Easy Questions?

[The Black Spot]

Arrr...

Why do biscuits turn soft when they're stale, but cakes turn hard?

[Sibling Zono (anon1mat0)]

Mmm... someone didn't read the thread? 

Concerning the cake and biscuit question
Clearly depends on the type. One thing is clear: there is a dynamic exchange between the atmosphere (gas with "dissolved" water) and the solid object. The cake could be seen as a sponge that already holds water while the biscuit is more like a dry sponge.
The atmosphere being big will have a more or less constant humidity (let's ignore the weather for a moment) while in the wrapping of the eatables there is not much air, i.e. only a fixed small amount of water. In the wrappers there will be an equilibrium between water in the eatable thing and the surroundings. In the open air a dry good will suck water from the surroundings until there is again an equilibrium. If on the other hand you have a humid good the not saturated air (unlimited capacity) will suck the water out until there is either an equilibrium or there is no accessible water left. Goods will dry often at the surface only because the water at the center cannot easily come to the outside (and the other way around in dry food, soggy on the outside but still dry inside).
That food gets soggy in the fridge is slightly different. Cold air can hold less water than hot air. The fridge permanently exchanges the air inside itself. The air from outside is cooled down and will lose some of its water content that will end either as condensation or will be sucked up by food stored in the fridge. In the oven you have the opposite effect. The hot air can hold more water and will suck it from the baking goods.
I have to admit that there are a lot of aspects concerning this that I do not completely understand. There is probably another effect of water tightly and loosely bound in the food. We will not notice it in the former case but if by some effect the tightly bound water becomes loose we will.
A nice experiment demonstrating something similar can be carried out at home too.
Get some copper sulphate. It is blue and apparently dry. Now heat it ( a gas stove would be best) and it will turn colorless/white. If you keep it under dry conditions it will stay that way but under humid conditions it will turn blue again without seemingly getting wet (at first). If you have precise scales you can show that the blue dry salt is slightly heavier than the white dry salt. In the blue version the water is bound so tightly that it cannot be detected.

Interestingly, not too long ago I placed a bag of burger buns (in their package) in the fridge and after a while some turned hard, the others where almost wet, inside the same bag (I believe the ones on top were hard and the ones below wet). Some interesting dynamics there.

[Swatopluk]

Water following the pull of gravity? I have made the experience that items in the lower parts of the fridge and at the back get more water even when proteced against condensed water dropping down. I can only guess that it is so because the lower parts are cooler (natural air distribution) and the back too (the cooling spirals are located there).

[Black Bart]

Arrr...

Why do biscuits turn soft when they're stale, but cakes turn hard?

Ye blaggard...you've got cake on board yer ship now!

[Swatopluk]

Could a moon of a planet always be in the shadows by having a period of exactly one year or would such an orbit be mechanically impossible?

[Sibling Lambicus the Toluous]

Unless my physics fails me, yes, it would be possible.

Going back to Kepler, assuming a circular orbit:

(P/2pi)²=a³/[G(M+m)]

where:
P = Period (i.e. time for one orbit)
a = semimajor axis (i.e. distance from the Sun to the planet, or from the planet to the moon
M, m = masses of the bodies in question

So, for the lunar orbit to be exactly one year, the two periods would be equal, or:

a_s-p³/[G(m_s+m_p)]=a_p-m³/[G(m_p+m_m)]

(note: s = sun, p = planet, m = moon)

Assuming that the masses and the distance from the sun to the planet are fixed,

a_p-m³ = a_s-p³*(m_p+m_m) / (m_s+m_p)

So, for any particular set of sun, planet and moon masses, there is always a distance that you can plunk the moon at that will cause an orbital period of once per year

... for a circular orbit, at least.  I'd attempt to figure it out for an eliptical orbit, but I'm worried that would make my brain explode.

