Night sky and supernova questions

Government$

Here are my two questions:

1) When i look up to a night sky, do i see mostly stars and galaxy's rather then planets?
Here is my logic: Since star is many many times more brighter then reflected light of planets i see brightest object and since we are looking form earth planets and stars would appear close to one another and the brightens of star doesn't allow us to see planets around that star.

2) If Supernova were to happen right now would i see it gradually or would it start shining immediately like when i turn on light in a dark room?

phinds

The ONLY planets you can see with the naked eye are some of the ones in our solar system, and these only when the conditions are right.

Something like 99.9% of everything you see with the naked eye is nearby stars in the Milky Way. I'm not sure there are any galaxies visible with the naked eye other than Andromeda, which is.

phyzguy

Supernovae brighten over a period of hours to days, then fade over a period of months to years. Here is a light curve of a recent supernova.

So it wouldn't be like turning on a light - it would turn on much more gradually than this.

Drakkith

You are mostly correct. The star is far to bright for us to see any planets that may orbit it. The other key thing is that at the distances stars are, any planets are simply too close to them to see at all. With the naked eye you simply don't have enough resolution to visually separate two objects that are as close together as all stars and their planets are. http://en.wikipedia.org/wiki/Angular_resolution

mfb

To extend Drakkith's answer: Even with the best telescopes, most planets around other stars cannot be observed directly. There are some exceptions, but most of the known ~780 exoplanets have been observed with indirect methods. See Wikipedia for an overview.

Drakkith

Yep. Even the few that have been observed by direct imaging are typically very close to us and the distance between the star and the planet is many AU's, giving them an angular separation of around 0.5 -a few arcseconds. An arcsecond is 1/60 of an arcminute, which is itself 1/60 of a degree, of which there are 360 degrees in a circle. The full moon has an angular diameter of about 0.5 degrees, or 30 arcminutes, or 1,800 arcseconds. So the few planets that we have directly imaged look as far apart as a tiny fraction of the diameter of a full moon. A very small angular distance that only large telescopes can resolve with anything resembling decent resolution.

DaveC426913

While it is true that planets are too close to their nearby stars to resolve, it is my assertion that the primary reason why we cannot see planets has more to do with how dim they are. Here's why:

The Moons of Jupiter are visibly quite distinct from Jupiter itself. The angle subtended by the 4 Jovian Moons is substantial (so wide that they barely fit in the viewing port of a low power eyepiece.) This is large enough to be quite easily resolvable to the naked eye, yet we cannot see them. The reason is that they are too dim. Ganymede, he largest has an apparent mag of 4.38, which is pretty good, but the rest of them are all above 6 (the limit of the human eye).

BTW, here's a chart showing a lot of common celestial sights and their apparent mag.
http://en.wikipedia.org/wiki/Apparent_magnitude

Drakkith

Dave I'm talking just about exoplanets and why they are not detectable by the human eye in that post. But you are correct. They are very dim, plus Jupiter is too bright, so the spot that it's light falls on at the retina is larger than it's moons orbits are.

mfb

With visible light, the sun's apparent magnitude is -26.7 (all numbers from here). Mercury, at a similar distance (actually a bit more here), has a maximal apparent magnitude of -2.5, I'll round the difference to 24. The scale is logarithmic, therefore this difference is the same for all observers. As a nice feature, it just depends on geometry and albedo of the planet - if the star is brighter, the planet will be brighter as well.

Hubble can observe objects up to a magnitude of 31.5. Using the difference of 24, mercury-like objects would be visible at stars with a magnitude of 6.5 or less, which corresponds to ~5000 stars, basically all stars visible to the human eye.

What about other planets?
The received light scales with [itex]\left(\frac{\rm{planet radius}}{\rm{orbital radius}}\right)^2[/itex]. Without the square and in units of 10_-5, Mercury has 3.5, venus has more with 5.6, earth is between both with 4.3. Jupiter is better with 8.9, Saturn has 4. Uranus is worse with 0.9, and neptune has 0.5.
Neglecting the outer two planets, all other planets have a better ratio, mercury is actually the hardest planet to spot in terms of brightness.

Summary: If absolute brightness would be the limiting factor for imaging exoplanets, we could observe most planets similar to those in our solar system at ~5000 stars.
Hot Jupiters are easier to see, a jupiter-sized planet with the orbit of mercury is brighter by a factor of ~800, which is equivalent to a difference of ~7 in magnitude. We could observe tham at all stars up to a magnitude of ~13, which corresponds to several million stars.

DaveC426913

I know.

I'm using a close example to point out why I think it's wrong*.

Regardless of how bright Jupiter is, they are below our threshold. If Jupiter were to disappear, we still wouldn't be able to see any but the largest moon with the naked eye. Because they're too dim.

The moons orbit many times Jupiter's diameter away - on the order of 4 arcminutes. That's about 1/7th of the Moon's diameter. The Moons' positions would be easily resolvable - but only if they were bright enough.


* sorry, not wrong. I should say "much less of a factor".