# Tag Info

3

Being struck by a piece of rock that is pebble sized or bigger could do damage to any base. Fortunately such objects are rare. Their danger could not be entirely avoided, but would be just part of the overall risk of such a mission. Micrometeorites would hit any exposed base, just as they have hit the ISS and the space shuttle in the past. The base would ...

6

It is possible but unlikely. Here is a really good 'Science 2.0' article about the possibility (http://www.science20.com/robert_inventor/could_you_see_moon_city_lights_or_a_greenhouse_from_earth_just_for_fun-157480). Essentially, you likely wouldn't see the light on Moon settlements because there would need to have many thousands of bright lights and ...

1

Upon some google-ing and wiki-ing I found this image of a gravity map of the moon: That scale up the top is measured in milli-Gal which is thousandths of a cm/s^2. For scale gravity is ~9.81m/s^2 which equals ~981000mGal. The difference between gravity at sea level and the top of mount everest is 2Gal, or 2000mGal, which is 0.2% of average gravity. On the ...

0

A few responses to my own question: (a) TDB time isn't critical because it only varies from TT by a few milliseconds. (b) The resulting vector was not ECI but the vector from the earth to the sun in the celestial frame. I needed to build the celestial to terrestrial rotation matrix using iauC2t00a() (assuming zero polar motion since it's generally less ...

1

The sun is an excellent source of light for all the planets and moons, although it gets a little dodgy way out near Pluto. The moon reflects about 10% of the sunlight that hits it, that's why we see it. Venus' albedo is about 0.75, what with its clouds and all. That means 75% of the sunlight that hits it is reflected back into space for us to see.

2

The strength of the light from the Sun scales with the inverse square of distance [note 1]. That means that we would need to have the Earth-Moon system at $\frac{1}{\sqrt{2}}$ (approx 0.7) AU for the full Moon to be twice as bright [note 2]. Note 1: The fact that the Sun is not a point source of light has only a very minor effect on the scaling. Note 2: ...

2

Moons are not always tidally locked. They are formed with a certain spin angular momentum that will be dragged by asymmetries on the hosting planetary body. Those asymmetries then backreact onto the moon and drag it's spin to become synchronized with the orbit angular momentum. How rapidly this process happens is roughly a function of the tidal forces ...

1

Many, many moons throughout our solar system are tidally locked. https://en.wikipedia.org/wiki/Tidal_locking This wikipedia page lists 34 moons throughout our solar system that are known to be tidally locked, and another 26 that are suspected. It is very common for moons within a certain distance of their planets to be tidally locked. If they are too far ...

0

There's a mistake in the book (see page 280 on google books): the values for $\lambda_{ecl}$ and $\phi_{ecl}$ should be swapped. Your result is correct: $$-6338.688^\circ\text{ mod } (360^\circ) = -218.688^\circ,$$ which is the value that is erroneously listed under $\phi_{ecl}$.

1

Yes, Frank Drake looked into this - see page 91 here http://www.lpi.usra.edu/lunar/strategies/objectives/ast_observations_moon.pdf He calculated that you could make a telescope with diameter 30 km using steel, and up to 60 to 90 km with stronger materials. And no problem with wind loading on the Moon.

Top 50 recent answers are included