[Swatopluk]

Same for me, the ellipses are beyond my capabilities (that are therefore elliptical )
A second question would be, whether that orbit could be stable, if one includes the gravity of the sun in the calculations. If not, would an orbit that effectively is always in the light be stable (i.e. moon permanently between planet and sun instead of planet between moon and sun)? Is it a Lagrangian point?

Edit: Yes, those are L Points but unstable ones.

[Sibling Lambicus the Toluous]

A second question would be, whether that orbit could be stable, if one includes the gravity of the sun in the calculations.

I don't see why not.

Another way of looking at it (and probably the way I should've used above) is to consider the planet-moon system as one body with its own mass and centre of gravity.  In terms of orbit around the sun, this system would behave just like a planet.

Then if you look at the moon-planet interaction, you can consider that as a separate system...

... oh, but hang on.  That system would be in an accelerating frame of reference.  That would make things... goofy.

If not, would an orbit that effectively is always in the light be stable (i.e. moon permanently between planet and sun instead of planet between moon and sun)? Is it a Lagrangian point?

Hmm... we're beyond my expertise, but would there be a Lissajous orbit* that satisfies the original problem?

*i.e. just outside Lagrange - I think ZZ Top wrote a song about the orbital mechanics involved. 

[Bluenose]

Umm.  I don't think it would work.  What you are proposing is in effect a tethered pair of objects - that is two objects with the same solar orbital period but at different orbital distances.  This can work if the two objects are tethered and the net result is tidal forces providing a stretching force along the tether with the closer (to the sun) object orbiting faster than it would on its own and the outer object orbiting slower than it would on its own.  This idea has been proposed as a way to provide artificial gravity on a satellite - two flat surfaces tethered and in orbit, with each feeliing "up" as being in the direction of the other.  In the case of a planet and moon, gravity alone is no where near strong enough to overcome the tidal forces that would ensue and so the pair would very quickly move apart.  Actually at the difference in their nominal orbital velocities.

Sibling Bluenose

[Griffin NoName]

What if the two objects were entangled?

[Swatopluk]

Another Oldie

A worm sits at one end (or near it) of a rod that permanently expands.
It begins to crawl towards the far end with a speed lower than the expansion of the rod.
Will it ever reach the end and, if yes, how long will it take?
Given parameters
L = length of the rod at the start
v1 = expansion speed of the rod
v2 = crawling speed of the worm
x0 = starting position of the worm

The end of the rod where the worm is starting is considered non-moving and the rod expands uniformly, i.e. the forward speed at a point x[0;L] is (x/L)*v1

[Sibling Lambicus the Toluous]

Ack!  Integrals!

[Griffin NoName]

I have an Easy Question.

Why is this thread called Easy Questions?

[Bluenose]

I couln't be bothered doing the math, and frankly I'm not sure if I remember how to do the math, but I'm pretty sure the answer is, it depends on the starting position.

(Ignoring relativistic effects) at any given point x_n the velocity of the worm (v3) is equal to its crawling speed v2 plus a proportion of the expansion of the rod v1 equal to the ratio of the worm's position x_n and the length of the rod L, in other words v3=v2+v1(x_n/L).  So for any starting position x0 where v3 is greater than v1, the worm will indeed reach the end.

Thus if v3>v1 or if (v2+v1(x0/L)>v1, then the worm will get to the end, otherwise it won't make it.

Hmm, looks like it did the math after all.

These are called easy questions because they are questions that look like they are simple, but often require complex or subtle answers to explain.  I guess we are saying that the question may well be easy, that does not say anything about the answer!

[The Meromorph]

If not, would an orbit that effectively is always in the light be stable (i.e. moon permanently between planet and sun instead of planet between moon and sun)? Is it a Lagrangian point?

Hmm... we're beyond my expertise, but would there be a Lissajous orbit* that satisfies the original problem?

*i.e. just outside Lagrange - I think ZZ Top wrote a song about the orbital mechanics involved. 

I think they also wrote one about a proposed solution "I'm a Fool for Your Stockings, I believe."