# Do planets have actual lumosity

1. Dec 23, 2004

### DB

Do planets have actual lumosity or are they just lit up by stars? Might be a stupid question but im just wondering.

Thnx

2. Dec 23, 2004

### The Bob

Personally I do not know but I assume they might have because of some of the chemical reactions that happen on some of them (like the Great Spot on Jupiter). I don't know, to be honest, but I would say yes.

Sorry.

The Bob (2004 ©)

3. Dec 23, 2004

### turbo

Well THIS particular planet is certainly luminous! It's a big pain in the butt, and I'm glad I live in a thinly populated place.

http://www.pha.jhu.edu/~atolea/second/page1.html [Broken]

Last edited by a moderator: May 1, 2017
4. Dec 23, 2004

### Staff: Mentor

Its virtually all reflected from the sun.

5. Dec 24, 2004

### Nereid

Staff Emeritus
If you had IR eyes, you'd still see the planets ... and some (most in some cases) of that IR would be from the planet itself.

You can tell how much of the light (or IR) that you see from a planet is 'native' vs reflected by looking at the planet at night ... i.e. looking at the side of the planet that is not facing the Sun. We can do this - from Earth - for Mercury and Venus, but not for the others. Do you know why DB?

Also, even if the Earth had no cities, fishing fleets, etc, it would still be quite visible 'at night' from out in space ... at least some of the time. Do you know why?

Finally, things like aurorae and lightning are 'local' in origin, so ... (But Mercury is pretty dark 'at night', why?)

6. Dec 24, 2004

### Chronos

That's Nereid. Always handing out homework.

7. Dec 24, 2004

### Orion1

Giant Jupiter...

Jupiter's Luminosity and Intensity are calculated by reference (3) as:
L_j = 1.05 * 10^17 W - wrong!
$$I_J = \frac{L_J}{4 \pi dr_c^2}$$ - wrong!
$$dr_c$$ - Cassini-Jupiter surface range
I_j = 9.83 * 10^(-5) W/m^2 - wrong!

$$L_J = \frac{L_\odot}{4} \left( \frac{r_j}{r_t} \right)^2$$
$$I_J = \frac{L_J}{4 \pi r_J^2}$$
$$r_J$$ - Jupiter radius
$$r_t$$ - Sol-Jupiter distance

Luminosity based upon reference 3 equation and information:
$$L_J = 8.122*10^{17} W$$
$$I_J = 12.678 W*m^{-2}$$

However, Arxiv information reference 4:
$$L_J = 8.363*10^{17} W$$
$$I_J = 13.054 W*m^{-2}$$

Reference:
Contemporary Astronomy (Jay M. Pasachoff, 1977)
The Cosmic Voyage (William K. Hartmann, 1992)
http://www.nap.edu/html/oneuniverse/energy_solution_12.html
http://arXiv.org/abs/astro-ph/9506055

Last edited: Dec 24, 2004
8. Dec 24, 2004

### DB

Yes, we can only see the sides of Mercury and Venus oposing the sun because they are in front of us in the solar system.

As for why we can see earth at night from space, im assuming it's due to the light of the stars, the light reflected from planets behind us, the moon at the right position would reflect some light on the dark side of the planet.

So know I'm thinking that rock planets generally do not emit their own lumosity, but gas planets can, due to their (as The Bob said) chemical reactions.

9. Dec 24, 2004

### DaveC426913

Inasmuch as they are not at absolute zero, all things, including plants and rocks emit heat, and woud be visible (at least against a background of empty space) if you could see in the IR band.

Now, the *origin* of that heat is mostly from the Sun, although a certain percentage is due to the Earth's internal molten core.

As for actually emitting its own EM radiation, Jupiter is the only planet in our Solar System that actually emits more than it absorbs from the Sun. It is a pretty strong emitter of radio frequencies.

But none of the planets are luminous in the *visible* band, unless you count local, discrete events such as aurorae and lightning.

10. Dec 24, 2004

### DaveC426913

You can see the back of Mars if you look at just the right time. But you need to lean waaaaaay out, and a mirror on a stick helps too.

:-)