence[itex]I_M=0.12 I_S[/itex], where I_M is the flux of light that comes from The Moon. So I get:

[itex]
\Delta m_M = -2.5 log_{10}(I_M/I_0)= -2.5 log_{10}(0.12 I_S/I_0)= -2.5 log_{10}(I_S/I_0)-2.5 log_{10}(0.12) = -27 -(-2)=-25
[/itex]

I'm sure that the apparent magnitude of The Moon is definitely not 25mag. It's around -13mag. What's wrong with my idea? Where did I make mistake?

Thank you for your answer. I will be very thankful for your help!
 
Astronomy news on Phys.org
  • #2
You have assumed that all of the sun's light reflected by the moon makes it to the Earth. Actually, the light is reflected in all directions from the moon's surface, and only a small fraction of the light intercepts the Earth. You need to take this into account.
 

1. How is the apparent magnitude of the Moon calculated?

The apparent magnitude of the Moon is calculated using the formula m = -2.5log(F), where m is the apparent magnitude and F is the Moon's flux. Flux is a measure of the amount of light reaching Earth from the Moon and is affected by the Moon's distance from Earth and its reflectivity.

2. What is the range of apparent magnitudes for the Moon?

The range of apparent magnitudes for the Moon is approximately -12.74 to -2.5. This means that on average, the Moon appears about 400,000 times fainter than the brightest star in the sky.

3. How does the Moon's phase affect its apparent magnitude?

The Moon's phase, or the amount of illuminated surface visible from Earth, does not have a significant effect on its apparent magnitude. However, a full moon appears brighter than a crescent moon because more of its surface is reflecting sunlight towards Earth.

4. Why is the Moon's apparent magnitude not constant?

The Moon's apparent magnitude is not constant because it orbits Earth in an elliptical path, meaning its distance from Earth varies. When the Moon is at its closest point (perigee), it appears brighter and has a lower apparent magnitude. On the other hand, when it is at its farthest point (apogee), it appears dimmer and has a higher apparent magnitude.

5. How does atmospheric conditions affect the apparent magnitude of the Moon?

Atmospheric conditions, such as humidity, pollution, and cloud cover, can affect the apparent magnitude of the Moon. These factors can block or scatter the light from the Moon, making it appear dimmer than its calculated apparent magnitude. Additionally, the Moon's apparent magnitude can also be affected by the amount of light pollution in a particular area.

Similar threads

  • Astronomy and Astrophysics
Replies
1
Views
2K
  • Astronomy and Astrophysics
Replies
4
Views
3K
  • Astronomy and Astrophysics
Replies
1
Views
4K
  • Astronomy and Astrophysics
Replies
6
Views
4K
Replies
4
Views
735
  • Astronomy and Astrophysics
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
17
Views
2K
  • Introductory Physics Homework Help
Replies
15
Views
1K
Back
Top