2
$\begingroup$

Stars give off light and are visible at nighttime but Planets do not give off light . So why are Planets visible at nighttime ?

$\endgroup$
2
  • 8
    $\begingroup$ They reflect sunlight, just like the moon. $\endgroup$ Oct 19, 2017 at 22:42
  • 2
    $\begingroup$ @AtmosphericPrisonEscape It's only seven words long, but that's the answer to the question as it currently appears :) . $\endgroup$
    – user10106
    Oct 20, 2017 at 7:47

2 Answers 2

4
$\begingroup$

Planets and moons reflect sunlight, as mentioned in this comment by AtmosphericPrisonEscape.

Now I'll add a bit more info to make this post more informative :)

That also means that just like the moon, the planets have phases, but they're just too subtle to see with the naked eye, and for any planet further from the Sun than the Earth is, we don't see all the phases. They're full when the planet is furthest from us (on the other side of the Sun), so it'd be hard to tell the phase just from how bright the planet appears.

But through a telescope, you can easily see the phases of Venus, and you can tell it's smaller when it's full (because it's farther away):

Phases of Venus
Source: Wikimedia Commons

$\endgroup$
1
-1
$\begingroup$

This is incorrect. All bodies give off light according to Planck's Law. The amount of emitted light may be small compared with reflected light, or small compared to the sensitivity of your detector.

This may seem like nitpicking, but failure to incorporate emittance when analyzing spectra will lead to sadness.

$\endgroup$
3
  • 1
    $\begingroup$ Technically true. But most planets have their emission peaks in the infrared, so that wouldn't amount to much to the human eye, which is a photometer and not a spectrometer. $\endgroup$ Oct 20, 2017 at 15:12
  • 1
    $\begingroup$ Unless the question gets updated, I'm fairly certain he just means why are the planets visible at night, with an optical telescope or just with my eyes? $\endgroup$
    – user10106
    Oct 20, 2017 at 15:25
  • 4
    $\begingroup$ Yes, but in the context of the original question, this is so "technically true" as to be misleading. $\endgroup$ Oct 20, 2017 at 19:34

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .