A friend claimed last night that there was a small part of the Moon that never saw darkness. His explanation involved it being a high point near the pole. Whilst this sounds vaguely plausible, I think it’s highly unlikely and have never heard of such a thing. I’ve googled but to no avail. Anyone know about this (cites would be good). Thanks people.
Thanks, Squink, who’d have thought it. I will now go eat humble pie for being wrong.
Don’t be too quick to apologize. The article says that MAYBE there’s such a place.
They are sure there are places on the Moon with continuous sunlight over the lunar summer (just like on Earth’s polar areas) but they are NOT certain the same effect lasts through the lunar winter.
One has to be careful with the words “never” or “always.” Even if there were such a place, it would not get any sunlight during a lunar eclipse, so it wouldn’t “always” be in the sun.
While I don’t know whether it’s true, I don’t see why it should be so unlikely. All it takes is a sufficient tilt of the axis. Look at Uranus.
YOU look at Uranus! **I’m ** certainly not gonna look at Uranus!
However, there is always a part of the moon that’s getting light from some star (a lot of them, actually), just not necessarily the closest one.
Well I think we can all agree that the moon is very useful.
Actually that has the opposite consequence. (Not to mention that the concept of “place” is a little slippery on a gas giant like Uranus, which has no solid surface, but I’m going to ignore that complication.)
Because of Uranus’s axial tilt, it has no place that’s always lit by the Sun. That will be true of any spherical body that (1) has a non-zero axial tilt and (2) isn’t tidally locked to the Sun. There might be an asteroid or two with those properties, but I don’t think they hold for any of the planets or moons.
I found that aspect of the story to be kind of puzzling. Of course there are places near the pole of the sunward hemisphere that experience continuous light. The only issue, one would think, would be whether the effect persists during winter–which they don’t answer.
Which hemisphere is the sunward one? Since the moon is tidally locked with the earth I would imagine that the “sunward hemisphere” changes over time…
By “sunward hemisphere” I mean the northern lunar hemisphere during northern-lunar-hemisphere spring and summer, and the southern lunar hemisphere during southern-lunar-hemisphere spring and summer. The “seasons” change over the course of a year just like on Earth, although this has almost no effect on lunar “weather”.
OK, but then the sunward hemisphere does change as a function of time. The lunar northern hemisphere is sunward during the lunar summer and the southern is sunward during the winter.
So the only way a part of a perfect sphere would recieve continual light while orbiting the sun and rotating is if it had 0 axial tilt. Any amount of axial tilt will destroy this case.
For an irregular object like the moon orbiting the sun, a mountain near one of the poles could concievably get continual light but only if the plane perpendicular to the axis of rotation and tangent to the surface at the pole intersects the peak. With the moon it may be more complicated yet as we do not know if the plane of the moons orbit is coplaner with the plane of the earth’s orbit. Wouldn’t this cause the axis of the moon’s rotation to precess relative to the sun?
ps sorry for any grammer and spelling errors, I am pressed for time…
Oops, another requirement is that the plane also intersects the sun when the pole is tilted away from the sun.
More Here:
Eternal Sunshine of the Lunar Kind
You can calculate how high a polar peak would have to be to still get sun when it’s tilted 1.5 degrees away from the sun.
The formula for distance to the horizon is d = ( h[sup]2[/sup] + 2hr )[sup]1/2[/sup], where h = height, and r = the moon’s radius (1,738 km).
A degree on the moon is 30.3 km, 1.5 degrees (see previous post) is 45.5 km.
solve for h:
45.5 = ( h[sup]2[/sup] + 3476h )[sup]1/2[/sup]
h[sup]2[/sup] + 3476h - 2070 = 0
Apply the quadratic formula
h = ( -3467 +/- 3465.8 )/2
h = .2/2 = 0.1 km
If I calculated correctly, a polar peak need only be 100 meters tall to get sun year round. It’ll still go dark during lunar eclipses.
Would the plane of the moons revolution not being co-planar with the plane of the earths revolution effect this?
I would think so. Taking the most extreme case, the axis of the moons revolution lying in the plane of the earth’s revolution would make it impossible (Think conservation of angular momentum here) for any part of the moon to have eternal sunlight.
Anyway, is the axis of the moon rotation or revoltution (1.5 degrees apart, I know) around the earth coaxial with the axis of the earth’s revolution around the sun? And if not, wouldn’t this effect also have to be added in to determine if a location of eternal sunlight exists on the moon (without setting up a camera to watch for years on end as the Nasa guys in Squink’s link did)?
Several relavant quotes from Squink’s link (bolding mine):
I’m kind’ve suprised, this should not be too hard to calculate. All you need is an elevation for the mountain/high spot, the size of the two spheres (sun and moon), distance between them, and the relevant rotation/revolution vectors… Take the worst case for the axis of moon rotation pointing away from the sun (which may be affected by the plane of the moons rotation around the earth) and place the mountain on the far side from the sun. Construct the line from the mountain tip to the pole and see if it intersects the sun. If you want to get more complicated you could add in other mountains around you eternal sunshine place and instead of a line use a plane as I mentioned above.
Good stuff, but it does not even need to be that tall. You also need to add in the angular extent of the sun from the moons surface. After all you only need to see a little of the sun to be in the sun, not the center.
Sure. The moon isn’t perfectly round either, so the calc will be a little off from that too, and any peaks over a hundred meters tall within 45.5 km of the pole could throw shade on a polar peak. Without great maps, the only way to tell for sure is to watch for several years.